powerpoint presentation - suffolk county community … presentation chapter 15 acids and bases ©...

3/6/2014

1

© 2014 Pearson Education, Inc.

Sherril Soman

Grand Valley State University

Lecture Presentation

Chapter 15

Acids and Bases


Stomach Acid and Heartburn

• The cells that line your stomach produce

hydrochloric acid.

– To kill unwanted bacteria

– To help break down food

– To activate enzymes that break down food

• If the stomach acid backs up into your

esophagus, it irritates those tissues, resulting in

heartburn.

– Acid reflux


Curing Heartburn

• Mild cases of heartburn can be cured by

neutralizing the acid in the esophagus.

– Swallowing saliva, which contains bicarbonate ion

– Taking antacids that contain hydroxide ions and/or

carbonate ions


GERD

• Chronic heartburn is a problem for some people.

• GERD (gastroesophageal reflux disease) is chronic leaking

of stomach acid into the esophagus.

• In people with GERD, the muscles separating the stomach

from the esophagus do not close tightly, allowing stomach

acid to leak into the esophagus.

• Physicians diagnose GERD by attaching a pH sensor to the

esophagus to measure the acidity levels of the fluids

over time.


Properties of Acids

• Sour taste

• Ability to dissolve many metals

• Ability to neutralize bases

• Change blue litmus paper to red


Common Acids

3/6/2014

2


Structures of Acids

• Binary acids have acid hydrogens attached

to a nonmetal atom.

HCl, HF


Structure of Acids

• Oxyacids have acid hydrogens attached to

an oxygen atom.

H2SO4, HNO3


Structure of Acids

• Carboxylic acids

have COOH group.

HC2H3O2, H3C6H5O7

• Only the first H in the

formula is acidic.

The H is on the

COOH.


Properties of Bases

• Taste bitter

alkaloids = plant product that is alkaline

often poisonous

• Feel slippery

• Ability to turn red litmus paper blue

• Ability to neutralize acids


Common Bases


Indicators

• Indicators are chemicals that change color

depending on the solution’s acidity or

basicity.

• Many vegetable dyes are indicators.

– Anthocyanins

• Litmus

– From Spanish moss

– Red in acid, blue in base

• Phenolphthalein

– Found in laxatives

– Red in base, colorless in acid

3/6/2014

3


Definitions of Acids and Bases

• Arrhenius definition

– Based on H+ and OH-

• Brønsted–Lowry definition

– Based on reactions in which H+ is transferred

• Lewis definition


Arrhenius Theory

• Acids: produce H+ ions in aqueous solution.

HCl(aq) → H+(aq) + Cl−(aq)


Hydronium Ion

• The H+ ions produced by the acid are so reactive

they cannot exist in water.

– H+ ions are protons!

• Instead, they react with water molecules to produce

complex ions, mainly hydronium ion, H3O+.

H+ + H2O H3O+

– There are also minor amounts of H+ with multiple water

molecules, H(H2O)n+.


Arrhenius Theory

• Bases: produce OH− ions in aqueous solution.

NaOH(aq) → Na+(aq) + OH(aq)


Arrhenius Acid–Base Reactions

• The H+ from the acid combines with the OH−

from the base to make a molecule of H2O.

– It is often helpful to think of H2O as H—OH.

• The cation from the base combines with the

anion from the acid to make a salt.

acid + base → salt + water

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)


Problems with Arrhenius Theory

• It does not explain why molecular substances, such as NH3, dissolve in water to form basic solutions, even though they do not contain OH– ions.

• It does not explain how some ionic compounds, such as Na2CO3 or Na2O, dissolve in water to form basic solutions, even though they do not contain OH– ions.

• It does not explain why molecular substances, such as CO2, dissolve in water to form acidic solutions, even though they do not contain H+ ions.

• It does not explain acid–base reactions that take place outside aqueous solution.

3/6/2014

4


Brønsted–Lowry Acid–Base Theory

• It defines acids and bases based on what happens

in a reaction.

• Any reaction involving H+ (proton) that transfers

from one molecule to another is an acid–base

reaction, regardless of whether it occurs in

aqueous solution or if there is OH− present.

• All reactions that fit the Arrhenius definition also fit

the Brønsted–Lowry definition.


Brønsted–Lowry Theory

• The acid is an H+ donor.

• The base is an H+ acceptor. – Base structure must contain an atom with an unshared

pair of electrons.

• In a Brønsted–Lowry acid–base reaction, the

acid molecule donates an H+ to the base

molecule.

H–A + :B :A– + H–B+


Brønsted–Lowry Acids

• Brønsted–Lowry acids are H+ donors. – Any material that has H can potentially be a

Brønsted–Lowry acid.

– Because of the molecular structure, often one H in the molecule is easier to transfer than others.

• When HCl dissolves in water, the HCl is the acid because HCl transfers an H+ to H2O, forming H3O

+ ions. – Water acts as base, accepting H+.

HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)

acid base


Brønsted–Lowry Bases

• Brønsted–Lowry bases are H+ acceptors. – Any material that has atoms with lone pairs can potentially

be a Brønsted–Lowry base.

– Because of the molecular structure, often one atom in the molecule is more willing to accept H+ transfer than others.

• When NH3 dissolves in water, the NH3(aq) is the base because NH3 accepts an H+ from H2O, forming OH–(aq). – Water acts as acid, donating H+.

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

base acid


Amphoteric Substances

• Amphoteric substances can act as either an

acid or a base because they have both a

transferable H and an atom with lone pair

electrons.

• Water acts as base, accepting H+ from HCl.

HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)

• Water acts as acid, donating H+ to NH3.

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)


Brønsted–Lowry Acid–Base Reactions

• One of the advantages of Brønsted–Lowry theory is that it illustrates reversible reactions to be as follows:

H–A + :B :A– + H–B+

• The original base has an extra H+ after the reaction, so it will act as an acid in the reverse process.

