solutions ch02.doc

Chapter 2 Aqueous Solutions and Solubility

Chapter 2 Problems -- Solutions--Gamma Version

Section 2.1. Substances in Solutions

2.1. Tasting samples is unsafe and forbidden in the laboratory. If you had a solution of a sugar, how could you determine if the solute is uniformly distributed throughout the solution without tasting samples? Explain your plan.

Answer to 2.1: Students answer will vary. An example of an answer follows. Using a pipette, collect samples from various places throughout the mixture. Each sample could then be analyzed for properties like density, boiling point, freezing point, or specific gravity. The results could be compared to each other and to an analysis of the entire solution.

2.2. Mixtures involving solids and liquids can be classified into two groups: homogeneous solutions and heterogeneous mixtures in which, after a time, the solid settles out of the mixture. When a solid does not truly dissolve in the liquid but forms a heterogeneous mixture, it is called a suspension. A suspension is usually shaken to redistribute the solid prior to use. Go to your local grocery and/or drug store and identify 5 products that are solutions and 5 products that are suspensions.

Answer to 2.2: Some examples of solutions might be: soft drinks, saline solutions for contact lenses, rubbing alcohol, liquid bleach, liquid soaps. Some examples of suspensions might be: Milk of Magnesia, Peptobismol, oral antibiotics like Amoxycilin, calamine lotion, paint.

2.3. Draw energy diagrams for these processes. Label each energy change as endothermic or exothermic and explain why you draw it this way.(a) A solvent changing from the liquid state to the gaseous state.(b) Gaseous solvent molecules solvating a solute to form a liquid solution.

Answer to 2.3:(a)

Gamma Draft ACS Chemistry 1


(b)

(b)

2.4. (a) When a certain liquid molecular substance dissolves in water, the solution feels cool. Sketch an energy diagram that shows the relationships among energy theoretically needed to separate the molecules in the liquid state, energy released when the molecules dissolve in water, and the net energy change in this solution process.(b) When a certain liquid molecular substance dissolves in water, the solution feels warm. Sketch an energy diagram that shows the relationships among energy theoretically needed to separate the molecules, the energy released when the molecules dissolve, and the net energy change in this solution process.

Answer 2.4:(a) The solution feels cool to the touch because the process is drawing

needed heat energy from your hand, providing the energy needed for the

overall solution process. This is a sketch of the energy diagram showing

this endothermic solution process.



Separated Molecules

Liquid

Solution

Net Energy Change

Energy addedEnergy released

(b) The solution feels warm to the touch because the process is releasing heat energy to your hand, releasing some of the energy from the overall solution process as heat energy. This is a sketch of the energy diagram showing this exothermic solution process.

Separated Molecules

LiquidSolution

Energy added Energy released

Net Energy Change

2.5. The overall solution process for solution of calcium chloride, CaCl2, in water is exothermic. Draw an energy diagram for this process.

Answer 2.5:



Section 2.2. Solutions of Polar Molecules in Water

2.6. Can you predict whether a substance will be soluble in water by looking at its line formula? Can you predict whether a substance will be soluble in water by looking at its structural formula? Explain your responses.

Answer 2.6: No, you can not predict whether a substance will be soluble in water by looking its line formula. However, by looking at its structural formula, you can tell whether a substance will dissolve in water or not. A structural formula will have regions of polar and non-polar groups that will allow you to predict whether the substance will dissolve in water or not.

2.7. Explain what the expression “like dissolves like” means. Illustrate your explanation with appropriate examples.

Answer 2.7: The like-dissolves-like expression reflects the fact that attractions between solute molecules and some attractions between solvent molecules must be replaced by a solute-solvent attractions when a solution forms. If the new attractions are similar to those replaced, very little energy is required for the solution to form.A polar liquid, such as water, is generally the best solvent for ionic and polar compounds. Non-polar liquids, such as hexane are better solvents for non-polar compounds like for example wax.

2.8. For each of the following compounds, write out the structural formula, using a reference handbook to find the structures you don’t know. Circle all the polar bonds found in each structure. Show the direction of the bond polarity for each one. Do not consider the C-H bond to be polar.(a) testosterone(b) acetylsalicylic acid



(c) methyl salicylateAnswer 2.8:

(a) testosterone structure

OH

O

H2C

CC H

C

C

H2 C

C H2

CH 2

CHCH CH

C

H2 C

H2C

C H2

CH 2

CH

CH 3

CH 3OH

O

and with dipoles shown

testosterone (with polar bonds circled and direction of dipoles shown)

O

O

H

(b)

acetyl salicylic acid (aspirin)

O OH

O CH3

O

(c)



methyl salicylate (oil of wintergreen)

O OH3C

OH

2.9. (a) Use Lewis structures to help explain why ethanol, CH3CH2OH, is miscible with water.(b) How do you predict the solubility of 1-pentanol, CH3(CH2)4OH, in water will compare to that of ethanol in water? Explain.

Answer 2.9: (a) The dashed lines represent the network of hydrogen

bonds that form between ethanol and water.

OH

H

OC

H

HC H

HHH

OH

H

O

HH

Note that there are two hydrogen bonds between nonbonded electron pairs on the oxygen of ethanol, and another hydrogen bond between the –OH group’s hydrogen and the covalently bonded oxygen in water. The water molecules have other nonbonded pairs on oxygen and covalently bonded hydrogen, all of which are capable of extending the network of hydrogen bonds. This opportunity for maximum hydrogen bonding accounts for the miscibility of ethanol in water.



(b) Pentanol is likely to be soluble in water, because its alcohol group can also hydrogen bond with water. The solubility of pentanol is expected to be less than that of ethanol because the nonpolar hydrocarbon fragment is larger.

2.10. Explain in terms of intermolecular attractions each of the following “like dissolves like” observations.(a) Methanol (CH3OH) is not miscible with cyclohexane (C6H12).(b) Naphthalene (C10H8) is insoluble in water.(c) Naphthalene is soluble in benzene (C6H6).(d) 1-Propanol (CH3CH2CH2OH) is miscible with water.

Answer 2.10: (a) Cyclohexane is a nonpolar solvent while methanol is a

polar one. The interaction between water and cyclohexane molecules are

very weak. In the mixed state, without interaction between methanol and

cyclohexane molecules, both water and cyclohexane molecules lose some

of their freedom of movement to make way for each other. This type of

reorganization of “unlike” molecules is unfavorable.

(b) Naphthalene does not contain any polar groups. Water molecules form

hydrogen bonding with each other and reorganization of water molecule in

order to accommodate naphthalene molecules would be unfavorable. In

this case water molecules would lose some of their freedom of movement

while making way for naphthalene molecules.

(c) Both naphthalene and benzene do not contain any polar groups. The

reorganization involved in mixing two nonpolar compounds favors the

mixed state.

(d) Molecules of water can form a hydrogen bonds with acteone, which favors the mixed state.

2.11. Predict the relative solubility of gasoline, C8H18 in water. Explain the reasoning for your prediction.

Answer 2.11: Since gasoline is a non-polar molecule, it will be insoluble in water. It will not interact with water molecules.



2.12. (a) Write the Lewis structures for 1-hexanol, CH3(CH2)5OH, and 1,6–hexanediol, HO(CH2)6OH, and, in each molecule, identify the region or regions where hydrogen bonding with water can occur.(b) Which compound in (a) do you predict is more soluble in water? Explain.

Answer 2.12: (a) These are the two Lewis structures with regions for

hydrogen bonding identified.

hexanol

CH

HH

CH

HCH

HC C C O

H

H H

HH

H

H

Hydrogen bondingwith water could takeplace in this region.

1,6–hexanediol, HO(CH2)6OH

CH

OH

CH

HCH

HC C C O

H

H H

HH

H

H

Hydrogen bondingwith water could takeplace in these regions.

H

(b) 1,6-hexanediol is predicted to be more soluble. Compounds with multiple polar groups present more opportunity for hydrogen bonding with water molecules.

2.13. Predict the relative water solubilities (from most soluble to least soluble) for each set of structures. Make models to help visualize the structures.(a) CH3CH2CH2CH2OH (b) CH3CH2CH2CH2NH2

CH3CH2CH2CH2CH2OH (CH3)2CHCH2CH2NH2

HOCH2CH2CH2CH2OH (CH3)3CCH2CH2NH2

HOCH2CH2CH2OH (CH3CH2)3CCH2NH2

Answer 2.13: Recall (Section 2.2) that as the hydrocarbon portion of an alcohol increases in size, the solubilizing and hydrogen bonding effect of the polar hydroxy (–OH) group becomes increasingly less important. To a first approximation, the solubility is related directly to the ratio of carbon atoms to polar groups. In part (a), the polar functional group is an alcohol and in part (b), the polar functional group is an amine. The ratios and rank orders of solubilities are shown (#1 is most soluble in each part):

Carbon/funct grp ratio solubility rank



(a) CH3—CH2—CH2—CH2—OH 4/1 3CH3—CH2—CH2—CH2—CH2—OH 5/1 4HO— CH2—CH2—CH2—CH2—OH 2/1 2HO— CH2—CH2—CH2—OH 1.5/1 1

(b) CH3CH2CH2CH2NH2 4/1 1(CH3)2CHCH2CH2NH2 5/1 2(CH3)3CCH2CH2NH2 6/1 3(CH3 CH2)3CCH2NH2 8/1 4

2.14. Sugars are natural products that are generally quite soluble in water. Sucrose, table sugar, is a common source of sugar in food. Fructose and lactose are two other sugars that are important in living systems. Use library resources or the internet to find the structural formulas for these three sugars. On the basis of their structures, explain their high solubility in water. (Your response should include the reference(s) to where you found the structures.)

Answer 2.14:

There are many web resources that have the structures of these sugars. Students

should have the following structures:


CCH2OH

OH

H

HOH2C

HO H

H CC

C

HO

O

C

C

CO

OH

H

OH

H

HO

H CC

CH2OH

HO

H

H

Fructose(pyranose form = major isomer)

Fructose(furanose form = minor isomer)



O

C

CC

H

OH

H

H

H

H

CC

O

C

CC

H

OH

H

H

H

HOHC

C

O

CH2OH

HO CH2OH

OH

OH

Lactose

C

CH2OH

H

CH2OH

HO

H

H

C

C

C

OH

O

O

C

CC

H

OH

H

H

H

CC

H

CH2OH

HO

OH

O

Sucrose


2.15. The fats in our bodies are composed of relatively nonpolar molecules that are almost insoluble in water. Vitamins may be classified as fat soluble or water soluble. Consider the structures of vitamins A and C shown here. Which of these vitamins do you expect to be more soluble in aqueous systems and which in fatty tissues of the body? Use the structures to explain your reasoning.

CC

CC

CC

CC

CC

OC

C

CC

C

CCH

H

HH H

H

H HH

HHH

C CH

HH HHH

H

H H

H

H

HH

H

H

H

CC

OC

COO

H

H

OC

HC

OH

OH

H

H

H

vitamin A vitamin C

Answer 2.15: Given: Structures of vitamins A and C.Asked For: To explain which of these vitamins should be more soluble in aqueous systems and which in fatty tissues in the body.Recall: The more polar groups present in the molecule the more soluble it will be in water.Plan: We will inspect the structures to figure out which one has the most polar groups.Explanation: Vitamin C has the most polar groups. Thus, it will be the most soluble in aqueous systems.

2.16. (a) Vitamin B is water soluble and vitamins D, E, and K are fat soluble. (See Problem 2.15.) Based on this information, explain which vitamins could be stored in your body and which should be included in your daily diet.(b) From your answers for Problem 2.15, would vitamins A and C be stored in your body or should they be included in your daily diet?(c) With the ready availability of vitamin supplements, cases of hypervitaminosis, an illness caused by an excessive amount of vitamins, are now being diagnosed by medical doctors. Explain for which vitamins hypervitaminosis is likely to occur?

Answer 2.16: (a) Vitamins D, E, and K can be stored in your body. Vitamin B should be included in your daily diet. (b) Vitamin A can be stored in your body while vitamin C should be included in your daily diet.(c) Since only fat soluble vitamins (vitamins A, D, E, and K) can be stored in your body, true hypervitaminosis has been observed only for these vitamins.



Section 2.3. Characteristics of Solutions of Ionic Compounds in Water

2.17. In Investigate This 2.10, what had to be present in the solution in order for the light bulb to glow?

Answer 2.17: Ions must be present.

2.18. (WEB) Chap 2, Sect 2.3.1 has movies representing the Fe3+(aq) and NO3–(aq) ions. How

many water molecules are shown in the hydration layer for each ion? What are the similarities and differences in the arrangement and orientation of the water molecules around each of the ions? How do you explain the similarities and differences?

Answer 2.18: Both ions have six water molecules around them. However, Fe3+(aq) is attracted to the negative (oxygen) ends of the water molecules. NO3

–(aq) is attracted to the positive (hydrogen) ends of the water molecules.

2.19. Solution A was prepared by mixing 0.5 g of ethanoyl chloride (acetyl chloride), CH3C(O)Cl, with 100 mL of water. Solution B was prepared by mixing 0.5 g of 2-chloroethanol, ClCH2CH2OH, with 100 mL of water. Solution A conducts an electric current but solution B does not. What can you conclude about the contents of each solution? Explain the reasoning for your answer.

Answer 2.19: Solution A contains ions while Solution B does not.

2.20. What ions are present in solution when these solids dissolve? Identify each type of ion as either a cation or an anion.(a) BaCl2 (e) NH4Cl(b) KCl (f) Na2S(c) Na3PO4 (g) MgSO4

Answer 2.20: (a) Ba2+ (cation) Cl- (anion); (b) K+ (cation) ) Cl- (anion); (c) Na+ (cation) PO4

3- (anion); (e) NH4+ (cation) Cl- (anion); (f) ) Na+ (cation) S2- (anion);

(g) Mg2+ (cation) SO42- (anion)

2.21. (a) Imagine that one of your friends who is not taking this chemistry course says that salt solutions must conduct electrons, just like wires, because you can replace part of an electric circuit (as shown in the pictures in Investigate This 2.10) with a salt solution and the current will still flow. How will you answer your friend and explain how electric charge continues to flow without a flow of electrons through the solution?(b) (WEB) Chap 2, Sect 2.3.2. Would this interactive molecular level representation of electrical conductivity in ionic solution help your explanation? Why or why not?


WEB

WEB


Answer 2.21: Electrons do not have to flow through the solution. Cations are attracted to the negatively charged cathode wire, while the anions are attracted to the positively charge anode wire, as shown in WEB Chap 2, Sect 2.3.2.

