ionic equilibrium

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Ionic equillibrium This type of equilibrium is observed in substances that undergo ionization easily, or in polar substances in which ionization can be induced. Ionic and polar substances are more easily soluble in polar solvents because of the ease of ionization taking place in the solvent medium. With the dissolution of ionic and polar substances in the solvent, these solutions become rich in mobile charge carriers (ions) and thus can conduct electricity. Substances, which are capable of conducting electricity are called as electrolytes while those substances which are non-conducting are called as non-electrolytes.

Ionization in electrolytes

Strong electrolytes are almost completely dissociated / ionized into the constituent ions in aqueous solution. Even at higher concentrations, very small amount of any strong electrolyte is present in the unionized form. Weak electrolytes are however, partially ionized and a dynamic equilibrium exist between the ionized and unionized forms. In 1887, S.Arrhenius postulated the first theory of electrolytic ionization. The basic postulates of his theory are:

When an electrolyte is dissolved in water, it ionizes to produce ions in the solution. These ions are free to move throughout the bulk of the solution.

The total number of positive charges is equal to the total number of negative charges in any solution of any electrolyte. The ions tend to recombine to form the unionized electrolyte. Hence, a dynamic equilibrium exists between the ionized and unionized form of the electrolyte. For example, an electrolyte AB ionizes in the solution to form A+ and B-. The equilibrium established in the solution is:

The ionization constant (K) is then given by,

The properties of an electrolyte in the solution are due to the ions it gives in the solution. For example, a solution of copper sulphate is blue due to the presence of Cu2+(aq) ions in it.

The fraction of the total number of molecules present as free ions in the solution is known as the degree of ionization (a). This is defined as,

The degree of ionization at any concentration (C) of AB in moles per litre is denoted as aC.

Ionization of weak electrolyte Ionization of an acid (weak)

An acid of the type HA can undergo ionization when dissolved in water as,


If 'n' moles of the acid are dissolved in 'V' units of volume (litres L) and 'a' is the degree of ionization, then the equilibrium amounts of various species and the concentration in moles per litre in the solution are,

where 'C' is the molar concentration of the acid. The ionization constant for the above reaction is given by,

where 'V' is the volume of the solution in litres containing one mole of the acid HA. As the degree of ionization increases with-dilution then, the hydronium ion or hydrogen ion concentration is given by,

Ionization of a base (weak)

The ionization of a weak base is characterized by the equilibrium,

If 'n' moles of the base are dissolved in 'V' units of volume (litres L) and a is the degree of ionization, then the equilibrium amounts of various species and the concentration in moles per litre in the solution are,

where 'C' is the molar concentration of the base. The ionization constant for the above reaction is given by,


If 'a' is small then 1 - a ≈ 1

Hence, Kb = Ca2

where 'V' is the volume of the solution in litres containing one mole of the base MOH. As the degree of ionization increases with dilution then, the hydroxide concentration is given by,

Protic acids

Acids which contain ionizable hydrogen are called protic acids. These are classified as:

Types of protic acids

Monoprotic acids

These acids contain only one ionizable hydrogen in its molecule. It is denoted by HA. Example: HCl, CH3COOH, HNO3 etc.

Diprotic acids

The diprotic acids contain two ionizable hydrogen in its molecule. They are denoted by H2A. Example: H2SO4, (COOH)2, H2CO3 etc.

Triprotic acids

These acids contain three ionizable hydrogen in its molecule. It is denoted by H3A. Example: H3PO4 H3A SO4 etc.

In all these acids, the primary ionization constant (K1) is stronger than the secondary (K2), which is much stronger than the tertiary (K3) ionization constant. This is because a proton (H3O+) would be released more readily from an uncharged molecule than from a mononegative ion, and more readily from a mononegative ion than from a binegative ion. Also the hydronium concentration from the first ionization will suppress the subsequent ionizations due to common ion effect. Some general observations on the behaviour of protic acids in aqueous solution are:

The protic acids which have very high value of the acid dissociation or ionization constant, ionize/dissociate almost completely in aqueous solution at ordinary dilutions. They are called strong acids.