• And the original acid has a lone pair of electrons after the reaction, so it will act as a base in the reverse process:

:A– + H–B+ H–A + :B

3/6/2014

5


Conjugate Acid–Base Pairs

• In a Brønsted-–Lowry acid–base reaction,

– the original base becomes an acid in the reverse reaction.

– the original acid becomes a base in the reverse process.

• Each reactant and the product it becomes is called

a conjugate pair.


Conjugate Pairs

A base accepts a proton and becomes a conjugate acid.

An acid donates a proton and becomes a conjugate base.


Arrow Conventions

• Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions.

• A single arrow indicates that all the reactant molecules are converted to product molecules at the end.

• A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products.

– in these notes


Strong or Weak

• A strong acid is a strong electrolyte. – Practically all the acid molecules ionize.

• A strong base is a strong electrolyte.

– Practically all the base molecules form OH– ions, either through dissociation or reaction with water.

• A weak acid is a weak electrolyte. – Only a small percentage of the molecules ionize, .

• A weak base is a weak electrolyte – only a small percentage of the base molecules form OH–

ions, either through dissociation or reaction with water, .


Strong Acids

• Strong acids donate practically all their H’s.

100% ionized in water

Strong electrolyte

• [H3O+] = [strong acid]

[X] means the molarity of X


Weak Acids

• Weak acids donate a small fraction of their H’s.

Most of the weak acid molecules do not

donate H to water.

Much less than 1% ionized in water

• [H3O+] << [weak acid]

3/6/2014

6


Examples of Strong Acids


Examples of Weak Acids


Strengths of Acids and Bases

• Commonly, acid or base strength is measured by determining the equilibrium constant of a substance’s reaction with water.

HAcid + H2O Acid− + H3O+

Base: + H2O HBase+ + OH−

• The farther the equilibrium position lies toward the products, the stronger the acid or base.

• The position of equilibrium depends on the strength of attraction between the base form and the H+. – Stronger attraction means stronger base or weaker acid.


Ionic Attraction and Acid Strength


General Trends in Acidity

• The stronger an acid is at donating H, the weaker

the conjugate base is at accepting H.

• Higher oxidation number = stronger oxyacid

– H2SO4 > H2SO3; HNO3 > HNO2

• Cation stronger acid than neutral molecule; neutral

stronger acid than anion

– H3O+ > H2O > OH−; NH4

+ > NH3 > NH2−

– Trend in base strength opposite


Autoionization of Water

• Water is amphoteric; it can act either as an acid or a base.

– Therefore, there must be a few ions present.

• About 2 out of every 1 billion water molecules form ions

through a process called autoionization.

H2O H+ + OH–

H2O + H2O H3O+ + OH–

• All aqueous solutions contain both H3O+ and OH–.

– The concentration of H3O+ and OH– are equal in water.

– [H3O+] = [OH–] = 10−7M at 25 °C

3/6/2014

7


Ion Product of Water

• The product of the H3O+ and OH– concentrations

is always the same number.

• The number is called the ion product of water

and has the symbol Kw.

– Also know as the dissociation constant of water

• [H3O+] × [OH–] = Kw = 1.00 × 10−14 at 25 °C

– If you measure one of the concentrations, you can

calculate the other.

• As [H3O+] increases the [OH–] must decrease so

the product stays constant.

– Inversely proportional


Acid Ionization Constant, Ka

• Acid strength is measured by the size of the equilibrium constant when it reacts with H2O.

• The equilibrium constant for this reaction is called the acid ionization constant, Ka.

– larger Ka = stronger acid


Table 15.5


Acidic and Basic Solutions

• All aqueous solutions contain both H3O+ and OH– ions.

• Neutral solutions have equal [H3O+] and [OH–].

– [H3O+] = [OH–] = 1.00 × 10−7

• Acidic solutions have a larger [H3O+] than [OH–].

– [H3O+] > 1.00 × 10−7; [OH–] < 1.00 × 10−7

• Basic solutions have a larger [OH–] than [H3O+].

– [H3O+] < 1.00 × 10−7; [OH–] > 1.00 × 10−7


Measuring Acidity: pH

• The acidity or basicity of a

solution is often expressed

as pH.

• pH = −log[H3O+]

Exponent on 10 with a

positive sign

pHwater = −log[10−7] = 7

Need to know the [H3O+]

concentration to find pH

• pH < 7 is acidic; pH > 7 is

basic. pH = 7 is neutral.

• [H3O+] = 10−pH


Sig., Figs., and Logs

• When you take the log of a number written in scientific

notation, the digits before the decimal point come

from the exponent on 10, and the digits after the

decimal point come from the decimal part of the

number.

log(2.0 x 106) = log(106) + log(2.0)

= 6 + 0.30303… = 6.30303...

• Because the part of the scientific notation number that

determines the significant figures is the decimal part,

the sig. figs. are the digits after the decimal point

in the log.

log(2.0 × 106) = 6.30

3/6/2014

8


What Does the pH Number Imply?

• The lower the pH, the more acidic the solution;

the higher the pH, the more basic the solution. 1 pH unit corresponds to a factor of 10 difference

in acidity.

• Normal range of pH is 0 to 14. pH 0 is [H3O

+] = 1 M; pH 14 is [OH–] = 1 M.

pH can be negative (very acidic) or larger than 14

(very alkaline).


pOH

• Another way of expressing the acidity/basicity of

a solution is pOH.

• pOH = −log[OH], [OH] = 10−pOH

pOHwater = −log[10−7] = 7

Need to know the [OH] concentration to find pOH

• pOH < 7 is basic; pOH > 7 is acidic; pOH = 7 is

neutral.

• pH + pOH = 14.0


Relationship between pH and pOH

• pH + pOH = 14.00 at 25 °C.

– You can use pOH to find the pH of a

solution.

[H3O ][OH ] K

w 1.0 1014

log [H3O ][OH ] log 1.0 1014

log [H3O ] log [OH ] 14.00

pH pOH 14.00


pK

• A way of expressing the strength of an acid or

base is pK.

• pKa = −log(Ka), Ka = 10−pKa

• pKb = −log(Kb), Kb = 10−pKb

• The stronger the acid, the smaller the pKa.