2.22. Solid sodium chloride, NaCl, does not conduct electricity but an aqueous solution of sodium chloride is a good conductor. Solid mercuric chloride, HgCl2, does not conduct electricity and neither will its aqueous solution even though the solid is soluble. Offer a possible explanation for the difference in behavior of the aqueous solutions.

Answer 2.22: Neither solid will conduct electricity because any ions present are not free to move and transport the charge. Sodium chloride(aq) is a good conductor because the polar water molecules surround the ions and pull them apart, releasing Na+(aq) and Cl–(aq) into solution. HgCl2 is soluble in water, but the aqueous solution does not conduct electricity. Evidently no ions are released when this compound goes into solution. The molecule itself stays together in solution. This is not the usual circumstance for what appears to be a salt, but mercury(II) chloride molecules stay bonded together even while going into aqueous solution.

2.23. Imagine that you are a positively charged ion surrounded by large numbers of polar molecules like our simple ellipsoids with positive and negative ends Chapter 1 Figures 1.15 and 1.16. Why would you have a problem feeling the attraction of a negatively charged ion or the repulsion of another positively charged ion? Explain your reasoning. Use drawings, if they help clarify your explanation.

Answer 2.23: Given: Imagine that you are a positively charged ion

surrounded by large numbers of polar molecules like our simple ellipsoids

with positive and negative ends in Chapter 1.

Asked For: Explanation of why you would not feel the attraction of a negatively

charged ion or the repulsion of another positively charged ion.

Explanation: Each ion will be surround by several water molecules which create a

hydration layer. This layer shields the ion from the attraction of an oppositely

charged ion or the repulsion of a similarly charged ion.

Picture:



+

–

++

+

+ +

+

+––

– ––

–

The solvent molecules in the "hydration sphrere" are constantly exchanching with the bullk solvent.

Nonetheless, the surface of the hydration sphere is partailly positively charged.

Section 2.4. Formation of Ionic Compounds

2.24. Which of the following do you predict to conduct electricity when dissolved in water? Explain your reasoning in each case.(a) MgBr2 (d) CH3OCH3

(b) CH3OH (e) KNO3

(c) NaOH (h) CH3CH2CH2CH3

Answer 2.24: (a), (c), and (e) would all form solutions that are electrolytes since these are soluble ionic compounds. The molecules in (b), (d), and (h) will not ionize to a degree sufficient to be detected by most conductivity apparatus available to our students.

2.25. Predict the most likely charge when the following elements form monatomic ions. Explain the rationale for your choice in each case.(a) alkali metals (c) alkaline earth metals(b) oxygen family (d) halogensAnswer 2.25: (a) 1+ (b) 1- (c) 2+ (d) 2- (e) 3- (f) 3+

2.26. Write the chemical formula for the ionic compound formed by the combination of the following ions.(a) magnesium cation + bromide anion(b) calcium cation + nitrate anion(c) magnesium cation + sulfate anion(d) potassium cation + oxide anionAnswer 2.26: (a) MgBr2 (b) Ca(NO3)2 (c) MgSO4 (d) K2O

2.27. Name these ionic compounds.



(a) Na2SO4 (c) (NH4)2CO3

(b) MgCl2 (d) Al2S3

Answer 2.27: (a) sodium sulfate; (b) magnesium chloride; (c) ammonium carbonate; (d) aluminum sulfide.

2.28. Write the formulas for these ionic compounds.(a) barium nitrate (c) calcium oxide(b) ammonium phosphate (d) potassium sulfateAnswer 2.28: (a) Ba(NO3)2; (b) (NH4)3PO4; (c) CaO; (d) K2SO4

2.29. Name these ionic compounds.(a) MgS (c) NH4NO3

(b) Na3PO4 (d) LiOHAnswer 2.29: (a) magnesium sulfide; (b) sodium phosphate; (c) ammonia nitrate; (d) lithium nitrate

2.30. Write the formulas for these ionic compounds.(a) calcium iodide (c) potassium carbonate(b) sodium fluoride (d) barium hydroxideAnswer 2.30: (a) CaI2; (b) NaF; (c) K2CO3; (d) Ba(OH)2

2.31. Complete this grid, giving both the formula and the name of the compound formed between each pair of ions.

CO32– PO43– F–

Mg2+

NH4+

Al3+

Na+

Answer 2.31:



CO32– HCO3– PO43– F–

Mg2+ MgCO3

magnesium carbonate

Mg(HCO3)2

magnesium hydrogen carbonate

Mg3(PO4)2

magnesium phosphate

MgF2

magnesium fluoride

NH4+ (NH4)2CO3 ammonium carbonate

NH4HCO3 ammonium hydrogen carbonate

(NH4)3PO4

ammonium phosphate

NH4Fammonium

fluoride

Al3+ Al2(CO3)3

aluminum carbonate

Al(HCO3) 3

aluminum hydrogen carbonate

AlPO4

aluminum phosphate

AlF3

aluminum fluoride

Na+ Na2CO3

sodium carbonate

NaHCO3

sodium hydrogen carbonate

Na3PO4

sodium phosphate

NaFsodium fluoride

2.32. Find from suitable references (or labels on containers) the chemical formulas and write chemical names for the following ionic substances.(a) Milk of Magnesia® (d) Caustic soda(b) Epsom salt (e) Soda ash(c) Plaster of ParisAnswer 2.32: (a) Milk of magnesia: Mg(OH)2 – magnesium hydroxide; (b) Epsom salt : MgSO4 - magnesium sulfate; (c) Plaster of Paris : CaSO4 - calcium sulfate; (d) Caustic soda : NaOH - sodium hydroxide; (e) Soda ash : Na2CO3 – sodium carbonate.

2.33. Equation (2.1) can be broken down into two steps: (1) loss of an electron by a sodium atom, Na Na+ + e–, and (2) gain of an electron by a chlorine atom, Cl + e– Cl–. Write the appropriate reaction equations for formation of the common cations or anions of these elements.(a) potassium (d) sulfur(b) calcium (e) bromineAnswer 2.33: (a) K(g) K+(g) + e-

(b) Ca(g) Ca2+(g) + 2e-

(c) S(g) + 2e- S2-(g)

(d) Br(g) + e- Br-(g)



2.34. In terms of electrical attraction [Coulomb’s Law, equation (2.2)], explain why the ionization energy for all elemental atoms always has a positive value. For example, the energy required for the reaction, Na Na+ + e–, is the ionization energy, Eionization = 496 kJ·mol–1, for sodium atoms.Answer 2.34: By definition, ionization always involves the separation of a negative charge (an electron) and a positive charge (the cation that remains after the electron has departed). If energy is released when opposite charges come together (energy has a negative value for the process), then the reverse process requires the input of energy (energy has a positive value for the process). Mathematically, Coulomb’s law expresses the energy of attraction of opposite charges. This attraction must be overcome (a reversal of the mathematical sign from negative to positive) in order to separate the opposite charges.

2.35. Examine the lattice energies in Table 2.3. Are these data consistent with Coulomb’s law? Explain the reasoning for your answer.Answer 2.35:

Given: Table of lattice energies.

Asked For: Are these data consistent with Coulomb’s Law?

Recall: Coulomb’s Law states that

This is Equation (2.3) found in Section 2.4.

Plan: Lattice energy is proportional to the charge on the cations and anions. Thus, the

higher the lattice energy should imply the higher the charges on the cations and

anions. So, we will randomly inspect charges on the cations and anions in some of

the above salts and see if we can observe if the data is consistent with Coulomb’s

Law. Let’s start by constructing another table to compare various salts, their

corresponding ions, their lattice energies, and q1q2. From this table, we should be

able to conclude if the data is consistent with Coulomb’s Law.

Salt Cation q+ Anion q- q+q- Lattice energy

NaBr Na+ 1+ Br- 1- 1 751 kJ mol-1

MgF2 Mg+2 2+ F- 1- 2 2961 kJ mol-1



MgO Mg+2 2+ O2- 2- 4 3406 kJ mol-1

Answer: By inspection of the above table, the lattice energy is directly proportional to q1q2. Thus, the data in this table is consistent with Coulomb’s Law.

2.36. (a) Based on Coulomb’s Law, in which crystal, KBr or CaBr2, would the greatest forces of attraction and repulsion be observed? Explain your reasoning. Assume that the distance separating the charges is the same for both crystals.(b) Do the data in Table 2.3 support your answer in part (a)? If so, explain how. If not, explain why not.

Answer 2.36: (a) The double charge of Ca2+ will result in a larger coulombic force. The formula for Coulomb’s Law shows that there is a direct relationship between the charge size and the force.(b) Yes, the data in Table 2.3 supports this answer. The lattice energy for CaBr2 is almost three times the lattice energy of KBr.

2.37. The lattice energy for calcium chloride, CaBr2, crystals is 2176 kJ·mol–1. The gaseous reaction forming the ions from the atoms,

Ca(g) + 2Br(g) Ca2+(g) + 2Br–(g),requires 966 kJ·mol–1. Draw an energy diagram, analogous to Figure 2.14, and use it to find Extal form for the formation of ionic crystals of CaBr2 from the gaseous atoms.

Answer 2.37:



2.38. Consider these lattice energies for some ionic solids. All values are in kJ·mol–1.

F– Cl– Br–

Li+ 1046 861 818Na+ 929 787 751K+ 826 717 689

(a) Use these data to discuss how the lattice energy changes with the size of the anion, keeping the size of the cation constant. Hint: The size of the ions in a group (column) of the periodic table increases as one goes down the group.(b) Use these data to discuss how the lattice energy changes with the size of the cation, keeping the size of the anion constant.(c) What generalization can be drawn about the size of ions and the lattice energies of their salts?(d) Use your generalization from part (c) to predict how the lattice energy of CsI compares with that of NaCl.Answer 2.38:

(a) For each Group IA cation listed, the lattice energy of its corresponding

Group VIIA salt decreases as the atomic number increases.

(b) For each Group VIIA anion listed, the lattice energy of its

corresponding Group IA cation salt decreases as the atomic number

increases.

(c) The lattice energy of a salt decreases as the size of its ions increases.

The lattice energies are largest when the ions are the smallest.

(d) The lattice energy of CsI, with even larger ions, should be considerably smaller

than that of NaCl.



2.39. Consider this energy diagram for the formation of one mole of ionic crystals of MgCl2.

Gaseous Ionsof Mg2+, Cl–

1.00 mole ofMgCl2 crystal

1490 kJ added 2524 kJ released

1034 kJ Released

Gaseous Atoms of Mg, Cl

(a) What is the lattice energy for this compound? Explain how you get your answer.(b) How much energy is required for the reaction: Mg(g) + 2Cl(g) Mg2+(g) + 2Cl–

(g)? Explain how you get your answer.(c) What is the energy associated with the formation of ionic crystals from the gaseous atoms? Explain how you get your answer.(d) How do the three energies associated with the formation for MgCl2 compare with those for NaCl, given in Figure 2.13?Answer 2.39:(a) The lattice energy is 2524 kJ.mol–1. (b) The energy required to change

gaseous atoms to gaseous ions is 1490.5 kJ.mol–1. (c) The energy associated

with the formation of ionic crystals from gaseous atoms is the difference

between these two energies, 1033.5 kJ.mol–1.

(d) Substance

Lattice Energy, kJ.mol–1

Energy, Gaseous Atoms to Gaseous Ions,

kJ

Energy, Formation of Ionic Crystals

kJ.mol–1

MgCl2 2524 1490.5 1033.5NaCl 787 145 642

2.40. (a) The energy required to remove electrons from gaseous silver atoms to form gaseous silver cations is 731 kJ·mol–1. When gaseous iodine atoms gain electrons to form gaseous iodide anions, 296 kJ·mol–1 of energy is released. Calculate the energy for this reaction:

Ag(g) + I(g) Ag+(g) + I–(g)



(b) The lattice energy for AgI(s) crystals is 887 kJ·mol–1. Draw an energy diagram analogous to Figure 2.14 and use it to find Extal form for the formation of ionic crystals of AgI from the gaseous atoms.

Answer 2.40a: Energy for this reaction = 435 kJ•mol-1.Answer 2.40b:

2.41. The lattice energy for potassium bromide, KBr, crystals is 689 kJ·mol–1. The formation of separate gaseous atoms of potassium, K(g), and bromine, Br(g), from the ionic crystal would require 594 kJ·mol–1. How much energy is required for the reaction of the gaseous potassium and bromine atoms to form ions in the gas phase?

K(g) + Br(g) K+(g) + Br–(g)

Explain how you get your answer and draw an energy diagram analogous to Figure 2.14 illustrating your answer.



Answer 2.41: 93 kJ•mol–1

(Branz -- PENDING -- need diagram)

2.42. What is the lattice energy for magnesium fluoride, MgF2? The formation of ionic crystals of MgF2 from the gaseous atoms releases 1424 kJ·mol–1. The reaction of the atoms to form ions in the gas phase, Mg(g) + 2F(g) Mg2+(g) + 2F–(g), requires 1533 kJ·mol–1. Explain how you get your answer and draw an energy diagram analogous to Figure 2.14 illustrating your answer.Answer 2.42: 2957 kJ•mol–1 (Branz -- PENDING -- need diagram)

Section 2.5. Energy Changes When Ionic Compounds Dissolve

2.43. When ammonium acetate, NH4C2H3O2 [= (NH4+)(C2H3O2

–)], is dissolved in water, the mixture becomes quite cold. (Ammonium acetate is the salt used in some cold packs.)(a) Is the dissolving of ammonium acetate endothermic or exothermic? Explain.(b) What are the relative magnitudes of the crystal lattice energy and hydration energy for ammonium acetate? Use an energy diagram to explain the reasoning for your answer.(c) Write formulas for the ions in solution using standard chemical notation.(d) Sketch the hydrated ions in the way we have tried to show molecular level interactions in Figure 2.9.Answer 2.43:

Given: Ammonium acetate, NH4C2H3O2(s) [= (NH4+)(C2H3O2

–)] dissolves in water.

The mixture feels quite cold.

(a)

Asked For: Explanation of the relative magnitudes of lattice energy and hydration

energies.

Recall: In Section 2.5, lattice energy and hydration energy are discussed.

Explanation:.The mixture turns cold when the solid dissolves, meaning that the

average energy of motion has decreased. Thus, this energy is now associated with

dissolving the salt in water, which involves both the lattice energy and the hydration

energy. In this case, more energy is used to break the crystal lattice than to hydrate

the individual ions.

(b)


Maureen Scharberg, 01/03/-1,

“crystal” should be “lattice”


Asked For: Is the dissolving of ammonium acetate described as endothermic or

exothermic?

Recall: In Section 2.5, the definitions and examples of endothermic and exothermic

reactions are given.

Explanation: This process is endothermic because thermal energy is absorbed from

the motion of molecules in solution.