Example: HCl, H2SO4, and HNO3.

Acids, which dissociate in aqueous solution to a smaller extent at ordinary dilution, to give low concentration of H+ ions in solution, are called weak acids.

Example: CH3COOH, (COOH)2, H3PO4, HCN, HF, etc. The Ka or K1 values of such acids are much smaller than one.

On dilution, the ionization of an acid increases. So, concentration of H+ ions also increases on dilution. Therefore, strength of the acid increases with dilution. It is for this reason that the acid strengths are compared at equal concentrations.

For di- and triprotic acids, first dissociation is stronger than the second, which in turn is much stronger than the third dissociation i.e., K1 » K2 » K3. This is because the removal of a proton from a negatively charged species is more difficult than from a neutral molecule.


Degree of ionization and ionization constant The equilibrium concentration of various species at equilibrium can be represented

in terms of degree of ionization as:

Then according to law of mass action,

where 'V' is the volume of the solution in litres containing one mole of the solute. From the above equation it follows that if the concentration decreases, the degree of ionization (a) must increase. Thus, at constant temperature, an electrolyte in solution gets ionized to a larger extent with increasing dilution. This is called Ostwald dilution law. This law is not applicable to solutions of strong electrolytes as they are completely dissociated and there is no equilibrium between the dissociated and undissociated molecules.

Strength of acids and base interm of Ka, Kb and pKa and pKb values

Problems 2. Which of the following acids is the strongest and which one is the weakest. Explain.

Solution

Thus, larger the Ka value stronger is the acid. Therefore,


3. Calculate the degree of ionization and [H3O+] of a 0.1 mol /L solution of acetic acid. Given: Ka(CH3COOH) = 1.8 x 10-5 mol / L.

Solution

Let 'a' be the degree of ionization. The concentration of various species involved in the equilibrium are as follows:

The equilibrium constant for the ionization of CH3COOH can be written as follows:

CH3COOH is very feebly ionized. So, 'a' may be ignored in comparison to 1. Then,

4. At 298 K, a 0.1 M solution of acetic acid is 1.34% ionized. What is the ionization constant (Ka) for the acid?

Solution

Acetic acid ionizes as follows:

Arrhenius concept of acids and bases Based on certain characteristic observable properties, electrolytes were classified into acids and bases.


Arrhenius Concept of Acids and Bases

Arrhenius in 1887 put forward this concept. Accordingly, an acid is a hydrogen-containing compound, which gives free hydrogen ions when dissolved in water. A base is a hydroxyl group containing compound which gives free hydroxyl ions (OH-) when dissolved in water.

Thus, according to the Arrhenius concept, hydrogen chloride, acetic acid, and sulphuric acid, are acids because all these compounds give free H+ ions in aqueous solutions.

Compounds such as NaOH and NH4OH are bases, because these compounds give free OH- ions in aqueous solutions.

Thus, according to Arrhenius concept of acids and bases, the neutralization of an acid with a base involves the reaction between H+(aq) and OH-(aq)i.e.,

However, the Arrhenius concept is applicable to the acid-base behaviour only in the aqueous medium. It does not provide any explanation to the acid-base behaviour in the absence of water. This concept defines acids and bases as compounds-containing hydrogen and hydroxyl group respectively. There are however, many compounds, which act as acid even when there is no hydrogen in their molecule. Similarly, there are many bases, which do not contain hydroxyl group.

For example, CO2 acts as an acid in its reaction with NaOH; and NH3 acts as a base although it does not contain OH- group.

In aqueous solutions, hydrogen ion exist as hydrated species (H9O4+) which is described by a simple formula H3O+


Bronsted lowrry concept of acids and bases In 1923, J.N.Bronsted and T.M.Lowry proposed a more general theory known as the Bronsted-Lowry proton transfer theory. According to this concept, any hydrogen containing species (a molecule, a cation or an anion), which is capable of donating one or more protons to any other substance, is called an acid.