– Larger Ka = smaller pKa

– Because it is the –log

• The stronger the base, the smaller the pKb.

– Larger Kb = smaller pKb


[H3O+] and [OH−] in a Strong Acid or

Strong Base Solution

• There are two sources of H3O+ in an aqueous

solution of a strong acid—the acid and the water.

• There are two sources of OH− in an aqueous

solution of a strong acid—the base and the water.

• For a strong acid or base, the contribution of the

water to the total [H3O+] or [OH−] is negligible.

– The [H3O+]acid shifts the Kw equilibrium so far that [H3O

+]water is

too small to be significant.

• Except in very dilute solutions, generally < 1 × 10−4 M


Finding pH of a Strong Acid or Strong

Base Solution

• For a monoprotic strong acid [H3O+] = [HAcid].

– For polyprotic acids, the other ionizations can generally

be ignored.

• For H2SO4, the second ionization cannot be ignored.

– 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00

• For a strong ionic base, [OH−] = (number OH−) ×

[Base]. – For molecular bases with multiple lone pairs available,

only one lone pair accepts an H; the other reactions

can generally be ignored.

– 0.10 M Ca(OH)2 has [OH−] = 0.20 M and pH = 13.30.

3/6/2014

9


Finding the pH of a Weak Acid

• There are also two sources of H3O+ in an

aqueous solution of a weak acid—the acid and

the water.

• However, finding the [H3O+] is complicated by the

fact that the acid only undergoes partial ionization.

• Calculating the [H3O+] requires solving an

equilibrium problem for the reaction that defines

the acidity of the acid.

HAcid + H2O Acid + H3O+


Percent Ionization

• Another way to measure the strength of an acid is to

determine the percentage of acid molecules that

ionize when dissolved in water; this is called the

percent ionization.

– The higher the percent ionization, the stronger the acid.

• Because [ionized acid]equil = [H3O+]equil


Relationship between [H3O+]equilibrium and

[HA]initial

• Increasing the initial concentration of acid results in increased [H3O

+] at equilibrium.

• Increasing the initial concentration of acid results in decreased percent ionization.

• This means that the increase in [H3O

+] concentration is slower than the increase in acid concentration.


Why Doesn’t the Increase in H3O+ Keep

Up with the Increase in HA?

• The reaction for ionization of a weak acid is as follows:

HA(aq) + H2O(l) A−(aq) + H3O

+(aq)

• According to Le Châtelier’s principle, if we reduce the

concentrations of all the (aq) components, the

equilibrium should shift to the right to increase the total

number of dissolved particles.

– We can reduce the (aq) concentrations by using a more

dilute initial acid concentration.

• The result will be a larger [H3O+] in the dilute solution

compared to the initial acid concentration.

• This will result in a larger percent ionization.


Finding the pH of Mixtures of Acids

• Generally, you can ignore the contribution of the

weaker acid to the [H3O+]equil.

• For a mixture of a strong acid with a weak acid,

the complete ionization of the strong acid provides

more than enough [H3O+] to shift the weak acid

equilibrium to the left so far that the weak acid’s

added [H3O+] is negligible.

• For mixtures of weak acids, you generally only

need to consider the stronger for the same

reasons, as long as one is significantly stronger

than the other and their concentrations are

similar. © 2014 Pearson Education, Inc.

Strong Bases

• The stronger the base, the

more willing it is to accept H.

Use water as the standard acid.

• For ionic bases, practically all

units are dissociated into OH–

or accept H’s.

Strong electrolyte

Multi-OH strong bases

completely dissociated

3/6/2014

10


Weak Bases

• In weak bases, only a small

fraction of molecules

accept H’s.

Weak electrolyte

Most of the weak base molecules

do not take H from water.

Much less than 1% ionization

in water

• [HO–] << [weak base]

• Finding the pH of a weak base

solution is similar to finding the

pH of a weak acid.


Base Ionization Constant, Kb

• Base strength is measured by the size of the

equilibrium constant when it reacts with H2O

:Base + H2O OH− + H:Base+

• The equilibrium constant is called the base

ionization constant, Kb.

– Larger Kb = stronger base


Table 15.8 page 721

Common Weak Bases


Structure of Amines


Acid–Base Properties of Ions and Salts

• Salts are water-soluble ionic compounds.

• Salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic. – NaHCO3 solutions are basic.

• Na+ is the cation of the strong base NaOH.

• HCO3− is the conjugate base of the weak acid H2CO3.

• Salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic. – NH4Cl solutions are acidic.

• NH4+ is the conjugate acid of the weak base NH3.

• Cl− is the anion of the strong acid HCl.


Anions as Weak Bases

• Every anion can be thought of as the conjugate base of an acid.

• Therefore, every anion can potentially be a base. – A−(aq) + H2O(l) HA(aq) + OH−(aq)

• The stronger the acid, the weaker the conjugate base.

• An anion that is the conjugate base of a strong acid is pH neutral.

Cl−(aq) + H2O(l) HCl(aq) + OH−(aq)

• An anion that is the conjugate base of a weak acid is basic.

F−(aq) + H2O(l) HF(aq) + OH−(aq)

3/6/2014

11


Relationship between Ka of an Acid and Kb

of Its Conjugate Base

• Many reference books only give tables of Ka values

because Kb values can be found from them.

When you add

equations, you

multiply the

K’s.


Cations as Weak Acids

• Some cations can be thought of as the conjugate acid of a weak base.

– Others are the counterions of a strong base.

• Therefore, some cations can potentially be acidic. – MH+(aq) + H2O(l) MOH(aq) + H3O

+(aq)

• The stronger the base, the weaker the conjugate acid.

– A cation that is the counterion of a strong base is pH neutral.

– A cation that is the conjugate acid of a weak base is acidic.

NH4+(aq) + H2O(l) NH3(aq) + H3O

+(aq)

• Because NH3 is a weak base, the position of this equilibrium favors the right.