(c)

Asked For: Write the formulas for the ions in solution using standard chemical

notation.

Recall: The ions present in the lattice are the same as in solution.

Answer: The two ions are NH4+ and C2H3O2

-.

(d)

Asked For: Sketch the hydrated ions in the way we have tried to show molecular

level interactions in the Figure 2.9.

Recall: In Section 2.3, ion-dipole attractions result from water molecules interacting

with ions. With respect to water, the partially negatively charged oxygen atom

interacts with cations, while the two partially positively charged hydrogen atoms

interact with anions.



Recap: The process of dissolving salts in water involves two types of energy. Lattice

energy is required to break the Coulombic attractions between cations and anions in the

lattice while hydration energy is released as water molecules surround these ions. The

difference between these two energies will determine whether thermal energy will be

absorbed (endothermic) or released (exothermic) by the dissolving process.

2.44. (a) When LiCl dissolves in water, is the process exothermic or endothermic? Use an energy diagram to explain the reasoning for your answer.(b) When KBr dissolves in water, is the process exothermic or endothermic? Use an energy diagram to explain the reasoning for your answer.Answer 2.44:(a) exothermic

(b) endothermic



2.45. Lithium sulfate, Li2SO4, is quite soluble in water (261 g·L–1) while calcium sulfate, CaSO4, is essentially insoluble (4.9 mg·L–1). What ion-ion and/or ion-dipole interactions are responsible for this difference? Be as specific as you can.Answer 2.45:Given: Lithium sulfate, Li2SO4, is very soluble in water (261 g L–1) while

calcium sulfate, CaSO4, is essentially insoluble (4.9 mg L–1).

Asked For: To identify what ion-ion and/or ion-dipole interactions are responsible for

this difference in solubilities.



Plan: We need to understand the process of dissolving a salt in water. This concept is

discussed in Section 2.5.

Explanation: Since Li2SO4 is very soluble in water, the attractive interactions between

Li+ and SO4-2 and water molecules must be much more stronger than the Coulombic

attractions in the Li2SO4 crystal lattice. To explain why CaSO4 is so insoluble in

water, the reverse must be true. The Coulombic attractions between Ca+2 and SO4-2

in the crystal lattice must be must stronger than the interactions between these ions

and water (ion-dipole interactions).

2.46. There are many ways to describe or represent what happens when sodium sulfate, Na2SO4, dissolves in water. First give an explanation in words. Next, write an ionic equation to represent the solution process. Then use a molecular level representation to illustrate what happens when sodium sulfate dissolves in water. (See Figures 2.9, 2.17, and 2.22.)Answer 2.46:Given: Sodium sulfate, Na2SO4, dissolves in water.

Asked For: We need to explain how Na2SO4 dissolves in water. Then we need to write a

net ionic equation and finally draw a particulate diagram to show this process.

Recall/Plan: We need to review how water molecules interact with salt crystals in order

to dissolve salts. This process involves both lattice energy and hydration energy.

This information will help us describe how Na2SO4 dissolves and draw a particulate

diagram. We also need to investigate solubility rules and writing net ionic

equations.

Explanation: Water molecules will be attracted to the Na+ and SO42- ions in the crystal

lattice. The negative oxygens in water molecules will be attracted to the Na+ and

positive hydrogens in water will be attracted to SO42-. These ion-dipole attractions

are greater than the lattice holds Na+ and SO42- in the crystal. Once the ions break

away from the crystal, water molecules surround each ion creating a hydration layer.

These hydration layers create a shield, making it difficult for hydrated ions that are

oppositely charged to get too close to each other. In our example, hydrated Na+ will

not be able to get close to hydrated SO42- and visa versa.

Net ionic equation:



Na2SO4(s) 2Na+(aq) + SO42-(aq)



Particulate Diagram:

2– 2–

2– 2– 2–

2–2–2–

+

+

+

+ +

+

+

+ + + +

++

+ +

+

2–

+

+

S

SS

SS

SS

SS S

S

S SS

+ = sodium cation

2– = sulfate dianion

S = solvent

Section 2.6. Precipitation Reactions of Ions in Solutions

NOTE: Whenever you write a chemical reaction equation, remember to include the appropriate state notation, (s), (l), (g), or (aq), for each species in your equation.2.47. When aqueous solutions of potassium phosphate, K3PO4, and calcium bromide, CaBr2,

are mixed, a white precipitate is formed. When tested for electrical conductivity, both starting solutions test positive. Following the mixing and precipitation, the product solution also tests positive for electrical conductivity.(a) Prepare a table modeled after Table 2.4 to summarize what has happened.



(b) What new combinations of cations and anions are possible following mixing? One of these new combinations is the precipitate and the other is soluble in water. Which is which? Explain your reasoning.(c) Draw a molecular level representation similar to Figure 2.17 to illustrate what happens when the two solutions are mixed.(d) Write a complete ionic equation describing the reaction that occurs when these solutions are mixed.(e) Write a net ionic equation describing the reaction that occurs when these solutions are mixed.Answer 2.47:(a)

Before Mixing After Mixing

Na3PO4 solution

CaBr2 solution

Na3PO4 and CaBr2

Positive ion(s)Na+ (aq)

Ca2+(aq) Na+(aq)+ Ca2+(aq)

Negative ion(s) PO4-(aq) Br–(aq) PO4

- (aq)+ Br– (aq)

Conductivity? yes yes yes

Precipitate? yes

(b) The new combinations are NaBr and Ca3(PO4)2. Since sodium bromide (NaBr) is a soluble compound (according to the solubility rules for ionic compounds (Section 2.7, page 2-38)), calcium phosphate (Ca3(PO4)2) must be the precipitate. The third solubility rule predicts that “ionic compounds of a multiply-charged cation and a multiply-charged anoin are likely to be insoluble.”(c) 6 Na+(aq) + 2 PO4

-(aq) + 3 Ca2+(aq) + 6 Br–(aq)—> 6 Na+(aq)+ 6 Br--(aq) + Ca3(PO4)2(s)

(d) 3 Ca2+(aq) + 2 PO4-(aq) —> Ca3(PO4)2(s)

2.48. (a) When aqueous solutions of potassium chloride, KCl, and sodium bromide, NaBr, are mixed, no precipitate is formed. What can you conclude about the water solubility of NaCl(s) and KBr(s)? Explain your reasoning.(b) Draw a molecular level representation similar to Figure 2.17 to illustrate what happens when the two solutions are mixed.



(b) Write a complete ionic equation describing what occurs when these solutions are mixed. What would you write for the net ionic equation? Explain.Answer 2.48:(a) Both are soluble in water, based on the solubility rules.(b)

(c) K+(aq) + Cl-(aq) Na+(aq) + Br-(aq) K+(aq) + Cl-(aq) Na+(aq) + Br-(aq); no net ionic equation.

2.49. Ba2+(aq) is extremely toxic to humans. However, when physicians need to x-ray the gastrointestinal (GI) tract – stomach and intestines, they fill the patient’s GI tract with barium sulfate and water. How can it be that the patient is not harmed by this procedure?Answer 2.49: The solubility rules indicate that multiple charged cations and anions tend to form insoluble ionic compounds or a precipitate. In this case, the mixture that doctors use contains a tiny amount of Ba2+ cation. This concentration is below the toxic level of Ba2+, so the patient is not harmed.



2.50. Differences in solubility can be used to help separate cations from solutions where they are mixed together. The process is called selective precipitation. Consider this table of solubilities, and then suggest a sequence of precipitation reactions to separate Ag+, Ba2+, and Fe3+ from solution. Explain your approach and write a net ionic equation for each reaction that takes place.

Cation

Test Solution Ag+(aq) Ba2+(aq) Fe3+(aq)

NaCl ppt no ppt no ppt

NaOH ppt no ppt ppt

Na2SO4 no ppt ppt no ppt

Answer 2.50:Given: Data from table of solubilities.

Asked For: Propose a sequence of precipitation reactions to separate these cations.

Explain your answer as well as write the net ionic equations.

Plan: We need to examine the above table very carefully, noting the consequences

of choosing one ion over another. We will determine the sequence as well as the

explanations justifying our approach. You should note that there could be more than

one correct approach.

Recall: This problem involves applying your knowledge of solubility rules.

Explanation: Let’s begin by writing the formulas for the precipitated salts.

CationTest Solution

Ag+ Ba2+ Fe3+

NaCl AgCl no ppt no pptNaOH AgOH no ppt Fe(OH)3

Na2SO4 no ppt BaSO4 no ppt

We will inspect the outcomes of the NaOH test solutions, noting that both AgOH and

Fe(OH)3 are solid precipitates. This solution is not a good place to begin our

separation since we would not be able to distinguish between Ag+ and Fe3+. So, we

will arbitrary begin with the results from the NaCl test solution and label this



strategy, “Scenario 1”. We will use flowcharts to illustrate our work. Spectator ions

will be omitted.

= precipitate = water soluble ions



Scenario 1:

Ag+, Ba2+, Fe3+

Add NaCl

AgCl Ba+2, Fe+3

Add NaOH

Fe(OH)3 Ba2+

Add Na2SO4

BaSO4



Scenario 2: In Scenario 1, we realize that after precipitation AgCl, we could have chosen

Na2SO4 before adding NaOH. Scenario 2 will show the outcome of this selection.

Ag+, Ba2+, Fe3+

Add NaCl

AgCl Ba+2, Fe+3

Add Na2SO4

BaSO4 Fe3+

Add NaOH

Fe(OH)3



Scenario 3: Instead of starting with NaCl, let’s start with Na2SO4.

Ag+, Ba2+, Fe3+

Add Na2SO4

BaSO4 Ag+, Fe3+

Add NaCl

AgCl Fe3+

Add NaOH

Fe(OH)3

Net Ionic Equations:

Formation of AgCl: Ag+(aq) + Cl-(aq) AgCl(s)

Formation of Fe(OH)3: Fe3+(aq) + 3OH-(aq) Fe(OH)3(s)

Formation of BaSO4: Ba2+(aq) + SO42-(aq) BaSO4(s)

2.51. Aluminum nitrate, Al(NO3)3, is soluble in water. So is sodium oxalate, Na2C2O4. When an aluminum nitrate solution is mixed with a sodium oxalate solution, a precipitate forms.(a) What is the precipitate? State the reasoning for your prediction.(b) Write a net ionic equation for the reaction that occurs.Answer 2.51:Given: Aluminum nitrate, Al(NO3)3, is soluble in water. So is sodium

oxalate, Na2C2O4. When an aluminum nitrate solution is mixed with a sodium

oxalate solution, a precipitate forms.

Since there are two parts for this problem. We will solve this problem in two parts.

Part(a):

Asked For: (a) What is the precipitate? State the reasoning for your prediction.

Plan: We can deduce this answer from analyzing the results from Investigate This 2.24.


Note: If we added NaOH, we would have precipated both Ag+ and


Explanation: When Al(NO3)3 is mixed with Na2C2O4, the solution contains Al3+, NO3–.,

Na+, and C2O42– ions. The new ionic compounds that could be formed by combining

the cations and anions to give neutral products are Al2(C2O4)3 and NaNO3. Our

solubility rules say that sodium and nitrate ionic compounds are soluble, so the

precipitate must be Al2(C2O4)3(s).

Part (b):

Asked For: Write a net ionic equation for what happens.

Plan: First, we will write the balanced chemical equation by inspection to satisfy the

conservation of mass.

From the balanced chemical reaction, we can determine the ions involved in the

reaction, the spectator ions, the precipitate, and subsequent net ionic reaction. The

total ionic equation will help us identify the spectator ions, so we can write the net

ionic equation.

Equations:

Balanced Chemical Reaction:

2Al(NO3)3(aq) + 3Na2C2O4(aq) Al2(C2O4)3(s) + 6NaNO3(aq)

Total Ionic Equation:

2Al3+(aq) + 6NO3-(aq) + 6Na+(aq) + 3C2O4

2-(aq) Al2(C2O4)3(s) + 6Na+(aq) +

6NO3-(aq)

Since Na+ and NO3- are present on both sides of the equation, they are the spectator ions

and cancel out.

Net Ionic Equation:

2Al3+(aq) + 3C2O42-(aq) Al2(C2O4)3(s)

2.52. A solution of lithium nitrate is mixed with a solution of sodium phosphate. A white precipitate is observed to form.(a) What is the white precipitate? State the reasoning for your prediction.(b) Write a net ionic equation for the reaction that occurs.Answer 2.52:(a) The white precipitate in this reaction must be lithium phosphate. The

solubility rules predict that all nitrates are soluble. Although compounds of

Group IA are generally expected to be soluble, note that that lithium is

discussed as an exception to the general rule.



(b) 3Li+(aq) + PO43–(aq) --> Li3PO4(s)

2.53. What does it mean if there is a forward arrow over a backward arrow, “æ,” in an equation?Answer 2.53: The reaction goes in both directions.

Section 2.7. Solubility Rules for Ionic Compounds

2.54. A solution of cadmium chloride, CdCl2, is mixed with a solution of ammonium sulfide, (NH4)2S. A yellow-orange precipitate is observed to form.(a) What is the orange-yellow precipitate? State the reasoning for your prediction.(b) Write a net ionic equation for the reaction that occurs.Answer 2.54:(a) CdS(s). Based on the solubility rules, CdS(s) is the likely choice.(b) Cd2+(aq) + S2-(aq) CdS(s)

2.55. Predict the products of each of the following reactions between aqueous solutions. If no visible change will occur, write NO APPARENT REACTION to the right of the arrow. Give the reasoning for your prediction in each case. Write the balanced complete ionic reaction equation and the net ionic reaction equation for each case where reaction occurs.(a) barium chloride(aq) + sodium sulfate(aq) (b) silver nitrate(aq) + magnesium chloride(aq) (c) strontium nitrate(aq) + potassium nitrate(aq) (d) ammonium phosphate(aq) + calcium bromide(aq) Answer 2.55:

Given: Four reactions in water.

Asked For: There are many items that we have to address in answering this problem. We

will organize them as follows:

Predict the products of each of the following reactions between aqueous solutions.

Give the reasoning for your prediction in each case.

Write the balanced molecular equation for each reaction.

Also write the net ionic reaction for each one. If no visible change will occur, write NO

APPARENT RXN to the right of the arrow. Put in the proper state symbols for each

compound.



Recall: This problems involves understanding the strategies needed for predicting the

correct products and writing chemical formulas (Section 2.4), balanced molecular

equations, net ionic equations (Section 2.7).

Plan: For each reaction we will need to predict the products and write the correct chemical

formulas for both reactants and products. We will need to explain why we chose these

products as well. Then we will write the balanced chemical equation followed by the net

ionic equation, following the format we used in previous problems.