Any species (molecule, cation or anion), which is capable of accepting one or more protons from an acid, is called a base. Thus, according to the Bronsted-Lowry concept, an acid is a proton-donor, and a base is a proton-acceptor.

The reaction of an acid with a base involves transfer of a proton from the acid to the base. So, an acid and a base should be present simultaneously in any system. The extent of an acid-base reaction is governed not only by the proton-donating ability of the acid, but also by the proton-accepting tendency of the base. Acids and bases classified on the basis of this concept are termed as Bronsted acids and bases.

In this reaction, HCl donates its one proton to become Cl-, and H2O accepts one proton to become H3O+. Thus, HCl is Bronsted acid and H2O is a Bronsted base. For the reverse reaction, H3O+is able to transfer its proton to Cl-. So, H3O+is a Bronsted acid and Cl- is a Bronsted base. Every acid must form a base on donating its proton, and every base must form an acid on accepting a proton. The base that is produced when an acid donates its proton is called the conjugate base of the acid. The acid that is produced when a base accepts a proton is called the conjugate acid of the base. The above reaction can be written as

In this Cl- is the conjugate base of the acid HCl and H2O is the conjugate base of the acid H3O+. The conjugate acid differs from conjugate base by one proton. A pair of an acid and a base which differ from one another by a proton constitute a conjugate acid base pair. Thus,

Although the Bronsted-Lowry concept of acids and bases is better than the Arrhenius concept, it cannot account for the acidic and basic character of compound not containing hydrogen. For example, acidic nature of oxides such as CO2, SO2 etc., and the basic nature of the compounds of the type CaO, Na2O etc.

Relative Strengths of Conjugate Acid-Base Pairs

A stronger Bronsted acid will have a higher tendency for donating a proton to the base, hence would tend to exist as its conjugate base. The conjugate base so formed will have very little tendency to pick up a proton, hence would act as a weak base.


Thus, there exists an interdependent relationship between the strengths of an acid and its conjugate base. 'The stronger an acid, weaker is its conjugate base, and stronger a base, weaker is its conjugate acid.'

In water, HCl acts as a strong acid. As the reverse reaction occurs to a very small extent Cl- ion acts as a weak base. So, the anion of a strong acid is a weak base. Similarly, the cation of a weak base acts as a strong acid, e.g., NH4+ ion in water is a strong acid.

Relative strengths of common conjugate acid-base pairs

Amphoteric substances

Substances which can act as an acid as well as a base are called amphoteric substances. For example, Al(OH)3, Zn(OH)2 behave both as acids and as bases in their reactions. The substance, which can donate or accept a proton to act as an acid or a base is termed amphiprotic. For example, the species such as HSO4 and HCO3 are amphiprotic, because of the following reactions.

Water is the most common solvent showing a unique behaviour. It can act as an acid as well as a base. In reactions like,

it acts as an acid, while it behaves like a base in a reaction of the type,


The dual role of water molecule may thus be represented by the equation,

Water is thus known as an amphiprotic solvent. Solvents which accept protons are called protophillic, while those which give up protons are called protogenic. Solvents which neither donate nor accept protons are called aprotic solvents. Thus, liquid ammonia is a protophillic solvent, while acetic acid is protogenic.

Relative Strength of Bronsted Acids and Bases

According to Bronsted-Lowry theory, an acid is a proton donor and base is a proton acceptor. Therefore the strength of an acid or a base is determined by its tendency to lose or gain protons. A strong acid is a substance which loses a proton easily to a base.

Relative strength of acids The relative tendency of acids to transfer a proton to a common base, generally water is expressed as the relative strength of bronsted acids. For example, HCl has a higher tendency to transfer a proton to H2O than CH3COOH. This means HCl is a stronger acid than acetic acid. Thus,

Similarly, a base having a higher tendency to accept a proton is stronger. For example, ammonia accepts a proton more readily from water molecule than a water molecule from another water molecule, and hence ammonia is more basic than water.