Metal Cations as Weak Acids

• Cations of small, highly charged metals are weakly acidic. – Alkali metal cations and alkali earth metal cations are

pH neutral.

– Cations are hydrated.

Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+

(aq) + H3O+(aq)


Classifying Salt Solutions as Acidic, Basic,

or Neutral

• If the salt cation is the counterion of a strong

base and the anion is the conjugate base of a

strong acid, it will form a neutral solution.

– NaCl Ca(NO3)2 KBr

• If the salt cation is the counterion of a strong


weak acid, it will form a basic solution.

– NaF Ca(C2H3O2)2 KNO2



or Neutral

• If the salt cation is the conjugate acid of a weak


strong acid, it will form an acidic solution.

– NH4Cl

• If the salt cation is a highly charged metal ion

and the anion is the conjugate base of a strong

acid, it will form an acidic solution.

– Al(NO3)3



or Neutral

• If the salt cation is the conjugate acid of a weak


weak acid, the pH of the solution depends on the

relative strengths of the acid and base.

– NH4F because HF is a stronger acid than NH4+, Ka of NH4

+

is larger than Kb of the F−; therefore, the solution will be

acidic.

3/6/2014

12


Ionization in Polyprotic Acids

• Because polyprotic acids ionize in steps, each H

has a separate Ka.

• Ka1 > Ka2 > Ka3

• Generally, the difference in Ka values is great

enough so that the second ionization does not

happen to a large enough extent to affect the pH.

– Most pH problems just do first ionization.

– Except H2SO4 uses [H2SO4] as the [H3O+] for the

second ionization.

• [A2−] = Ka2 as long as the second ionization is

negligible.



Ionization in H2SO4

• The ionization constants for H2SO4 are as follows:

H2SO4 + H2O HSO4 + H3O

+ strong

HSO4 + H2O SO4

2 + H3O+ Ka2 = 1.2 × 10−2

• For most sulfuric acid solutions, the second

ionization is significant and must be accounted for.

• Because the first ionization is complete, use the

given [H2SO4] = [HSO4−]initial = [H3O

+]initial.


Strengths of Binary Acids

• The more d+ H─X d− polarized the bond, the more acidic the bond.

• The stronger the H─X bond, the weaker the acid.

• Binary acid strength increases to the right across a period. – Acidity: H─C < H─N < H─O < H─F

• Binary acid strength increases down the column. – Acidity: H─F < H─Cl < H─Br < H─I


Relationship between Bond Strength and

Acidity

Acid

Bond

Energy

kJ/mol

Type of

Acid

HF 565 weak

HCl 431 strong

HBr 364 strong


Strengths of Oxyacids, H–O–Y

• The more electronegative the Y atom, the stronger the oxyacid.

– HClO > HIO

– Acidity of oxyacids decreases down a group.

• Same trend as binary acids

– Helps weaken the H–O bond.

• The larger the oxidation number of the central atom, the stronger the oxyacid.

– H2CO3 > H3BO3

– Acidity of oxyacids increases to the right across a period.

• Opposite trend of binary acids

• The more oxygens attached to Y, the stronger the oxyacid.

– Further weakens and polarizes the H–O bond

– HClO3 > HClO2

3/6/2014

13


Relationship between Electronegativity and

Acidity

Acid

H─O─Y

Electronegativity

of Y Ka

H─O─Cl 3.0 2.9 × 10−8

H─O─Br 2.8 2.0 × 10−9

H─O─I 2.5 2.3 × 10−11


Relationship between Number of Oxygens

on the Central Atom and Acidity


Lewis Acid–Base Theory

• Lewis acid–base theory focuses on transferring an electron pair. – Lone pair bond

– Bond lone pair

• Does NOT require H atoms

• The electron donor is called the Lewis base. – Electron rich; therefore nucleophile

• The electron acceptor is called the Lewis acid. – Electron deficient; therefore electrophile


Lewis Bases

• The Lewis base has electrons it is willing to give away to or share with another atom.

• The Lewis base must have a lone pair of electrons on it that it can donate.

• Anions are better Lewis bases than neutral atoms or molecules.

– N: < N:−

• Generally, the more electronegative an atom, the less willing it is to be a Lewis base.

– O: < S:


Lewis Acids

• They are electron deficient, either from being attached to electronegative atom(s) or not having an octet.

• They must have an empty orbital willing to accept the electron pair.

• H+ has empty 1s orbital.

• B in BF3 has empty 2p orbital and an incomplete octet.

• Many small, highly charged metal cations have empty orbitals they can use to accept electrons.

• Atoms that are attached to highly electronegative atoms and have multiple bonds can be Lewis acids.


Lewis Acid–Base Reactions

• The base donates a pair of electrons to the acid.

• It generally results in a covalent bond forming

H3N: + BF3 H3N─BF3

• The product that forms is called an adduct.

• Arrhenius and Brønsted–Lowry acid–base

reactions are also Lewis.

3/6/2014

14


Examples of Lewis Acid–Base Reactions


Examples of Lewis Acid–Base Reactions

Ag+(aq) + 2 :NH3(aq) Ag(NH3)2

+(aq)

Lewis

Acid Lewis

Base Adduct


Solution a. Because H2SO4 donates a proton to H2O in this reaction, it is the acid (proton donor). After H2SO4 donates the

proton, it becomes HSO4−, the conjugate base. Because H2O accepts a proton, it is the base (proton acceptor).

After H2O accepts the proton, it becomes H3O+, the conjugate acid.

H2SO4(aq) + H2O(l) ⟶ HSO4−(aq) + H3O

+(aq)

In each reaction, identify the Brønsted–Lowry acid, the Brønsted–Lowry base, the conjugate acid, and the

conjugate base.

a.

b.

Example 15.1 Identifying Brønsted–Lowry Acids and Bases and

Their Conjugates


Solution b. Because H2O donates a proton to HCO3

− in this reaction, it is the acid (proton donor). After H2O donates the

proton, it becomes OH−, the conjugate base. Because HCO3− accepts a proton, it is the base (proton acceptor).