Equation (a):

barium chloride(aq) + sodium sulfate(aq)

Let’s write the formulas for the reagents:

Barium chloride: BaCl2

Sodium sulfate: Na2SO4

To figure out what the products are, we know that two new salts will form. So, Ba+2 must

go with SO42- and Na+ must go with Cl-. The products are then BaSO4 and NaCl.

We now must consider the solubilities of these salts. We know that all sodium salts

are soluble, but BaSO4 is insoluble in water.

The balanced molecular equation is

BaCl2(aq) + Na2SO4(aq) BaSO4(s) + 2NaCl(aq)

We will now write the total ionic equation, so we can determine the spectator ions.

Ba+2(aq) + 2Cl-(aq) + 2Na+(aq) + SO42-(aq) BaSO4(s) + 2Na+(aq) + 2Cl-(aq)

To figure out which ions are the spectator ions, we need to examinie the total ionic equation

and find out which ions are present in soloution in both the reactants and the products. In

this case, the spectator ions are Na+ and Cl-. When we write the net ionic equation, we

cancel these ions out. The net ionic equation for this reaction is

Ba2+(aq) + SO42-(aq) BaSO4(s)

We will repeat this strategy for solving the rest of the equations.

Equation (b):

(a) silver nitrate(aq) + magnesium chloride(aq)

Let’s write the formulas for the reagents:

Silver nitrate: AgNO3

Magnesium chloride: MgCl2



To figure out what the products are, we know that two new salts will form. So, Mg+2

must go with NO3- and Ag+ must go with Cl-. The products are then Mg(NO3)2 and

AgCl. We now must consider the solubilities of these salts. We know that all NO3-

salts are soluble, but AgCl is insoluble in water.

The balanced molecular equation is

2AgNO3(aq) + MgCl2(aq) 2AgCl(s) + Mg(NO3)2(aq)

We will now write the total ionic equation, so we can determine the spectator ions.

2Ag+(aq) + 2NO3-(aq) + Mg2+(aq) + 2Cl-(aq)

2AgCl(s) + Mg2+(aq) + 2NO3-(aq)

To figure out which ions are the spectator ions, we need to examine the total ionic equation

and find out which ions are present in the reactants and the products. In this case, the

spectator ions are Mg2+ and NO3-. When we write the net ionic equation, we cancel these

ions out. The net ionic equation for this reaction is

2Ag+(aq) + 2Cl-(aq) 2AgCl(s)

Look at this net ionic equation carefully. A “2” is in front of both reactants and the product.

If we “divide” the reaction by 2, we arrive at the preferred net ionic equation for this type of

situation.

Ag+(aq) + Cl-(aq) AgCl(s)

Equation (c):

(b) strontium nitrate(aq) + potassium nitrate(aq)

Before we even write the formulas for these two salts, we observe that they have the

same anion, the nitrate ion. We know that all nitrates are soluble, but in this case, we do

not form any new salts as products. What does this mean? This means that there is no

net ionic equation and no chemical reaction. Let’s investigate this situation further.

The formulas for the reactants are:

Strontium nitrate: Sr(NO3)2

Potassium nitrate: KNO3

We’ll write the balanced chemical equation next.

Sr(NO3)2(aq) + KNO3(aq) Sr(NO3)2(aq) + KNO3(aq)

The products and the reactants are the same!! What happen’s when we write the total

ionic equation?


Maureen Scharberg, 01/03/-1,

Steve pointed out that the insolubility of silver halides was not discussed in the chapter.


Sr2+(aq) + 3NO3-(aq) + K+(aq) Pb2+(aq) + 3NO3

-(aq) + K+(aq)

All the ions are spectator ions. Therefore, they all cancel out and there is no net ionic

reaction.

Equation (d):

Balanced complete ionic reaction equation:

6NH4+(aq) + 2PO4

3-(aq) + 3Ca2+(aq) + 6Br-(aq) Ca3(PO4)2(s) + 6NH4+(aq) +

2PO43-(aq)

Net ionic:

3Ca2+(aq) + 2PO43-(aq) Ca3(PO4)2(s)

2.56. Write balanced net ionic equations for reactions that would be suitable for laboratory preparation of the following solid ionic compounds. Suggest compounds whose aqueous solutions you could use to carry out these preparations.(a) BaSO4 (c) Ca3(PO4)2

(b) AgCl (d) CaC2O4

Answer 2.56:(a) Ba2+(aq) + SO4

2-(aq) ---> BaSO4(s). One may mix solutions of Ba(NO3)2

and Na2SO4 to produce BaSO4.

(b) Ag+(aq) +Cl-(aq) AgCl(s). One may mix solution of AgNO3 and

KCl to produce AgCl.

(c) 3Ca2+(aq) + 2PO43- (aq) Ca3(PO4)2 (s). One may mix solutions of

CaCl2 and K3PO4 to produce Ca3(PO4)2.

(d) Ca2+ (aq) + C2O42-(aq) CaC2O4 (s). One may mix solutions of K2C2O4

and CaCl2 to produce CaC2O4.

Section 2.8. Concentrations and Moles

2.57. You have prepared 1 L of a 0.1 M solution of NaOH. Next, you accidentally spilled about 200 mL of this solution. What has happened to the concentration of the remaining solution?Answer 2.57: Since the concentration of the solution is uniform throughout the entire volume, spilling will not change the concentration.



2.58. You have been asked to assist with a chemical inventory of a General Chemistry stockroom and have found a 0.5-L bottle about half full of a solution labeled, "0.5 M CaCl2."(a) What does this label tell you about the solution?(b) Can you tell about how many moles of CaCl2 are in the bottle? If so, show how. If not, tell what further information you need to answer the question.(c) Can you tell about how many grams of CaCl2 are in the bottle? If so, show how. If not, tell what further information you need to answer the question.Answer 2.58:(a) The solutions contains 0.5 moles of CaCl2(aq) per liter of solution. There are 0.5 moles of Ca2+(aq) per liter of solution and 1 mole of Cl-(aq) per liter of solution.(b) If the 0.5-L bottle is about half full, then it contains approximately 0.250 L of 0.5 M CaCl2 solution.

Number of moles of CaCl2 = = 0.125 mol of CaCl2

(c) 0.125 mol of CaCl2 = 13.9 g of CaCl2

2.59. Refer to the molecular structure of vitamin C, shown in Problem 2.15.(a) What is the molecular formula of vitamin C?(b) What is a molar mass of vitamin C? Explain your work.(c) How many moles of vitamin C are present in a 500-mg tablet of the vitamin? Explain your reasoning.(d) How many molecules of vitamin C are present in each 500-mg tablet? Explain your reasoning.Answer 2.59:(a) C6H8O6 (b) Molar mass = (612.011) + (81.008) + (615.999) = 176.124 g(c) 0.003 moles(d) 1.7 x1021 molecules

2.60. Calculate the mass (in grams) of the following. Show your reasoning clearly.(a) 2.5 mole of the artificial sweetener aspartame, C14H18N2O5

(b) 0.040 mole of aspirin, C9H8O4

(c) 2.5 x1023 molecules of cholesterol, C27H46O(d) 1.2 x1022 molecules of caffeine, C8H10N4O2



Answer 2.60:(a) 1 mole of aspartame weighs: (1412.011) + (181.008) + (214.007) + (215.999) = 246.310 g; 2.5 mole of aspartame weighs 2.5246.310 =615.8 grams.(b) 1 mole of aspirin weighs (912.011) + (81.008) + (415.999) = 180.159 g; 0.04 mole of aspirin weighs 0.04180.159 g = 7.21grams.(c) 1 mole of cholesterol contains 6.02x 1023 molecules and it weighs (2712.011) + (46 1.008) + 15.999 = 386.664 g; 2.5x1023 molecule = 0.42 moles of cholesterol therefore its mass is 0.42 x 386.664 g = 162.49 grams.(d) 1 mole of caffeine contains 6.021023 molecules and it weighs

(812.011) + (101.008) + (414.007) + (215.999) = 194.194 g;

1.21022 molecules =0.02 mole therefore its mass is 0.02194.194 g =

3.88 grams.

2.61. How many atoms of carbon are there in 5 mg of niacin? Show your reasoning clearly.

NC

CC

C

C

H

H C

HH

N

O

H

H

niacinAnswer 2.61:

Given: 5 mg of niacin as well as the structure of niacin.

Asked for: Number of carbon atoms in 5mg of niacin.Recall: This problem involves understanding and applying the

concept of the mole. Remember, one mole of anything = molar mass of anything = 6.02 X 1023 particles of anything.

Plan: We will use the structure of niacin to calculate its molar mass and then use unit factors to calculate the number of atoms of carbon in 5mg of niacin. We will need to convert from mg to g, since the units of molar mass are grams/mole.

Calculations: Given the structure of niacin, we can figure out its molecular formula, which is C6H6N2O. Note, if you wrote its formula as ON2C6H6 or another way, that’s OK for now. The



format we used in C6H6N2O has been agreed upon by chemists as the appropriate format. List carbon first, hydrogen second, then all other elements in alphabetical order.

Now, from the formula, C6H6N2O, we know that each molecule of niacin contains 6 carbons. We will keep this unit factor handy for later in the calculation.

Remember, we can write the above unit factor using moles:

For the next step, we will calculate the molar mass for niacin.

We now sum the individual masses. The result is the molar mass of niacin.

Since the units of “mole” are in “grams per mole”, we need to

convert from mg to We now are ready to set up an expression to calculate the number of C. Let’s

summarize our data before proceeding:

6 C per molecule of C6H6N2O definition of a mole molar mass of C6H6N2O (152.14 g per 1 mol C6H6N2O)



0.005 g niacin

However we arrange our data, the bottom line for our answer will be the numerical answer plus the appropriate units. So we must arrange our data to see if we can

calculate the number of C in 5 mg of C6H6N2O.

Recap: It seems that it took a lot of time to calculate this answer, but the strategy involved carefully examining our data to make sure we had all the pieces. Note that the problem did not give us the unit factor for converting to g from mg, or information on the mole. We had to develop a strategy for solving this problem.

2.62. Bacteria generally contain a single molecule of DNA that encodes all their genetic information. What is the concentration, mol·L–1, of DNA in a spherical bacterium that has a diameter of 10–6 m = 1 m? Clearly explain how you arrive at your answer.Answer 2.62:Given: Bacteria generally contain a single molecule of DNA that encodes all

their genetic information.

Asked For: We need to calculate the concentration, mol·L–1, of DNA in a spherical

bacterium that has a diameter of 10–6 m = 1 m. Also, we will explain how we

arrived at our answer.

Recall: The volume of a sphere is 4/3r3. We also need to remember that 1cm3 is equal

to 1 mL.

Plan: We will work on the unit conversions for volume. Since the diameter of the cell is

given in meters, we will convert to centimeters. This will allow us to use the

conversion of 1 cm3 is equal to 1 mL. This will allow us to convert to L. Then, we

2DNA.

Calculations: The diameter of the bacterium is 10-6 meters. This means the radius is

equal to 5 x 10-7 m. We will convert to centimeters before calculating the volume of

the cell.



Now, we will calculate the volume of the cell:

We can now convert to L. This will be the volume of the bacterium in L.

We will now calculate the molarity of DNA in the cell:

2.63. Blood serum is typically about 0.14 M in NaCl. Calculate the number of sodium ions in 50 mL of blood serum. Show your reasoning clearly.Answer 2.63:Given: The concentration of NaCl in blood serum (0.14 M NaCl).

Asked For: The number of sodium ions in 50 mL of blood serum.

Plan: From the concentration of NaCl in blood serum, we know that there are 0.14 moles

of NaCl per liter of blood serum. We will need to recall the concepts of the mole

and solution stoichiometry. We note that there is one Na+1 per empirical formula of

NaCl.

Calculation: We first need to calculate the number of moles of NaCl in 50mL of blood

serum. Before we can set up this calculation, we will need to convert from mL to L.

Now, we will calculate the number of moles of NaCl in 0.050 L blood serum:

To find the number of Na+ in this sample, we set up the following calculation:

Recap: Instead of performing this calculation in a step-by-step fashion, we could have also set up one calculation as follows:



Section 2.9. Mass-Mole-Volume Calculations

2.64. Calculate the number of grams of solute present in each of the following solutions. Show your reasoning clearly.(a) 350 mL of 0.105 M K2Cr2O7

(b) 50 mL of 1.0 M FeCl3·6H2O(c) 0.3 L of 1.70 M KClAnswer 2.64:(a) The molecular weight of K2Cr2O7 is 294.22 g/mole. 10.81 g of solute is present in this solution.(b) The molecular weight of FeCl3·6 H2O is 270.34g/mole. 13.52 g of solute is present in this solution.(c) The molecular weight of KCl is 74.56 g/mole. 38.03 g of solute is present in this solution.

2.65. Calculate the molar concentration of solute present in each of the following solutions. Show your reasoning clearly.(a) 120 mL containing 4.5 grams of NaCl(b) 0.25 L containing 1.3 g of NH4Cl(c) 1.3 L containing 1.85 g of AgNO3

Answer 2.65:(a) The molecular weight of NaCl = 58.45. 0.08 mole is present in 120ml. The solution is 0.64M.(b) The molecular weight of NH4Cl = 53.50. 0.03 mole is present in 0.25L. The solution is 0.1 M.(c) The molecular weight of AgNO3 = 169.89. 0.01 mole is present in 1.3 L. The solution is 0.01M.

2.66. (a) 5.405 g glucose, C6H12O6, is dissolved in enough water to make 1.000 L of solution at 20 ˚C. What is the molarity of glucose in this solution? Show your reasoning clearly.(b) How many mL of the solution prepared in part (a) will you need in order to obtain 0.950 millimoles of glucose at 20 ˚C? Show your reasoning and work clearly and completely.Answer 2.66:(a)



Given: 5.405 g glucose, C6H12O6, is dissolved in enough water to make 1.000 L of solution at 20 C.Asked For: Calculate the molarity of glucose in this solution.Recall: The definition of molarity, M, is moles of solute per liter of solvent.

Plan: We are given the formula for glucose from which we can calculate its molar mass. Remember, the units of molar mass are grams per mole. Since we know how much glucose we have in grams, we simply convert grams of glucose to mole of glucose. The volume is given as 1 liter of solvent, so we can just divide the calculated moles of glucose by 1 liter of solution.

Calculations:

Note: “5.405 g” contains 4 significant figures. This means we will have to round up to 3.000 x 10-2 mol glucose.