The ability of an acid to lose a proton (acid strength) is described by its acid ionization constant. The larger the value of the acid ionization constant Ka, higher is the concentration of H3O+ in the solution, stronger is the acid. Knowing the ionization constants of acids one can get the relative strengths of different acids at a particular temperature.


Since the ionization constant of HF is larger than the ionization constant of CH3COOH, hydrofluoric acid (HF) is a stronger acid than acetic acid, (CH3COOH).

Relative strength of bases

The ability of a base to accept a proton (basic strength) is described by its base ionization constant (Kb). The larger the value of Kb, higher is the concentration of OH-, stronger is the base. For example,

Since the value of Kb for aniline is much less than that of ammonia, hence aniline (C6H5NH2) is a weaker base than ammonia (NH3).

Problem 1. Give the conjugate acids of the following: (a) OH-, HCO3-, HPO4-2, CH3NH2, CO3-2, NH3,CH3COOH

(b) Give the conjugate bases of the following: HS-, H3O+, H2PO4-, HSO4

-, HF, CH3COOH, [Al(H2O)6]3+.

Solution


Ionization of water, pH and pH scale Pure water being a weak electrolyte under goes self ionization to a small extent as follows:

The equilibrium constant for this reaction is:

The concentration of unionized water is taken as constant because the degree on ionization of water is very small. So we can write this equation as:

where Kw is a constant and is known as the ionic product of water whose value is 1.008 x 10-14 mol2 L-2at 298 K. In pure water the concentration of H3O+ and OH- are equal and so we can write,

[H3O+] = [OH-]

If, Kw = [H3O+] [OH-] = 1.008 x 10-14 mol2 L-2 then,

[H3O+] [OH-] = 1.008 x 10-14

[H3O+]2 = 1.008 x 10-14


Thus in pure water [H3O+] = [OH-] = 1.0 x 10-7 mol L-1 at 298 K

Effect of temperature on K The value of Kw varies with the change in temperature. The values of [H3O+] and [OH-] are always equal to each other at all temperatures but the values of Kw are different at different temperatures. The value of Kwincreases with the rise in temperature. This is because increase in temperature will shift the equilibrium in the forward direction producing large concentrations of [H3O+] and [OH-] ions (Le Chatelier's principle).

Hence, Kw increases with rise in temperature.

In acidic solution

When an acidic solution of HCl is added to a pure neutral solution of water, the concentration of [H3O+] becomes larger than 1.0 x 10-7 mol L-1. The dissociation equilibrium of water shifts in the reverse direction (Le Chatelier's principle). The excess [OH-] ions combine with hydronium ions to form undissociated water molecules so that the value of Kw remains constant in the solution. The concentration of [OH-] ions will then be equal to

The concentration of [H3O+] is more than the concentration of the [OH-] ions in acidic solution. When a few drops of a base like NaOH is added to pure water, the concentration of [OH-] increases and that of hydrogen ions decreases. The concentration of [H3O+] can be calculated as:

Thus in basic solution the concentration of [OH-] will be greater than of [H3O+].

It can be concluded that the hydronium and hydroxyl ions are always present in solution whether they are acidic or basic. However their concentrations differ.

Problem

6. Calculate the hydronium and hydroxyl ion concentrations in (i) 0.01 M HCl (ii) 0.001 M NaOH solution at 298 K.


Solution (i) HCl completely ionizes as:

The concentration of hydronium ions is equal to that of hydrochloric acid, [H3O+] = [HCl]

[HCl] = 0.01M = 1 x 10-2 mol L-1

The ionic product of water is Kw = [H3O+] [OH-]

Thus [H3O+] = 1 x 10-2 mol L-1 and [OH-] = 1 x 10-12 mol L-1

(ii) NaOH ionizes completely as:

NaOH(aq) Na+(aq) + OH-

(aq)

The concentration of [OH-] is equal to that of NaOH i.e., [NaOH] = [OH-]

[NaOH] = 0.001 M = 1 x 10-3 mol L-1

[OH-] = 1 x 10-3 mol L-1

Kw = [H3O+] [OH-]

Thus, [OH-] = 1 x 10-3 mol L-1and [H3O+] = 1 x 10-11 mol L-1.