After HCO3− accepts the proton, it becomes H2CO3, the conjugate acid.

HCO3−(aq) + H2O(l) H2CO3(aq) + OH−(aq)

For Practice 15.1 In each reaction, identify the Brønsted–Lowry acid, the Brønsted–Lowry base, the conjugate acid, and the

conjugate base.

a.

b.

Continued

Example 15.1 Identifying Brønsted–Lowry Acids and Bases and

Their Conjugates


Solution a. To find [OH−] use the ion product constant. Substitute the given value for [H3O

+] and solve the equation for [OH−].

Since [H3O+] > [OH−], the solution is acidic.

b. Substitute the given value for [H3O+] and solve the acid ionization equation for [OH−].

Since [H3O+] < [OH−], the solution is basic.

Calculate [OH−] at 25 °C for each solution and determine if the solution is acidic, basic, or neutral.

a. [H3O+] = 7.5 × 10−5 M b. [H3O

+] = 1.5 × 10−9 M

c. [H3O+] = 1.0 × 10−7 M

Example 15.2 Using Kw in Calculations


Solution c. Substitute the given value for [H3O

+] and solve the acid ionization equation for [OH−].

Since [H3O+] = 1.0 × 10−7 and [OH−] = 1.0 × 10−7, the solution is neutral.

For Practice 15.2 Calculate [H3O

+] at 25 °C for each solution and determine if the solution is acidic, basic, or neutral.

a. [OH−] = 1.5 × 10−2 M b. [OH−] = 1.0 × 10−7 M

c. [OH−] = 8.2 × 10−10 M

Continued

Example 15.2 Using Kw in Calculations

3/6/2014

15


Solution a. To calculate pH, substitute the given [H3O

+] into the pH equation.

Since pH < 7, this solution is acidic.

Calculate the pH of each solution at 25 °C and indicate whether the solution is acidic or basic.

a. [H3O+] = 1.8 × 10−4 M b. [OH−] = 1.3 × 10−2 M

Example 15.3 Calculating pH from [H3O+] or [OH−]


For Practice 15.3 Calculate the pH of each solution and indicate whether the solution is acidic or basic.

a. [H3O+] = 9.5 × 10−9 M b. [OH−] = 7.1 × 10−3 M

Continued

Example 15.3 Calculating pH from [H3O+] or [OH−]

b. First use Kw to find [H3O+] from [OH−].

Then substitute [H3O+] into the pH expression to find pH.

Since pH > 7, this solution is basic.


Solution To find the [H3O

+] from pH, start with the equation that defines pH. Substitute the given value of pH and then

solve for [H3O+]. Since the given pH value was reported to two decimal places, the [H3O

+] is written to two

significant figures. (Remember that 10log x = x (see Appendix I ) . Some calculators use an inv log key to represent

this function.)

For Practice 15.4 Calculate the [H3O

+] for a solution with a pH of 8.37.

Calculate the [H3O+] for a solution with a pH of 4.80.

Example 15.4 Calculating [H3O+] from pH


Procedure For… Finding the pH (or [H3O

+]) of a Weak Acid Solution

To solve these types of problems, follow the outlined procedure.

Solution Step 1 Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE

table showing the given concentration of the weak acid as its initial concentration. Leave room in

the table for the changes in concentrations and for the equilibrium concentrations. (Note that the

[H3O+] is listed as approximately zero because the autoionization of water produces a negligibly small

amount of H3O+).

Find the [H3O+] of a 0.100 M HCN solution.

Example 15.5 Finding the [H3O+] of a Weak Acid Solution


Step 2 Represent the change in the concentration of H3O+ with the variable x. Define the changes in the

concentrations of the other reactants and products in terms of x. Always keep in mind the

stoichiometry of the reaction.

Step 3 Sum each column to determine the equilibrium concentrations in terms of the initial

concentrations and the variable x.

Continued



Step 4 Substitute the expressions for the equilibrium concentrations (from step 3) into the expression

for the acid ionization constant (Ka). In many cases, you can make the approximation that x is small

(as discussed in Section 14.8). Substitute the value of the acid ionization constant (from Table 15.5)

into the Ka expression and solve for x. Confirm that the x is small approximation is valid by

calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should

be less than 0.05 (or 5%).

Therefore the approximation is valid.

Continued


3/6/2014

16


Step 5 Determine the [H3O+] from the calculated value of x and calculate the pH if necessary.

[H3O+] = 7.0 × 10−6 M

(pH was not asked for in this problem.)

Step 6 Check your answer by substituting the computed equilibrium values into the acid ionization

expression. The calculated value of Ka should match the given value of Ka. Note that rounding errors

and the x is small approximation could result in a difference in the least significant digit when

comparing values of Ka.

Since the calculated value of Ka matches the given value, the answer is valid.

For Practice 15.5 Find the H3O

+ concentration of a 0.250 M hydrofluoric acid solution.

Continued



Procedure For… Finding the pH (or [H3O

+]) of a Weak Acid Solution

To solve these types of problems, follow the outlined procedure.

Solution Step 1 Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE

table showing the given concentration of the weak acid as its initial concentration. Leave room in

the table for the changes in concentrations and for the equilibrium concentrations. (Note that the

[H3O+] is listed as approximately zero because the autoionization of water produces a negligibly small

amount of H3O+).

Find the pH of a 0.200 M HNO2 solution.

Example 15.6 Finding the pH of a Weak Acid Solution


Step 2 Represent the change in the concentration of H3O+ with the variable x. Define the changes in the

concentrations of the other reactants and products in terms of x. Always keep in mind the

stoichiometry of the reaction.

Step 3 Sum each column to determine the equilibrium concentrations in terms of the initial

concentrations and the variable x.

Continued



Step 4 Substitute the expressions for the equilibrium concentrations (from step 3) into the expression

for the acid ionization constant (Ka). In many cases, you can make the approximation that x is small

(as discussed in Section 14.8). Substitute the value of the acid ionization constant (from Table 15.5)

into the Ka expression and solve for x. Confirm that the x is small approximation is valid by

calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should

be less than 0.05 (or 5%).