To calculate the molarity, simply divide by 1 liter:

(b)Given: From Problem 2.38, we calculated the molarity of solution which was 3.000 x 10-

2 M.Asked For: Number of mL of the solution prepared in Problem 2.38 will you need in order to obtain 0.950 millimoles of glucose at 20 oC.Recall: This problem involves solution stoichiometry as well as unit factor conversions. Remember, molarity is given is moles of solute per liter of solution.Plan: Since we need to calculate mL of solution, we will have to use the unit factor of:



We are also need to convert moles to millimoles (mmol). The unit factor here is:

Now, we have to arrange our data for our calculation. We need to solve for the volume in mL that contains 0.950 millimoles of glucose given 3.000 x 10-2 M glucose.Calculations:We will perform this calculation in one step, following our plan described above.

Recap: You may have chosen a different route to calculate this volume. That’s OK. This calculation demonstrates that we can set up all the unit factors and perform the calculation in just one step. You may have chosen more than steps and that’s perfectly acceptable—as long as you understand the concepts and arrive at the correct answer.

2.67. Two students were asked to prepare a 1.00 M solution of CuSO4. One student found a bottle, labeled "CuSO4·5H2O". He weighed 159.60 g of this hydrated copper sulfate, transferred it to a 1-L volumetric flask and dissolved it in a small quantity of water. Then, he added more water until the solution just reached the calibration mark etched on the neck of the flask and thoroughly mixed the contents of the flask. The second student followed exactly the same procedure, but she used the anhydrous salt of copper sulfate, CuSO4. Which student prepared the solution with the correct concentration? Calculate the molar concentration of CuSO4 in each solution.Answer 2.67: Molar mass of hydrated copper (II) sulfate, CuSO4·5H2O, is 249.50 g. He weighted only 159.60 g of this hydrated copper sulfate. Thus, he used too little solute because his final molar concentration is 0.64 M. To prepare a 1.00 M solution, 159.60 g of the anhydrous copper sulfate, CuSO4, is required. The second student prepare the correct solution.

2.68. You need about 170 mL of 0.10 M NaOH for an experiment. The concentration of this solution has to be fairly exact. Describe how to prepare the solution.Answer 2.68: Since you need 170 mL of solution , you can use 200 mL volumetric flask to prepare the solution.



(a) Calculate the mass of pure NaOH required to prepare the solution: Mass of NaOH = (0.1mole x 0.2L x 39.99 g/mole)/1L = 0.80g(b) Weigh out the required mass of substance, and place it in the volumetric flask(c) Add some water and agitate until the substance is dissolved.(d) Add water up to mark and agitate again to ensure a uniform concentration throughout.

2.69. Normal saline, a solution given by intravenous injection, is a 0.90% (mass to volume %) sodium chloride solution. How many grams of sodium chloride are required to make 250. mL of normal saline solution? What is the molarity of this solution? The density of this solution is the same as water, 1.00 g·mL–1. Show your reasoning clearly.Answer 2.69:Given: A 0.90% (mass to volume %) sodium chloride solution.Since there are two questions to answer, we will categorize the first question as “Part(a)” and the second question as “Part(b).Part (a): Asked For: Calculate the number of grams of sodium chloride required to make a 250.0 mL normal saline solution.Recall: We need to understand that this is a concentration problem, involving solution stoichiometry. Plan: The concentration of this saline solution is given as a percent. This is different from the concept of molarity. So, let’s discuss what this percent concentration means. We know that percent has to do with a fraction of 100. “0.90% (mass of solute to volume of solution %)” means that in 100 mL of water, 0.90 grams of NaCl are found. The problem requests us to calculate how much NaCl is needed to make a 250.0 mL solution. Calculation:

Recap: We need 2.3 grams of NaCl to prepare a 250 mL normal saline solution. Part(b):Asked For: Molarity of this solution.



Recall: The definition of molarity of a solution is moles of solute per liter of solution.Plan: We know that 2.3 grams of NaCl is required to prepare a 250 mL solution (answer to Part (a). We will need the molar mass of NaCl to convert from grams of NaCl to moles of NaCl. Also, mL will need to be converted to L.Calculations:

Recap: We could have performed this calculation in just one step:

You should note that using 0.9 grams of NaCl in 100 mL of aqueous solution to calculate the molarity of this solution would give the same answer.

2.70. When urine is analyzed, the normal range for urea, (NH2)2CO, one of the solutes in urine, is 13-40 g·(24 hr)–1. (For urinalysis, a patient’s urine is collected over a 24-hour period to be sure that the sample is representative.) A patient’s laboratory tests show a urea content of 25 g·(24 hr)–1. Suppose the normal output of urine for a patient of this age is 2.5 L·(24 hr)–1. What is the molarity of the urea in the patient’s urine? Explain your reasoning.Answer 2.70:Given: The normal range for urea, (NH2)2CO, is 13-40 g/24 hr. . A patient’s laboratory tests show a urea content of 25 g/24 hr. The normal output of urine for a patient of this age is 2.5 L/24 hr. Asked For: The molarity of the urea in the patient’s urine.Recall: We need to understand that this is a concentration problem, involving solution stoichiometry.



Plan: In a 24 hr period, the patient excretes 25 g of urea in 2.5 L of urine. To convert to molarity, we will need to covert 25 g of urea to moles of urea using urea’s molar mass (60.0 g/mol). The units for volume do not need to be changed.

Calculation:

2.71. One of the ionic compounds in sports drinks is potassium dihydrogen phosphate, KH2PO4. The label on one of these drinks tells us that 240 mL of the solution contains 30 mg of potassium. KH2PO4 is the only ingredient in the solution that can provide this potassium. How many grams of KH2PO4 are dissolved in 240 mL of the solution? What is the molarity of the KH2PO4, in this solution? Clearly show and explain, all the work you do to solve this problem.Answer 2.71:Given: One of the electrolytes in sports drinks is monopotassium phosphate, KH2PO4. The label on one of these drinks tells us that 240 mL of the solution contains 30 mg of potassium. KH2PO4 is the only ingredient in the solution that can provide this potassium.Asked For: There are two parts of this problem. We will label the first question as “(a)” and the second question as “(b)”.How many grams of KH2PO4 are dissolved in 240 mL of the solution? What is the molarity of the KH2PO4, in this solution?Recall: We will begin with Question (a). We are told that there are 30 mg of potassium in 240 mL of this sports drink. The only source of potassium is KH2PO4. Therefore, this question involves stoichiometry using the information contained in the chemical formula as well as the fact that we are given 30 mg of potassium. Also, “potassium” refers to “K+”.Plan: We know that for every one mole of KH2PO4 produces one mole of potassium. Since we are given 30 mg of potassium, we can calculate the number of moles of potassium which will be equal to the number of moles of KH2PO4. We will need the molar mass of potassium as well as the unit factor for the conversion from mg to g. Then, given the molar mass of KH2PO4, we can convert from moles of KH2PO4 to grams of KH2PO4.Calculations:



Now, let’s use the molar mass of KH2PO4 to convert from mol of KH2PO4 to grams of KH2PO4:

Part (b)Plan: We know how many moles of KH2PO4 as well as the volume (240mL). We need to remember that the units of molarity are in moles per liter, so we will have to convert mL to L, using the appropriate unit factor.Calculation:

Section 2.10. Reaction Stoichiometry in Solutions

2.72. How many moles each of carbon, hydrogen, and oxygen atoms are present in two moles of ammonium acetate, NH4C2H3O2? What is the total number of moles of atoms in two moles of the compound? What is the total number of moles of ions in two moles of the compound?Answer 2.72:Given: Two moles of ammonium acetate, NH4C2H3O2.Part (a):Asked For: Number of moles of carbon, hydrogen, and oxygen.Plan: We will assume that one molecule of ammonium acetate contains two carbons, 7 hydrogens, 1 nitrogen, and 2 oxygens. Therefore, we can assume that one mole of NH4C2H3O2 contains two moles of carbon, seven moles of hydrogen, one mole of nitrogen, and 2 moles of oxygen. We need to set up a unit factor that expresses this relationship for each of the atoms requested.



Calculations:

Part (b):Asked For: Total number of moles of atoms.Plan: Using the information given in Part(a), we can calculate the total number of

moles of atoms. We know that 1 mole of NH4C2H3O2 has 12 moles of atoms, so we can create a unit factor that shows this information.Calculation:

Part (c):Asked For: Total number of moles of ions.Plan: Each molecule of NH4C2H3O2 produces two ions, NH4

+ and C2H3O2-.

Therefore, one mole of NH4C2H3O2 produces two moles of ions.Calculation:

Recap: There can be more than one method used to solve these types of problems. You might have solved them differently, but you still calculated the correct answers. You might have solved them differently, but if you calculated the correct answers, that’s OK.

2.73. A student is trying to prepare artificial kidney stones in the laboratory. How many grams of calcium phosphate can he make by mixing 125. mL of 0.100 M calcium chloride with 125. mL of 0.100 M sodium phosphate? Explain your reasoning and any assumptions you make in solving this problem.Answer 2.73:



Given: 125.0 mL of 0.100M calcium chloride mixed with 125.0 mL of 0.100M sodium phosphate.Asked For: Number of grams of calcium phosphate are formed.Recall: This problem involves many concepts discussed in Chapter 2. We will need to review the nomenclature of salts as well as the definition of molarity and solution stoichiometry. Plan: First and foremost, we will need to write the formulas for calcium chloride, sodium phosphate, and calcium phosphate. Then, we will write a balance chemical equation for the reaction. We will need to determine the additional product, besides calcium phosphate, that is formed from mixing calcium chloride and sodium phosphate. Next, we will need to calculate how many moles of calcium chloride and sodium phosphate we have. We need to know the molar mass for each salt. Our goal at this step is to figure out which of the two starting materials is consumed completely and which one is in excess. Once we have this information, we can use the balanced chemical reaction to calculate the number of moles of calcium phosphate formed. Then, using the molar mass of calcium phosphate, we can easily convert to grams of calcium phosphate. Calculations: Let’s begin with writing the correct formulas for calcium chloride, sodium phosphate, and calcium phosphate:Calcium chloride: CaCl2

Sodium phosphate: Na3PO4

Calcium phosphate: Ca3(PO4)3

Now, we can write the balanced chemical equation. We know that Ca3(PO4)2 is one of the products. But what is the other product? Let’s start by writing what we know so far:

CaCl2(aq) + Na3PO4(aq) Ca3(PO4)2(s) + ?

So, Na+ and Cl- are left over and will be soluble ions. We can write this as NaCl(aq). The balanced chemical reaction is:

3CaCl2(aq) + 2Na3PO4(aq) Ca3(PO4)2(s) + 6NaCl(aq)

Next, we need to determine the number of moles of CaCl2 and Na3PO4. Remember, we are given the same quantities of each starting material (125 mL of 0.100M CaCl2 and 125 mL of 0.100M Na3PO4). We will need to convert from mL to L.



We have now determined that we have 1.25x10-2 mol of CaCl2 and Na3PO4.In our balanced chemical equation, we do not have a 1:1 ratio of CaCl2 and Na3PO4. Instead, for every 2 moles of CaCl2, we need 3 moles of Na3PO4. In our reaction, one of these reagents will be completely consumed and one will have some material remaining. We will now figure out which one.Let’s consider this possible scenario. We will assume that all of the CaCl2 is completely used in the reaction. So we will calculate how much Na3PO4 is required. Our calculation is:

But we only have 1.25 x 10-2 mol of Na3PO4. This means that CaCl2 cannot be completely used up this reaction, because we simply do not enough Na3PO4.We will now calculate the second scenario in which we assume that all the Na3PO4 is completely consumed in this reaction.

Thus, all of Na3PO4 will be consumed in the reactions, but only 8.33 x 10-3 mol of CaCl2 will be used. This means that 4.17 x 10-3 mol of CaCl2 will remain in solution and not participate in the reaction.Now we know how much of Na3PO4 and CaCl2 are used in this reaction. We arbitrary choose Na3PO4 to calculate the number of moles of Ca3(PO4)2 produced in the reaction. We know from the balanced chemical reaction, 2 mol of Na3PO4 produce 1 mol of Ca3(PO4)2. Our calculation is:

We now need the molar mass of Ca3(PO4)2 to convert from mol to grams of Ca3(PO4)2. This is our final calculation for this problem and its results will give us our answer to this problem.



Recap: There are a variety of strategies that you can use for these types of problems. You might have chosen a different strategy, but still arrived at the same answer. With these types of problems, the major hurdle is remembering to figure out which reagent limits the amount of product(s) formed.

2.74. What volume of 0.100 M SO32–(aq) is needed to react exactly and completely with 24.0

mL of 0.200 M Fe3+(aq)? The equation that represents the reaction that occurs is:2Fe3+(aq) + SO3

2–(aq) + 3H2O(l) 2Fe2+(aq) + SO42–(aq) + 2H3O+(aq)

Answer 2.74:The coefficients in the balanced ionic equation give the relative number of moles of each reactant and product; this ratio can also be expressed in a ratio of millimoles.

2.75. Assume that you mix 50.0 mL of a solution that is 0.45 M Na2SO4 with 50.0 mL of a solution that is 0.36 M BaCl2.(a) How many moles of each of the four ions, Na+, SO4

2–, Ba2+, and Cl–, are present in the mixture? Explain your reasoning clearly.(b) If the SO4

2–(aq) in the mixture reacts with Ba2+(aq) to give BaSO4(s), how many moles of Ba2+(aq) are required to react with all the SO4

2–(aq) in the mixture? Explain your reasoning clearly.(c) If the Ba2+(aq) in the mixture reacts with SO4

2–(aq) to give BaSO4(s), how many moles of SO4

2–(aq) are required to react with all the Ba2+(aq) in the mixture? Explain your reasoning clearly.(d) Is Ba2+(aq) or SO4

2–(aq) the limiting reactant in this mixture? Explain how you make this choice.Answer 2.75:

Given: A mixture of 50.0 mL of a solution that is 0.45 M Na2SO4 with 50.0 mL of a

solution that is 0.36 M BaCl2.

Asked For: . How many moles of each of the four ions, Na+, SO42–, Ca2+, and Cl–, are

present in the mixture?

Recall: This problem requires knowledge in solution stoichiometry and the concepts

associated with molarity. The balanced chemical equation for the precipitation of

BaSO4.



Ba2+(aq) + SO42-(aq) CaSO4(s)

From this net ionic reaction, we know that Na+ and Cl- are the spectator ions.

Plan: We need to calculate the number of moles present in each solution. This

information will help us determine which solution contains the limiting reagents.

Let’s begin by arbitrarily choosing Na2SO4. Remember, we will need to convert

from mL to L.

For every one mole of Na2SO4, two moles of Na+ and one mole of SO42- are produced.