Therefore the approximation is valid (but barely so).

Continued



Step 5 Determine the [H3O+] from the calculated value of x and calculate the pH if necessary.

Step 6 Check your answer by substituting the computed equilibrium values into the acid ionization

expression. The calculated value of Ka should match the given value of Ka. Note that rounding errors

and the x is small approximation could result in a difference in the least significant digit when

comparing values of Ka.


For Practice 15.6 Find the pH of a 0.0150 M acetic acid solution.

Continued



Solution Step 1 Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table

showing the given concentration of the weak acid as its initial concentration. (Note that the H3O+

concentration is listed as approximately zero. Although a little H3O+ is present from the autoionization

of water, this amount is negligibly small compared to the amount of HClO2 or H3O+ formed by

the acid.)

Find the pH of a 0.100 M HClO2 solution.

Example 15.7 Finding the pH of a Weak Acid Solution in Cases

Where the x is small Approximation Does Not Work

3/6/2014

17


Step 2 Represent the change in [H3O+] with the variable x. Define the changes in the concentrations of the

other reactants and products in terms of x.

Step 3 Sum each column to determine the equilibrium concentrations in terms of the initial concentrations

and the variable x.

Continued




Step 4 Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the

acid ionization constant (Ka). Make the x is small approximation and substitute the value of the acid

ionization constant (from Table 15.5) into the Ka expression. Solve for x.

Continued




Check to see if the x is small approximation is valid by calculating the ratio of x to the number it was

subtracted from in the approximation. The ratio should be less than 0.05 (or 5%).

Therefore, the x is small approximation is not valid.

Continued




Step 4a If the x is small approximation is not valid, solve the quadratic equation explicitly or use the method of

successive approximations to find x. In this case, we solve the quadratic equation.

Since x represents the concentration of H3O+, and since concentrations cannot be negative, we reject

the negative root. x = 0.028

Continued




Step 5 Determine the H3O+ concentration from the calculated value of x and calculate the pH (if necessary).

Step 6 Check your answer by substituting the calculated equilibrium values into the acid ionization expression.

The calculated value of Ka should match the given value of Ka. Note that rounding errors could result

in a difference in the least significant digit when comparing values of Ka.


For Practice 15.7 Find the pH of a 0.010 M HNO2 solution.

Continued




Solution Use the given pH to find the equilibrium concentration of [H3O

+]. Then write the balanced equation for the ionization of

the acid and use it as a guide to prepare an ICE table showing all known concentrations.

A 0.100 M weak acid (HA) solution has a pH of 4.25. Find Ka for the acid.

Example 15.8 Finding the Equilibrium Constant from pH

3/6/2014

18


Use the equilibrium concentration of H3O+ and the stoichiometry of the reaction to predict the changes and

equilibrium concentration for all species. For most weak acids, the initial and equilibrium concentrations of the

weak acid (HA) are effectively equal because the amount that ionizes is usually very small compared to the initial

concentration.

Substitute the equilibrium concentrations into the expression for Ka and calculate its value.

Continued



For Practice 15.8 A 0.175 M weak acid solution has a pH of 3.25. Find Ka for the acid.

Continued



Solution To find the percent ionization, you must find the equilibrium concentration of H3O

+. Follow the procedure in Example

15.5, shown in condensed form here.

Therefore, [H3O+] = 0.034 M.

Find the percent ionization of a 2.5 M HNO2 solution.

Example 15.9 Finding the Percent Ionization of a Weak Acid


Use the definition of percent ionization to calculate it. (Since the percent ionization is less than 5%, the x is small

approximation is valid.)

For Practice 15.9 Find the percent ionization of a 0.250 M HC2H3O2 solution at 25 °C.

Continued

Example 15.9 Finding the Percent Ionization of a Weak Acid


Solution The three possible sources of H3O

+ ions are HF, HClO, and H2O. Write the ionization equations for the three sources

and their corresponding equilibrium constants. Since the equilibrium constant for the ionization of HF is about

12,000 times larger than that for the ionization of HClO, the contribution of HF to [H3O+] is by far the greatest. You

can therefore just calculate the [H3O+] formed by HF and neglect the other two potential sources of H3O

+.

Write the balanced equation for the ionization of HF and use it as a guide to prepare an ICE table.

Find the pH of a mixture that is 0.150 M in HF and 0.100 M in HClO.

Example 15.10 Mixtures of Weak Acids


Substitute the expressions for the equilibrium concentrations into the expression for the acid ionization constant

(Ka). Since the equilibrium constant is small relative to the initial concentration of HF, you can make the x is small

approximation. Substitute the value of the acid ionization constant (from Table 15.5) into the Ka expression and

solve for x.

Confirm that the x is small approximation is valid by calculating

the ratio of x to the number it was subtracted from in the

approximation. The ratio should be less than 0.05 (or 5%).

Therefore, the approximation is valid (though barely so).

Continued


3/6/2014

19


Determine the H3O+ concentration from the calculated value of x and find the pH.

For Practice 15.10 Find the ClO− concentration of the above mixture of HF and HClO.

Continued



Solution a. Since KOH is a strong base, it completely dissociates into K+ and OH− in solution. The concentration of OH−

will therefore be the same as the given concentration of KOH. Use this concentration and Kw to find [H3O+].

Then substitute [H3O+] into the pH expression to find the pH.

What is the OH− concentration and pH in each solution?

a. 0.225 M KOH b. 0.0015 M Sr(OH)2

Example 15.11 Finding the [OH−] and pH of a Strong Base Solution


b. Since Sr(OH)2 is a strong base, it completely dissociates into 1 mol of Sr2+ and 2 mol of OH− in solution. The

concentration of OH− will therefore be twice the given concentration of Sr(OH)2.

Use this concentration and Kw to find [H3O+].

Substitute [H3O+] into the pH expression to find the pH.

For Practice 15.11 Find the [OH−] and pH of a 0.010 M Ba (OH)2 solution.

Continued

Example 15.11 Finding the [OH−] and pH of a Strong Base Solution


Solution Step 1 Write the balanced equation for the ionization of water by the base and use it as a guide to prepare an

ICE table showing the given concentration of the weak base as its initial concentration. Leave room in

the table for the changes in concentrations and for the equilibrium concentrations. (Note that you should

list the OH− concentration as approximately zero. Although a little OH is present from the autoionization

of water, this amount is negligibly small compared to the amount of OH− formed by the base.)

Find the [OH−] and pH of a 0.100 M NH3 solution.

Example 15.12 Finding the [OH−] and pH of a Weak Base Solution


Step 2 Represent the change in the concentration of OH− with the variable x. Define the changes in the

concentrations of the other reactants and products in terms of x.


and the variable x.

Continued



Step 4 Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the

base ionization constant. In many cases, you can make the approximation that x is small (as discussed

in Chapter 14).

Substitute the value of the base ionization constant (from Table 15.8) into the Kb expression and

solve for x.

Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was


Therefore, the approximation is valid.

Continued


3/6/2014

20


Step 5 Determine the OH− concentration from the calculated value of x.

Use the expression for Kw to find [H3O+].

Substitute [H3O+] into the pH equation to find pH.

For Practice 15.12 Find the [OH−] and pH of a 0.33 M methylamine solution.

Continued



Solution a. From Table 15.3, we can see that NO3

− is the conjugate

base of a strong acid (HNO3) and is therefore pH-neutral.

b. From Table 15.5 (or from its absence in Table 15.3), we

know that NO2− is the conjugate base of a weak acid

(HNO2) and is therefore a weak base.

Classify each anion as a weak base or pH-neutral:

a. NO3− b. NO2

− c. C2H3O2−

Example 15.13 Determining Whether an Anion Is Basic or pH-Neutral


c. From Table 15.5 (or from its absence in Table 15.3), we know that C2H3O2− is the conjugate base of a weak acid

(HC2H3O2) and is therefore a weak base.

Continued



For Practice 15.13 Classify each anion as a weak base or pH-neutral:

a. CHO2− b. ClO4

−

Continued



Solution Step 1 Since the Na+ ion does not have any acidic or basic properties, you can ignore it. Write the balanced

equation for the ionization of water by the basic anion and use it as a guide to prepare an ICE table

showing the given concentration of the weak base as its initial concentration.

Find the pH of a 0.100 M NaCHO2 solution. The salt completely dissociates into Na+(aq) and CHO2−(aq), and

the Na+ ion has no acid or base properties.

Example 15.14 Determining the pH of a Solution Containing an Anion

Acting as a Base


Step 2 Represent the change in the concentration of OH− with the variable x. Define the changes in the

concentrations of the other reactants and products in terms of x.


and the variable x.

Continued


Acting as a Base

3/6/2014

21


Step 4 Find Kb from Ka (for the conjugate acid).

Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for Kb.

In many cases, you can make the approximation that x is small.

Substitute the value of Kb into the Kb expression and solve for x.

Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was


Continued


Acting as a Base


Therefore, the approximation is valid.

Continued


Acting as a Base


Step 5 Determine the OH− concentration from the calculated value of x.

Use the expression for Kw to find [H3O+].

Substitute [H3O+] into the pH equation to find pH.

For Practice 15.14 Find the pH of a 0.250 M NaC2H3O2 solution.

Continued


Acting as a Base


Solution a. The C5H5NH+ cation is the conjugate acid of a weak base and is therefore a weak acid.

b. The Ca2+ cation is the counterion of a strong base and is therefore pH-neutral (neither acidic nor basic).

c. The Cr3+ cation is a small, highly charged metal cation and is therefore a weak acid.

For Practice 15.15 Classify each cation as a weak acid or pH-neutral.

a. Li+ b. CH3NH3+ c. Fe3+

Classify each cation as a weak acid or pH-neutral.

a. C5H5NH+ b. Ca2+ c. Cr3+

Example 15.15 Determining Whether a Cation Is Acidic or pH-Neutral


Solution a. The Sr2+ cation is the counterion of a strong base (Sr(OH)2) and is pH-neutral. The Cl− anion is the conjugate

base of a strong acid (HCl) and is pH-neutral as well. The SrCl2 solution is therefore pH-neutral (neither acidic

nor basic).

b. The Al3+ cation is a small, highly charged metal ion (that is not an alkali metal or an alkaline earth metal) and

is a weak acid. The Br− anion is the conjugate base of a strong acid (HBr) and is pH-neutral. The AlBr3 solution

is therefore acidic.

Determine if the solution formed by each salt is acidic, basic, or neutral.

a. SrCl2 b. AlBr3 c. CH3NH3NO3

d. NaCHO2 e. NH4F

Example 15.16 Determining the Overall Acidity or Basicity of Salt

Solutions


c. The CH3NH3+ ion is the conjugate acid of a weak base (CH3NH2) and is acidic. The NO3

− anion is the conjugate

base of a strong acid (HNO3) and is pH-neutral. The CH3NH3NO3 solution is therefore acidic.

d. The Na+ cation is the counterion of a strong base and is pH-neutral. The CHO2− anion is the conjugate base of a

weak acid and is basic. The NaCHO2 solution is therefore basic.

Continued


Solutions

3/6/2014

22


e. The NH4+ ion is the conjugate acid of a weak base (NH3) and is acidic. The F− ion is the conjugate base of a weak

acid and is basic. To determine the overall acidity or basicity of the solution, compare the values of Ka for the

acidic cation and Kb for the basic anion. Obtain each value of K from the conjugate by using Ka × Kb = Kw.

Since Ka is greater than Kb, the solution is acidic.

Continued


Solutions


For Practice 15.16 Determine if the solution formed by each salt is acidic, basic, or neutral.

a. NaHCO3 b. CH3CH2NH3Cl c. KNO3 d. Fe(NO3)3

Continued


Solutions


Solution To find the pH, you must find the equilibrium concentration of H3O

+. Treat the problem as a weak acid pH problem

with a single ionizable proton. The second proton contributes a negligible amount to the concentration of H3O+ and

can be ignored. Follow the procedure from Example 15.6, shown in condensed form here. Use Ka1 for ascorbic acid

from Table 15.10.