Therefore, we have:

We will now calculate the number of moles present in the BaCl2 solution.

One mole of Ba2+ and two moles of Cl- are produced from one mole of BaCl2. So,

We will assume that the formation of BaSO4 goes to completion. In this problem, the

limiting reagent is Ba2+ with SO42- in excess. So, all 0.018 mol Ba2+ and 0.018 mol

SO42- out of 0.023 mol of SO4

2- is consumed. That means there will be 0.0045 mol of

SO42- are remaining in the solution, along with Na+ and Cl-.



(b) If the SO42–(aq) in the mixture reacts with Ba2+(aq), how many moles of Ba2+(aq) are

required to react with all the SO42–(aq) in the mixture?

You don't have enough Ba2+(aq) to react with all the SO42–(aq).

(c) If the Ba2+(aq) in the mixture reacts with SO42–(aq), how many moles of SO4

2–(aq) are

required to react with all the Ba2+(aq) in the mixture?

(see above in part (a))

(d) Is Ca2+(aq) or SO42–(aq) the limiting reagent in this mixture? Explain why you make

this choice.

The limiting reagent in the mixture is Ba2+, as explained in part (a).

2.76. Predict what precipitate will form when each of the following aqueous solution mixings is carried out. Determine the limiting reagent for each reaction and the mass of the precipitate (assuming that all precipitation reactions go to completion). If there is no precipitate, then write NO APPARENT REACTION and explain your reasoning.(a) Mix 125 mL of 0.15 M BaBr2 with 125 mL of 0.15 M Na3PO4.(b) Mix 85 mL of 0.40 M NH4Cl with 65 mL of 0.50 M KNO3.(c) Mix 85 mL of 0.40 M (NH4 )2S with 65 mL of 0.50 M ZnCl2.(d) Mix 15.0 mL of 0.20 M AgNO3 with 15.0 mL of 0.40 M NaBr.Answer 2.76: (a) AgBr (ppt) -- The limiting reagent is Ag+ because the total amount of Ag+ is 3.0 mmol (15.0 mL • 0.20 mmol/mL), and the total amount of Br– is 4 mmol (10.0 mL • 0.40 mmol/mL). Equimolar amounts of Ag+ and Br– are required to form AgBr, so the yield of product is limited by the Ag+, which “runs out” before the Br–.(b) Ba3(PO4)2 (ppt) -- The limiting reagent is Ba2+. There are 18.75 mmol each of Ba2+

and PO42–, but three equivalents of Ba2+ are required for every two equivalents of PO4

2– so the Ba2+ is depleted first.(c) NO APPARENT REACTION -- The two possible “cross products” (NH4NO3 and KCl) are both water soluble ionic compounds.(d) CdS (ppt) -- The limiting reagent is Cd2+.

2.77. When 50. mL of an aqueous 0.1 M SrCl2 solution are mixed with 50. mL of an aqueous 0.1 M Na3PO4 solution, a white precipitate is formed.(a) How many moles of chloride anion remain in solution when the precipitation is complete? How many grams of chloride is this? Explain the reasoning for your answers.



(b) How many moles of each of the other ions remains in solution when the precipitation is complete? Explain the reasoning for your answers.(c) Write a complete ionic reaction equation for the reaction in the mixture. Use this equation and your results from parts (a) and (b) to show that the solution is electrically neutral after the precipitation is complete. Show your reasoning clearly.Answer 2.77:<this question should have part (c) first because you can't do the calculations for part (a) and (b) without the net ionic equation--skip for now>

Section 2.11. Solutions of Gases in Water

2.78. Are gases very soluble in water? Explain your reasoning.Answer 2.78:To consider the solubilities of gases in water, we need to examine the polarity of the gas molecules as well as the temperature. Polar gas molecules, like ammonia, tend to be more soluble in water, while non-polar gas molecules, like nitrogen and oxygen, tend to be not soluble in water. Also, in both cases, the higher the temperature, the less soluble the gas will be.

2.79. Predict whether the noble gases (He, Ne, Ar, Kr, and Xe) have a low solubility in water (less than 1 g·L–1) or a high solubility in water (greater than 10 g·L–1). Explain clearly.Answer 2.79: Recall from Sect 2.11 that only when water reacts with a gas will it have a very high solubility. Since the noble gases all have complete valence shells of electrons (the “octet rule”), they show no tendency to bond with water. Water does dissolve polar molecules reasonably well. However, the electron clouds of the noble gases are perfectly symmetrical, rendering them non-polar and therefore highly insoluble.

2.80. (a) Use the data in Table 2.6 for this problem. How many moles of nitrogen gas, N2(g), dissolve in 10.0 L of water when the temperature is 25 oC and the pressure of the gas is 101 kPa (one atmosphere)?(b) How many moles of oxygen gas, O2(g), dissolve in 0.100 L of water when the temperature is 25 oC and the pressure of the gas is 101 kPa (one atmosphere)?Answer 2.80:(a)



(b)

2.81. The solubility of H2(g) in water at 25 C is 7.68 10–4 mol L–1. When the temperature is decreased to 0 C, the solubility of hydrogen is 9.61 10–4 mol L–1. How do you account for the greater solubility at the lower temperature?Answer 2.81:Given: Solubility of H2 in water at 25 C is 7.68 10–4 mol L–1. The solubility of hydrogen at 0 C is 9.61 10–4 mol L–1. Asked For: Explanation of observations.Plan: We need to consider what we know about the solubility behavior of gases at different temperatures, as discussed in Section 2.3. We know, too, that H2 is a non-polar molecule.Explanation: Since there is very little attraction between H2 and H2O, we have to examine the motion of H2 at different temperatures. At 25 C, H2 are moving faster than 0 C. Thus, more molecules will have enough energy to escape into the gas phase at 25 C than at 0 C. This means that the solubility of H2 will be greater at 0 C than at 25 C.

2.82. Hydrogen bromide gas, HBr(g), dissolves in water to form an acidic solution. What is the name of this aqueous solution? Hint: What is the analogous solution of HCl(g) called?Answer 2.82: Hydrobromic acid and hydrochloric acid, respectedly.

2.83. If HBr(g) is bubbled into water until the solution is saturated, the resulting solution is approximately 8.9 M in HBr(aq). The density of the solution is about 1.5 kg·L–1. What is the solubility expressed in g·kg–1 (as in Table 2.6 for other gases)? Clearly explain the reasoning for your answer.Answer 2.83: Approximately 480 g•kg–1.

2.84. (a) HCl(g) is quite soluble in diethyl ether (CH3CH2OCH2CH3). Various reason(s) for this solubility are given: dipole-dipole interactions, hydrogen bonding, or HCl ionization. Draw a picture to illustrate each of these potential interactions of HCl and ether.

(i) dipole-dipole (ii) hydrogen bonding (iii) HCl ionization

(b) What experiment could be done to eliminate or confirm one or more of these.



Answer 2.84: (a) For simplicity, diethyl ether, C2H5O C2H5, is written as R–O–R in these sketches showing (i) dipole-dipole, (ii) hydrogen bonding, and (iii) ionization of HCl in ether solution:

(b) The electrical conductivity of the solution could be tested, If it conducts, then we

know that at least some HCl ionization, interaction (iii), must occur. This does not rule

out contributions to the solubility from the other interactions. If the solution does not

conduct, then ionization is ruled out as a contributor to the solubility. The other

interactions are hard to distinguish exerimentally.

2.85. Would you expect hydrogen chloride gas, HCl(g), to be more or less soluble in hexane than in water. Explain your reasoning.

Answer 2.85: Hydrogen chloride can react with waer to form hydronium chloride (equation (2.16)). This increases the solubility of HCl in water. There is no parallel reaction of HCl with hexane, so its hexane solubility is much, much less.Assume that the amount of a gas that dissolves in water is directly proportional to its pressure over the solution; the lower the pressure, the less gas dissolved.

2.86. (a) Use the data in Table 2.6 to figure out the masses of nitrogen and oxygen that dissolve in 1.0 L of water at 25 ˚C when air (80% nitrogen and 20% oxygen — mole percents) at a total pressure of 101 kPa dissolves in the water. State all your assumptions explicitly and explain clearly the method you use to arrive at your answer.(b) What percent of the dissolved mass of gas is oxygen? Show how you get your answer.(c) Is the mass percent of oxygen in the air greater than, less than, or the same as its mass percent in the gases dissolved in water? Explain your reasoning.Answer 2.86:(a) The data in Table 2.6 are for the solubility of the gases at 101 kPa (one atmosphere) pressure. In air, the nitrogen pressure is 80% of one atmosphere, so only 80% as much nitrogen will dissolve: (0.80) (0.018 g·(kg water)–1) = 0.014 g·(kg water)–1. Oxygen



pressure is 20% of one atmosphere: (0.20) (0.039 g·(kg water)–1) = 0.008 g·(kg water)–

1.(b) In one kilogram of water saturated with gases from the air, there are 0.014 g of nitrogen and 0.008 g of oxygen, or 0.022 g total of these gases. The mass percent of oxygen is [(0.008 g)/(0.022 g)] 100% = 36%.(c) In a mole of air (Avogadro’s number of nitrogen and oxygen molecules), 80% of the molecules are nitrogen and 20% are oxygen. The mass of nitrogen in a mole of air is (0.80 mol) (28 g·mol–1) = 22.4 g. The mass of oxygen is (0.20 mol) (32 g·mol–1) = 6.4 g. The mass percent of oxygen in air is [(6.4 g)/(28.8 g)] 100% = 22%. The mass percent of oxygen from the air dissolved in water is greater than the mass percent in air.

Section 2.12. The Acid-Base Reaction of Water with Itself

2.87. What is an acid?Answer 2.87: Aqueous solutions that have a pH below 7.

2.88. What is a base?Answer 2.88: Aqueous solutions that have a pH above 7.

2.89. If a solution of acid A has a pH of 1 and a solution of acid B has a pH of 3, what can you tell about the two acid solutions?Answer 2.89: Acid A is 100 times stronger or more acidic than Acid B.

2.90. Identify aqueous solutions with these properties as acidic or basic or neither. Explain your reasoning in each case.(a) pH < 7 (e) [H3O+(aq)] > 1.0 10-7 M(b) [H3O+(aq)] = 1.0 10-7 M (f) [OH–(aq)] < 1.0 10-7 M(c) [OH–(aq)] > 1.0 10-7 M (g) [H3O+(aq)] < 1.0 10-7 M(d) pH > 7 (h) [OH–(aq)] = 1.0 10-7 MAnswer 2.90: (a) acidic; (b) neutral; (c) basic; (d) neutral; (e) basic; (f) acidic; (g) acidic; (h) basic; (i) neutral

2.91. Calculate the pH of each of the following solutions.(a) [H3O+(aq)] = 1.0 10-2 M (c) [H3O+(aq)] = 5.0 10-4 M(b) [H3O+(aq)] = 1.0 10-10 M (d) [H3O+(aq)] = 5.0 10-8 MAnswer 2.91: (a) 2.00 (b) 10.00 (c) 3.30 (d) 7.30



2.92. Assuming that the reaction of HCl(g) and water goes to completion to form H3O+(aq) and Cl–(aq), what is the molar concentration of HCl(aq) that will result in solutions having(a) pH = 4? (b) pH = 2?Answer 2.92: (a) 110-4 M (b) 110-2 M

2.93. HCl(g) is named hydrogen chloride, but HCl(aq) is named hydrochloric acid. By analogy, what are the names of HI(g) and HI(aq)? of H2S(g) and H2S(aq)?Answer 2.93: HI(g) is hydrogen iodide. HI(aq) is hydroiodic acid. H2S(g) is hydrogen sulfide. H2S(aq) is hydrosulfuric acid.

2.94. (WEB) Chap 2, Sect 2.12.3. Write a brief essay describing the relationship of the two movies to the figure at the bottom of the page (which is similar to Figure 2.26).

Section 2.13. Acids and Bases in Aqueous Solutions

2.95. Phosphorus pentoxide, P2O5(s), is a nonmetal oxide which reacts with water to form a solution of phosphoric acid, (HO)3PO(aq) (or H3PO4(aq)).(a) Write the balanced chemical reaction equation for the reaction of phosphorus pentoxide with water.(b) If 1.42 g of phosphorus pentoxide is mixed with 250. mL of water, what is the molarity of the resulting phosphoric acid solution? Show your reasoning clearly.Answer 2.96:(a) P2O5(s) + 3H2O(l) 2H3PO4(aq)

(b) 1.42 g = 2.00 10-2 mol H3PO4(aq)

molarity = = 8.0 10-2 M H3PO4(aq)

2.96. Give a name for each of the following ionic compounds with oxyanions (shown with their conventional formulas). See Table 2.7 for the names of oxyanions. Hint: Arsenic, As, is in the same family as P and forms many analogous compounds.(a) Ca(HSO4)2 (d) Ce2SO4

(b) Na2CO3 (e) KHCO3

(c) Al2(HPO4)3 (f) Na3AsO4


WEB


Answer 2.96: (a) calcium hydrogen sulfate (b) sodium carbonate (c) aluminum hydrogen phosphate (d) cesium sulfate (e) potassium hydrogen carbonate (f) sodium arsenate

2.97. Draw Lewis structures (showing all nonbonding electron pairs as a pair of dots and all covalent bonds as lines) for the nitrate, ethanoate (acetate), and hydrogen sulfate oxyanions.Answer 2.98:

ON

OC

CO

O

H

H H

O S O

O

O

H

O2+

2.98. Identify each Brønsted-Lowry acid and base in the following reactions. If necessary, write out the complete balanced ionic equation before identifying the acids and bases. Place an A below each acid and a B below each base.(a) H2S(g) + H2O(l) æ HS–(aq) + H3O+(aq)

(b) NaOH(aq) + HCl(aq) æ NaCl(aq) + H2O(l)

(c) NH3(g) + HCl(g) æ NH4+Cl–(s)

Answer 2.98:(a) A: H2S (g) and H3O

+(aq); B: H2O (l) and HS-

(aq).

(b) A: HCl(aq) and H2O(l); B: NaOH(aq) and NaCl(aq).