Confirm that the x is small approximation is valid by calculating

the ratio of x to the number it was subtracted from in the

approximation. The ratio should be less than 0.05 (or 5%).

Calculate the pH from H3O+ concentration.

Find the pH of a 0.100 M ascorbic acid (H2C6H6O6) solution.

Example 15.17 Finding the pH of a Polyprotic Acid Solution


The approximation is valid. Therefore,

For Practice 15.17 Find the pH of a 0.050 M H2CO3 solution.

Continued

Example 15.17 Finding the pH of a Polyprotic Acid Solution


Solution Sulfuric acid is strong in its first ionization step and weak in its second. Begin by writing the equations for the two

steps. As the concentration of an H2SO4 solution becomes smaller, the second ionization step becomes more

significant because the percent ionization increases (as discussed in Section 15.6). Therefore, for a concentration of

0.0100 M, you can’t neglect the H3O+ contribution from the second step, as you can for other polyprotic acids. You

must calculate the H3O+ contributions from both steps.

The [H3O+] that results from the first ionization step is 0.0100 M (because the first step is strong). To determine

the [H3O+] formed by the second step, prepare an ICE table for the second step in which the initial concentration

of H3O+ is 0.0100 M. The initial concentration of HSO4

− must also be 0.0100 M (due to the stoichiometry of the

ionization reaction).

Find the pH of a 0.0100 M sulfuric acid (H2SO4) solution.

Example 15.18 Dilute H2SO4 Solutions


Substitute the expressions for the equilibrium concentrations (from the table just shown) into the expression for Ka2.

In this case, you cannot make the x is small approximation because the equilibrium constant (0.012) is not small

relative to the initial concentration (0.0100).

Substitute the value of Ka2 and multiply out the expression to arrive at the standard quadratic form.

Continued


3/6/2014

23


Solve the quadratic equation using the quadratic formula.

Since x represents a concentration, and since concentrations cannot be negative, we reject the negative root.

x = 0.0045

Determine the H3O+ concentration from the calculated value of x and calculate the pH. Notice that the second step

produces almost half as much H3O+ as the first step—an amount that must not be neglected. This will always be

the case with dilute H2SO4 solutions.

Continued



For Practice 15.18 Find the pH and [SO4

2−] of a 0.0075 M sulfuric acid solution.

Continued



Solution To find the [C6H6O6

2−], use the concentrations of [HC6H6O6−] and H3O

+ produced by the first ionization step (as

calculated in Example 15.17) as the initial concentrations for the second step. Because of the 1:1 stoichiometry,

[HC6H6O6−] = [H3O

+]. Then solve an equilibrium problem for the second step similar to that of Example 15.6,

shown in condensed form here. Use Ka2 for ascorbic acid from Table 15.10.

Find the [C6H6O62−] of the 0.100 M ascorbic acid (H2C6H6O6) solution in Example 15.17.

Example 15.19 Finding the Concentration of the Anions for a Weak

Diprotic Acid Solution


Since x is much smaller than 2.8 × 10−3, the x is small approximation is valid. Therefore,

[C6H6O62−] = 1.6 × 10−12 M.

For Practice 15.19 Find the [CO3

2−] of the 0.050 M carbonic acid (H2CO3) solution in For Practice 15.17.

Continued

Example 15.19 Finding the Concentration of the Anions for a Weak

Diprotic Acid Solution


U.S. Fuel Consumption

• Over 85% of the energy use in the United States comes

from the combustion of fossil fuels.

– Oil, natural gas, coal

• Combustion of fossil fuels produces CO2.

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

• Natural fossil fuels also contain small amounts of S that

burn to produce SO2(g).

S(s) + O2(g) → SO2(g)

• The high temperatures of combustion allow N2(g) in the

air to combine with O2(g) to form oxides of nitrogen.

N2(g) + 2 O2(g) → 2 NO2(g)


3/6/2014

24


What Is Acid Rain?

• Natural rain water has a pH of 5.6.

– Naturally slightly acidic due mainly to CO2

• Rain water with a pH lower than 5.6 is called

acid rain.

• Acid rain is linked to damage in ecosystems and

structures.


What Causes Acid Rain?

• Many natural and pollutant gases dissolved in the

air are nonmetal oxides.

– CO2, SO2, NO2

• Nonmetal oxides are acidic.

CO2(g) + H2O(l) H2CO3(aq)

2 SO2(g) + O2(g) + 2 H2O(l) 2 H2SO4(aq)

4 NO2(g) + O2(g) + 2 H2O(l) 4 HNO3(aq)

• Processes that produce nonmetal oxide gases as

waste increase the acidity of the rain.

– Natural – volcanoes and some bacterial action

– Man made – combustion of fuel


pH of Rain in Different Regions

Figure 15.16 pg 740


Weather Patterns

• The prevailing winds in the United States travel

west to east.

• Weather patterns may cause rain to be

acidic in regions other than where the

nonmetal oxide is produced.

• Much of the Northeast United States has rain of

very low pH, even though it has very low sulfur

emissions, due in part to the general weather

patterns.


Damage from Acid Rain

• Acids react with metals and materials that contain

carbonates.

• Acid rain damages bridges, cars, and other

metallic structures.

• Acid rain damages buildings and other structures

made of limestone or cement.

• Acidifying lakes affects aquatic life.

• Soil acidity causes more dissolving of minerals

and leaching more minerals from soil, making it

difficult for trees.


Damage from Acid Rain

3/6/2014

25


Acid Rain Legislation

• 1990 Clean Air Act attacks acid rain.

– Forces utilities to reduce SO2

• The result is acid rain in the Northeast stabilized

and beginning to be reduced.

powerpoint presentation - suffolk county community … presentation chapter 15 acids and bases ©...

Documents