(c) A: HCl(aq) and NH4+(aq); B: NH3 (g) and Cl-

(aq)

2.99. Identify each Brønsted-Lowry acid and base in the following reactions. If necessary, write out the complete balanced ionic equation before identifying the acids and bases.(a) NO2

-(aq) + H3O+(aq) æ HNO2(aq) + H2O(l)

(b) 2H3O+(aq) + 2ClO4–(aq) + Mg2+(OH–)2(s) æ Mg2+(aq) + 2ClO–

4(aq) + 2H2O(l)

(c) HNO3(aq) + Al3+(OH–)3(s) æ (d) HCN(aq) + NaOH(aq) æ Answer 2.99:Completed equations:(c) 3HNO3(aq) + Al(OH)3(aq) æ Al(NO3)3(aq) + 3 H2O(l)

(d) HCN(aq) + NaOH(aq) æ NaCN(aq) + H2O(l)

Acids and bases: (a) Acids: H3O+(aq) and HNO2-(aq); Bases: NO2

-(aq) and H2O(l); (b) Acids: HClO4(aq) and H2O(l); Bases: Mg(OH)2(s) and Mg(ClO4)2(aq);



(c) Acids: 3HNO3(aq) and H2O (l); Bases: Al(OH)3(aq) and Al(NO3)3(aq)

(d) Acids: HCN(aq) and H2O(l); Bases: NaOH(aq) and NaCN(aq)

2.100. The Lewis structures of HOCO2–, (HO)2PO2

–, and HOPO32– are omitted from Table 2.7.

Draw their Lewis structures (showing all nonbonding electron pairs as a pair of dots and all covalent bonds as lines).Answer 2.100:

OC

O

O

H O P O

O

O

H H O P O

O

O

H

Section 2.14. Extent of proton-transfer reactions: Le Chatelier's principle

2.101. When ammonia dissolves in water, it does so as the result of an acid-base reaction. Two possible acid-base reactions of ammonia and water are:

H2O(l) + NH3(g) æ H3O+(aq) + NH2–(aq)

H2O(l) + NH3(g) æ OH–(aq) + NH4+(aq)

(a) Identify the Brønsted-Lowry acids and bases in each reaction by placing an A below each acid and a B below each base.(b) Use reasoning based on the relative electronegativities of nitrogen and oxygen to predict which equation represents the actual acid-base reaction when ammonia gas dissolves in water. (You can check your prediction by recalling that, in Investigate This 2.63, you discovered that an aqueous ammonia solution has a pH > 7.)Answer 2.101: (a)

H2O(l) + NH3(g) —> H3O+(aq) + –NH2(aq)

B A A B

H2O(l) + NH3(g) —> HO–(aq) + +NH4(aq)

A B B A

(b) The second equation represents the correct acid-base reaction. Oxygen is more electronegative than nitrogen. The first equation generates a very unfavorable (high energy) nitrogen anion, whereas the second equation generates the more favorable



hydroxide anion. The fact that the pH > 7 indicates that there is hydroxide present in concentrations greater than found in neutral water (pH = 7).

2.102. When methylamine, CH3NH2(g), dissolves in water, a weak electrical conductivity is observed. Explain this observation using a balanced equation in your answer. Omit any ions/molecules that do not directly participate in the reaction.Answer 2.102:Given: When methylamine is dissolved in water, a weak electrical conductivity is observed.Asked For: We need to explain why this solution conducts electricity as well as to write the net ionic equation for this observation.Recall: If the solution conducts electricity weakly, this means that the reaction does not proceed very far to the right and only a few ions are produced.Plan: In examining the structure of methylamine, we observe that its structure is similar to ammonia, NH3 in which one H from NH3 has been replaced with a methyl (CH3-) group. For review, we need to now examine the reaction of NH3 in water, as shown in Section 2.13 and apply it to methylamine. Explanation: In this problem, the balanced chemical reaction is the same as the total ionic equation which is also the same as the net ionic equation.CH3NH2(aq) + H2O æ CH3NH3

+(aq) + OH-(aq)

We observe that two ions are produced in this reaction, CH3NH3+ and OH-.

Since the solution weakly conducts electricity, only a few ions are produced and the reaction does conduct electricity. We indicate this with a two directional equilibrium arrow in the balanced equation (as shown above).

2.103. Ethylene glycol, HOCH2CH2OH (used in automobile antifreeze products), is miscible with water in all proportions. Will the resulting solution be basic, acidic, or neutral? Explain. Will the resulting solution display electrical conductivity? Explain.Answer 2.103:Given: Ethylene glycol, CH2(OH)CH2OH, is miscible with water in all proportions. Asked For: Explanation if the resulting solution will be basic, acidic, or neutral as well as if it will conduct electricity.Recall: In Section 2.2, solutions of polar molecules are discussed. In section 2.13, reactions of acids and bases are discussed.



Plan: We need to closely examine the structure of ethylene glycol and to recognize that it contains the same functional groups as glucose.Explanation: Ethylene glycol contains two alcohol functional groups. These two alcohol groups create hydrogen bonds with water. Just like glucose dissolving in water, ethylene glycol simply dissolves in water. It does not react with water. Thus, the resulting solution is neutral and does not conduct electricity.

2.104. Esterification (which is discussed in Chapter 6) is one of the most important reactions of carboxylic acids in biological systems. A simple example is the reaction of acetic (ethanoic) acid with ethanol to form ethyl acetate (a common solvent found in fingernail polish remover) and water in this equilibrium reaction:

CH3C(O)OH + HOCH2CH3 æ CH3C(O)OCH2CH3 + H2Oacetic acid ethanol ethyl acetate

Use Le Chatelier’s principle to predict and clearly explain the outcome of these reaction conditions:(a) Starting with 0.1 mole of acetic acid and 0.1 mole of ethanol, would more, less, or the same amount of ethyl acetate be formed, if water is added to the reaction mixture?(b) Would a mixture of 0.2 mole of acetic acid and 0.1 mole of ethanol form more, less, or the same amount of ethyl acetate as a mixture of 0.1 mole of acetic acid and 0.1 mole of ethanol?Answer 2.104: (a) The addition of water disturbs this equilibrium in a way that will use up the additional water molecules by increasing the extent of the reverse reaction. Of course, the reverse reaction uses up ethyl acetate in addition to water, so the overall yield of ester product would be reduced.

(b) The mixture containing 0.2 mol acetic acid will produce more ethyl acetate. Imagine adding 0.1 mol of (additional) acetic acid to the second equimolar mixture. The excess acetic acid would disturb the equilibrium by increasing the extent of the forward reaction.

2.105. Use explanations based on Le Chatelier’s principle to explain or make predictions in each of the following cases.(a) Consider This 2.79 examined the solubility of carbon dioxide in water. Why is the solubility of carbon dioxide greater in an aqueous sodium hydroxide solution than in water itself?



(b) Calcium sulfate is slightly soluble in water. If sodium sulfate (solid) is added to a saturated aqueous solution of calcium sulfate, what will happen to the concentration of calcium cation, [Ca2+(aq)]?Answer 2.105:(a) Consider the following equations:

CO2(g) + H2O æ (HO)2CO(aq)

(HO)2CO(aq) + H2O æ H3O+(aq) + HOCO2-(aq)

H3O+(aq) + OH-(aq) æ 2H2OOverall: CO2(g) + OH-(aq) æ HOCO2

-(aq)

Adding more hydroxide, OH-(aq), will shift the equilibrium to the right, causing more CO2(g) to react with OH-(aq) and produce more HOCO2

-(aq). Thus, CO2(g) is more soluble in aqueous sodium hydroxide.

(b) If sodium sulfate (solid) is added to a saturated aqueous solution of calcium sulfate, more solid calcium sulfate will form, lowering the concentration of calcium cation, Ca2+, in solution.

Section 2.15. EXTENSION — CO 2 and Le Chatelier’s Principle

2.106. Represent each of these statements as a complete balanced chemical equation.(a) Carbonic acid is formed when carbon dioxide reacts with water.(b) Calcium carbonate (limestone) reacts with carbonic acid to form an aqueous solution of calcium hydrogen carbonate.(c) Calcium hydrogen carbonate reacts with calcium hydroxide to form calcium carbonate precipitate.(d) Calcium hydrogen carbonate reacts with sodium hydroxide to form calcium carbonate precipitate and the water-soluble salt sodium carbonate.(e) The mixing of aqueous solutions of sodium hydrogen carbonate and sodium hydroxide is exothermic. A reaction has occurred but there is no precipitate.Answer 2.106: (a) Equation (2.27).(b) CaCO3(s) + (HO) 2CO(aq) —> Ca(HOCO2)2(aq)

(c) Ca(HOCO2)2(aq) + Ca(OH)2(aq) —> 2 CaCO3(s) + 2 H2O(d) Ca(HOCO2)2(aq) + 2 NaOH(aq) —> CaCO3(s) + Na2CO3(aq) + 2 H2O



2.107. The names stalactite and stalagmite for the structures that grow, respectively, down from the ceiling and up from the floor of a limestone cave (Figure 2.31) are derived from the Greek word meaning “to drip.” Indeed, if you examine the tip of a stalactite, you will often find a drop of liquid. The liquid is an aqueous solution containing calcium cations and hydrogen carbonate anions. See Check This 2.92.(a) As the water evaporates from this drop, what happens to the concentration of calcium cations? of hydrogen carbonate anions? Explain your reasoning.(b) What reaction does Le Chatelier’s principle predict will occur in the evaporating drop of solution in part (a)? Clearly explain your choice.(c) Does your answer in part (b) help explain the growth of stalactites? How about stalagmites (which grow directly under stalactites)? Give our reasoning clearly.Answer 2.107:(a) As the water droplet evaporates the concentrations of the solutes, calcium cations and hydrogen carbonate anions, increase. The volume of solution is less, so the number of ions per unit volume is greater, that is, a higher concentration.(b) Reaction (2.37) will occur in reverse:

Ca2+(aq) + 2HOCO22–(aq) CaCO3(s) + H2O(l) + CO2(g)

This happens because the concentrations of both reactants in this reaction increase as the droplets evaporate (see part (a)). The increased concentrations of the products of reaction (2.37) are a disturbance to the system. Le Chatelier’s Principle states that the system will respond by minimizing the disturbance, that is, by reacting in a way that decreases the concentrations. The reaction that does this is the reverse of reaction (2.37), which is written above.

(c) The result of the reaction in part (b) is to precipitate some solid calcium carbonate, which causes the stalactite to grow downward (very slowly). If the drop of liquid should drop off the tip of the stalactite, it will end up directly below and the same evaporation and precipitation process on the floor of the cave will build stalagmites upward toward the stalactites. If they meet, they form a column, which is another common structure in limestone caves. (e) Na(HOCO2)(aq) + NaOH(aq) —> Na2CO3(aq) + H2O



2.108. A gas is evolved when calcium carbonate is placed in an aqueous solution of HCl(g).(a) What is the gas? Explain how you reached this conclusion.(b) Write the net ionic equation for the reaction that produces the gas.(c) Show how Le Chatelier’s principle and your knowledge of gas solubilities explain the observed results.(a) Given: A gas is evolved when calcium carbonate is placed in an aqueous solution of HCl(g).Part (a): Ask For: What is the gas? How did you reach this conclusion? Answer:

We know that:HCl(g) + H2O(l ) H3O+(aq) + Cl-(aq)

Then,CaCO3(s) + H3O+(aq) æ Ca2+(aq) + HOCO2

-(aq)

The bicarbonate ion, HOCO2-(aq) is a Bronsted-Lowry base, that can then react with

HOCO2-(aq) + H3O+(aq) æ (HO)2CO(aq) + H2O(l)

Carbonic acid, (HO)2CO(aq) will decompose to yield water and dissolved carbon dioxide (Reaction (2.24)). Some of this carbon dioxide can leave the solution as a gas.Part (b):Asked For: Write a net ionic equation for the reaction that produces the gas.Recall: We will apply the strategy used for solving net ionic equations.Equations:(b) Balanced Chemical Reaction:CaCO3(s) + 2H3O+(aq) Ca2+(aq) + (HO)2CO(aq) + H2O Note: This balanced chemical reaction is also the total ionic and net ionic equation. Since carbonic acid, (HO)2CO(aq) decomposes to give carbon dioxide and water. We can substitute carbonic acid with carbon dioxide and water. Thus, the reaction can be written as:

CaCO3(s) + 2H3O+(aq) Ca2+(aq) + CO2(g) + 2H2O

Recap: In this problem, we need to recognize that carbonic acid, H2CO3, is very unstable and decomposes to water and dissolved carbon dioxide.



(c) The reactions that occur to produce the gas are:CaCO3(s) + 2H3O+(aq) Ca2+(aq) + 2H2CO3(aq) + 2H2O(l) H2CO3(aq) CO2(g) + H2O(l)

The reaction of calcium carbonate with hydrochloric acid, the first reaction, produces carbonic acid in the solution. Carbonic acid is formed by the combination of carbon dioxide with water (the reverse of the second reaction). We know that carbon dioxide is only sparingly soluble in water. Production of a lot of carbonic acid by the first reaction will disturb the second equilibrium, which will reduce the disturbance, LeChatelier’s principle, by reacting to reduce the amount of carbonic acid, thus forming carbon dioxide gas that escapes from the solution. The first reaction produces a lot of carbonic acid because the presence of a high concentration of hydronium ion (in the hydrochloric acid solution) is also a disturbance the system acts to reduce by reacting to lower the amount of hydronium ion present.

General Problems

2.109. (WEB) Sugar (sucrose) crystals are hard and they crunch and break when you put pressure on them. Grease is soft and easily smeared on a surface using little pressure. How do the Web Companion Chap 2, Sect 2.2.3 and 5, molecular level representations of sucrose and grease explain the observed macroscopic behavior of these substances? Clearly relate your explanation to the structures shown.

Answer 2.109: Grease contains long chains of hydrocarbons that have their electron clouds spread evenly over each molecule. Thus, the hydrocarbons can slide over each other because there is no polarity. Sugar molecules are polar and cannot slide of each other because of their uneven distribution of electrons within each sugar molecules.

2.110. (a) About 2 g of calcium sulfate, CaSO4(s), dissolve in a liter of water. What are the molarities of Ca2+(aq) and SO4

2–(aq) in a saturated solution of calcium sulfate? Is seawater saturated with calcium sulfate? (See Consider This 2.54 for the composition of seawater.) Explain your response.(b) An ionic compound is less soluble in a solution that already contains either its cation or anion. Is this effect consistent with Le Chatelier’s principle? Explain why or why not.(c) Does the effect described in part (b) influence your response to part (a)? How?(d) Does the effect described in part (b) help explain why the calcium carbonate in seashells (see the chapter opening) does not redissolve in the sea? Explain your response.


WEB


Answer 2.110:(a) The molar mass (formula mass) of CaSO4 is 136 g·mol–1:

Ca 40 g·mol–1

S 32 g·mol–1

4 O 4(16 g·mol–1) = 64 g·mol–1

The number of moles that dissolve in water is (2 g·L–1)/(136 g·mol–1) = 1.5 10–2 mol·L–

1. For every mole of the solid that dissolves, the solution contains a mole of calcium ions and a mole of sulfate ions. Thus, a saturated solution of calcium sulfate is 1.5 10–2 M in both ions. Seawater contains 1.0 10–2 M calcium cation and 2.8 10–2 M sulfate anion. There is a little less calcium ion than in a saturated solution of calcium sulfate, but almost twice as much sulfate ion as in the saturated calcium sulfate solution. Le Chatelier’s Principle would suggest that increasing the amount of the sulfate in a saturated solution of calcium sulfate (by adding some sodium sulfate, for example) should cause the dissolution reaction to be reversed to try to decrease the amount of sulfate, which, of course, also has the effect of decreasing the amount of calcium ion in the solution. When the high concentration of sulfate is accounted for, we are probably safe in saying that the seawater is close to saturation in calcium sulfate.

(b) The effect of cation or anion already present in a solution on the solubility of an ionic

compound that contains one or the other of these ions, was discussed in part (a) from the

point of view of addition of such an ion to a saturated solution of the ionic compound.

The conclusion was that the ionic compound will be less soluble than it would be in pure

water because Le Chatelier’s Principle favors the precipitation (or decreased solubility).

[From part (a), we find that a saturated solution of CaSO4 contains about 0.015 M concentrations of the calcium cation and sulfate anion. Seawater has [Ca2+] = 0.010 M and [SO42–] = 0.028 M, so it is not easy to tell, given where we are in the text, whether seawater is saturated or not. Ksp from the solubility is 0.00022 and the ion product in seawater is 0.00028, so the seawater is saturated.](c) Seawater contains more sulfate ion than is present in the saturated solution of calcium sulfate we calculated in part (a). Since seawater contains more of the anion than could have been present from CaSO4 alone, we can imagine that the CaSO4 dissolves in a solution that



already contains some of the anion. From part (b), we find that the solubility of the CaSO4 will be lower in such a solution. This is the direction of the effect we observe; there is less calcium cation in seawater than would be present in a saturated solution in pure water. It is likely that the seawater is saturated with CaSO4, in the presence of extra sulfate anion from some other source. In Chapter 9, we will find that the product of the concentrations of the cation and anion characterizes the solubility equilibrium for a saturated solution of a sparingly soluble salt like CaSO4. For the saturated solution in water, we have [Ca2+][SO42–] = (0.015 M)(0.015 M) = 0.00022 M2. For seawater, we have [Ca2+][SO42–] = (0.010 M)(0.028 M) = 0.00028 M2. These two values are almost the same and reinforce the conclusion that seawater is saturated with CaSO4, in the presence of extra sulfate anion.

(d) Since seawater already contains a substantial concentration of calcium anion, the solubility of calcium carbonate will be lower than it would be in pure water. The solubility of calcium carbonate is very low in pure water, so lowering it even further helps explain why seashells do not redissolve (at least not rapidly) in the sea.

2.111. A sample of salt water with a density of 1.02 g·mL–1 contains 17.8 ppm (by mass) of nitrate, NO3

–(aq). Calculate the molarity of nitrate ion in the sample of salt water. (ppm = parts per million).Answer 2.111:Given: Density of sea water (1.02 g·mL–1 which contains 17.8 ppm (by mass) of nitrate, NO3

–.Asked For: Molarity, M, of NO3

- in the seawater sample.Recall: This problem involves the concept of molarity.Plan: We need to focus on converting the density of seawater to reflect the grams of NO3

-

in a sample of sea water. Since we are told that sea water contains 17.8 ppm NO3-, we

can state that for every one million (1 x 106) grams of sea water, there are 17.8 grams of



NO3-. Now, we can express this relationship in terms of molarity by converting g of NO3

-

to moles of NO3- and mL to L.

Calculation: We use the density of sea water and the concentration of NO3- to calculate

the g of NO3- per mL.

We will now convert from g of NO3- to mol of NO3

- in seawater:To arrive at the molarity of NO3

- in our sample of seawater, we simply convert from mL seawater to L seawater:

Recap: Instead of a step-by-step calculation, we could have set up the calculation in one step:

2.112. (a) Assume that seawater may be represented by a 3.50% by weight aqueous solution of NaCl which has a density of 1.025 g·mL–1. What is the molarity of sodium chloride in this “seawater”? (b) Is your result in part (a) consistent with your answer to Consider This 2.51(b)? Explain why or why not.Answer 2.112:To find the molarity, it is necessary to find the ratio of the number of moles of NaCl in a liter of NaCl solution. It is convenient to break this calculation into two parts, finding first the number of moles of NaCl in 100 g of solution, and then the volume of 100 g of solution. Finding the number of moles of NaCl in 100 g from the weight percent:

Finding the volume of 100 g of solution:



Finding the ratio of moles of NaCl per liter of solution:

Therefore, the molarity of seawater is 0.613 M.

Note: We cannot assume that 1 g of solution has a volume of 1 mL. The density of the solution is given, and should be used. The answer makes sense because there are 35.0 g of NaCl in 1 L of water; this is about 0.6 mol NaCl.

2.113. A student prepared a solution for her biochemistry laboratory by weighing 5.15 grams of a compound and dissolving it in 10.0 grams of water. The concentration of this solution was 2.7 M, and its density was 1.34 g·mL–1. Which of the following compounds did the student use to prepare the solution? Explain your reasoning clearly.(a) (NH4)2SO4 (c) CsCl(b) KI (d) Na2S2O3

Answer 2.113: If we can find out how many moles of compound are dissolved, we can determine its molar mass and compare the result with the molar masses of the possible solutes. The known molarity of the solution, 2.7 M, is equal to the number of moles dissolved divided by the volume, in liters, of the solution. To get the volume of solution, we divide the total mass of the solution 15.15 g (= 5.15 g + 10.0 g) by its density 1.34 g·mL–1, and get a volume of 11.3 mL or 11.3 10–3 L.. Therefore, we have:

(moles of compound)/11.3 10–3 L = 2.7 Mmoles = 3.05 10–2 mol

It is not really legitimate to carry the third significant figure in this result, since the molarity has only two significant figures (the value is good to about 3%), but we’ll carry it along and check whether it makes a difference later. Now we know that 5.15 g of the compound is 3.05 10–2 mol of the compound, so:molar mass = (5.15 g)/(3.05 10–3 mol) = 169 g·mol–1 The molar masses of the four possible compounds are: (NH4)2SO4

, 132 g·mol–1; KI, 166

g·mol–1; CsCl, 168 g·mol–1; and Na2S2O3, 158 g·mol–1. The molar mass we calculated has

about a 3% uncertainty from the uncertainty in the second significant figure of molarity,



so the range of possibilities is about 164 to 174 g·mol–1. The possibilities, therefore, are

KI or CsCl for the compound the student dissolved. The calculation fits CsCl best, but the

uncertainty makes this identification unsure. We would need a better value for the

molarity to be sure.

2.114. What percentage (approximate) of the water molecules is protonated in aqueous solutions with these pH’s? Explain your reasoning. Hint: The molarity of water in pure water and dilute aqueous solutions is about 55.5 M.(a) 7 (b) 6 (c) 4Answer 2.114: (a) Approximately 10–7 molecules of water per liter are protonated and the concentration of pure water is approximately 55 moles per liter, the ratio is 10–7/55 = 1.8 x 10–9. Multiply this value by 100 to find that the percentage = 1.8 x 10–7 %.(b) 1.8 x 10–6 % (c) 1.8 x 10–4 % (or approximately 0.0002 %)

2.115. Figure 2.24 shows the concentration of hydroxide ion, [OH–(aq)], as well as the concentration of hydronium ion, [H3O+(aq)], correlated with the pH of solutions.(a) When the pH is 3, what are the concentrations of the hydroxide and hydronium ions? What is the numeric value of the mathematical product [H3O+(aq)]·[OH–(aq)] at this pH?(b) For any pH you choose, what is the mathematical product [H3O+(aq)]·[OH–(aq)]? Can you think of a reason for this result?

Answer 2.115: (a) 10-14 (b) 10-14

2.116. (a) The sulfur dioxide, SO2(g), molecule has a permanent dipole moment. What is the shape of the molecule? Explain the reasoning for your answer.(b) When sulfur dioxide, a nonmetal oxide, dissolves in water, the resulting solution conducts electricity. How can this be explained? Be sure to include an appropriate chemical reaction equation to justify your answer.Answer 2.116:(a) Sulfur dioxide has a bent structure, similar to water.(b) Asked For: Explanation of this observation, including the appropriate chemical equation.Explanation: When SO2 dissolves in water, ions must be formed since the resulting solution conducts electricity. From Section 2.13, we know that the oxides of nonmetals often dissolve and react with water to give acidic solutions. The reaction is as follows:



SO2(g) + H2O(l) æ (HO)2SO(aq)

The product, sulfurous acid, can transfer one or both protons to water to form ions:(HO)2SO(aq) + H2O(l) æ H3O+(l) + HOSO2

–(aq) HOSO2

–(aq) + H2O(l) æ H3O+(l) + SO32–(aq)

Sulfurous acid transfers essentially all of its first proton to water, so many ions will be present in the solution to conduct electricity.

2.117. (a) Table 2.7 lists the name of the HOCO2– ion (or HCO3

–) as hydrogen carbonate ion. This ion also has the common name “bicarbonate,” used in substances such as bicarbonate of soda, NaHCO3. Explain how this name can be rationalized.(b) TSP is the common name for a cleaning product containing sodium and phosphate ions. What is the chemical formula for the major ingredient in TSP and what do the letters TSP represent?Answer 2.117:(a) The name “bicarbonate” differentiates the HCO3

– ion from the carbonate ion, CO32–.

It may be that the “bi” refers to the two different cations that are present in common compounds containing this ion. For example, both sodium and hydrogen are present in “bicarbonate of soda”, NaHCO3. (b) TSP stands for “trisodium phosphate”. The correct formula is Na3PO4.

2.118. What volume of 0.075 M sulfuric acid, (HO)2SO2(aq), solution will be required to reach the equivalence point of the reaction with each of the following basic solutions? Hint: Each sulfuric acid molecule can provide two hydronium ions to the solution.(a) 1.00 g of KOH(s) dissolved in 75 mL of water(b) 1.00 g of KOH(s) dissolved in 150 mL of waterAnswer 2.118: Volume of based does not matter, just the actual number of moles of base in solution, which is 1.8 10-2 mol KOH(s). However, 0.075 M (HO)2SO2(aq)

produces 0.15 M H3O+(aq). Thus, 120 mL of (HO)2SO2(aq) are required to neutralize the base.

2.119. Assume that you have a one pound (454 g) container of drain cleaner, mostly solid sodium hydroxide, that you wish to get rid of by reaction with vinegar, about 0.9 M ethanoic (acetic) acid.(a) Write a balanced chemical reaction equation for the reaction between the drain cleaner and vinegar.



(b) What is the minimum volume of vinegar required to react completely with the drain cleaner? Explain your reasoning clearly.Answer 2.119:(a) HC2H3O2(aq) + NaOH(aq) NaC2H3O2(aq) + H2O(aq)

(b) 454 g = 11.4 mol NaOH(aq)

11.4 mol NaOH(aq) = (x L) (0.9 M HC2H3O2(aq)); 12.7 L of vinegar is the minimum volume.

2.120. Rain drops dissolve gaseous oxides of nitrogen and sulfur (formed by both natural processes and by burning fossil fuels) and form acidic solutions (acid rain), such as nitric and sulfuric acids [see equation (2.28)]. Acid rain has caused a small pond to become so acidic that most of its aquatic life has died. A community group has made a proposal to restore the pond by adding enough lime, CaO(s) (quicklime), to react with the hydronium ion by this reaction stoichiometry:

2H3O+(aq) + CaO(s) 3H2O(l) + Ca2+(aq)

They have asked for your help to figure out how much lime to use.(a) The volume of the pond is about 4.5 104 m3 (1 m3 = 1000 L) and the H3O+(aq) concentration is 5.0 10–5 M (pH = 4.30). How many moles of hydronium ion does the pond contain? Explain clearly the procedure you use.(b) How many moles of lime have to be added to react with 90% of the hydronium ion present? How many kilograms of lime is this? If lime is purchased in 50 pound bags, how many bags will be needed? Explain clearly, so the community group can understand.(c) Estimate what the pH of the pond will be after the lime is added.Answer 2.120:

(a) 4.5103 5.010-5 molL-1 = 2.25103 mol of hydronium ions in the pond.(b) 90% of 2.25103 mol = 2.03103 mol of hydronium ions

2.03103 mol of H3O+(aq) = 5.67104

g of CaO(s)

5.67104 g CaO(s) = 2.5 bags



(c) 5.010-6 M H3O+(aq); pH - 5.3

2.121. To cool themselves, many animals sweat when the weather is warm (see Chapter 1, Section 1.10). Chickens, however, do not have sweat glands, so they pant to help cool themselves. In hot weather, chickens pant a lot and lose more carbon dioxide than normal (they hyperventilate). The level of dissolved carbon dioxide in their blood decreases and the hens lay eggs with thinner and more fragile shells.(a) Why are the shells less sturdy than usual? What reaction(s) is(are) being affected?(b) Egg farmers use a simple and inexpensive method to keep their hens’ dissolved-carbon-dioxide levels normal in hot weather. What do you think they do?Answer 2.121:The shells are less sturdy than usual because they contain too little calcium carbonate, which is what make an eggshell hard. The precipitation of calcium carbonate is being affected because the concentration of carbonate in the chicken’s bloodstream goes down when more than the normal amount of carbon dioxide is excreted. The directional flow of the reactions is:

CaCO3(s) Ca2+(aq) + CO32–(aq)

CO32–(aq) + H2O(l) HOCO2

–(aq) + OH–(aq) HOCO2

–(aq) + H2O(l) (HO)2CO(aq) + OH–(aq) (HO)2CO(aq) CO2(g) + H2O(l)

As the chicken pants and more CO2(g) leaves, CaCO3(s) tends to dissolve, as shown, rather than precipitate to form strong eggshells. To counteract the loss of CO2(g) in the chickens’ breath, the farmers give them carbonated water, seltzer water, to drink. Chickens apparently like the seltzer water better than plain water and drink even more than usual, so the method is even more effective than might have been expected.

2.122. (WEB) The chemical equations we write to represent precipitation reactions usually look like this one:

Ag+(aq) + Cl–(aq) æ AgCl(s)

This representation gives us little clue about what might be happening at the molecular level during the precipitation. The animated movie, Web Companion Chap 2, Sect 2.6.3, showing the interaction of chloride and silver ions provides one way to visualize the precipitation process. Describe the steps in the process and illustrate your description


WEB


with chemical reaction equations. That is, try to translate the molecular level representation to a symbolic representation.Answer 2.122:


solutions ch02.doc

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