Ph and ph scale

In 1909, Sorensen introduced a term for expressing the concentration of hydrogen ions, which give an idea about the acidic and basic characters of the aqueous solution. This term was called 'pH' which means the 'power of hydrogen ions'. The pH is defined as "the negative logarithm of the H3O+ion concentration in moles per litre".

For neutral solution at 298 K,

[H3O+] = [OH-] = 1.0 x 10-7 mol L-1

so that, pH = -log [H3O+] = -log (1.0 x 10-7) = 7

Substituting different values for [H3O+] in the above relation we have, For acidic solution pH <>

For basic solution pH > 7

For neutral solution pH = 7

A scale called as the pH scale is devised to express the acidic and basic properties of solution in terms of the pH value.


Fig: 8.1 - The pH scale

From the scale it is clear that for solutions with

pH between 0 to 2 strongly acidic

pH between 2 to 4 moderately acidic

pH between 4 to 7 weakly acidic

pH between 7 to 10 weakly basic

pH between 10 to 12 moderately basic

pH between 12 to 14 strongly basic.

Problems

6. Calculate the pH value of (i) 0.001 M HCl and (ii) 0.01 M NaOH

Solution

(i) Since HCl is a strong acid, it completely ionizes and therefore, H3O+ions concentration is equal to that of the acid itself i.e., [H3O+] = [HCl] = 0.001 M = 1 x 10-3 M

now, pH = -log [H3O+]

pH = -log [1 x 10-3]

= -(-3) log 10 = 3 (log 10 =1)

(ii) Since NaOH is a strong base, it completely ionizes and therefore, OH-ions concentration is equal to that of the base itself i.e.,

[OH-] = [NaOH] = 0.01 M = 1 x 10-2 M

Kw = [H3O+] [OH-]

pH = -log [H3O+]

pH = -log [1 x 10-12]

= -(-12) log 10 = 12

7. Calculate the pH of a solution whose hydronium ion concentration is 6.2 x 10-

9 mol L-1.

Solution

[H3O+] = 6.2 x 10-9 M

pH = -log [H3O+]

pH = -log [6.2 x 10-9]

= -(log 6.2 - 9 log 10)


= -log 6.2 + 9 x 1 (log 6.2 = 0.79)

= 9 - 0.79 = 8.21

8. Acid A, B, C and D have the following pKa values: A = 1.5, B = 3.5, C = 2.0, D = 5.0. Arrange these acids in the increasing order of acid strength.

Solution

We know that, pKa = -log Ka or Ka = 10-pKa

Therefore, for the given acids, Ka (A) = 10-1.5 Ka (B) = 10-3.5

Ka (C) = 10-2.0 Ka (D) = 10-5.0

Since, 10-5.0 <>-3.5 <>-2.0 <>-1.5

Hence, the strength of acids follows the order, D <>

9. The value of Kw is 9.55 x 10-14 at a certain temperature. Calculate the pH of water at this temperature.

Solution

Kw = 9.55 x 10-14

For water [H3O+] = [OH-]

If, Kw = [H3O+] [OH-] = 9.55 x 10-14 then, [H3O+] [H3O+] = 9.55 x 10-14

[H3O+]2 = 9.55 x 10-14

pH = -log [H3O+]

pH = -log [3.09 x 10-7]

= -(log 3.09 + log 10-7)

= -(0.49 - 7) = 6.51

10. What is the pH of a solution whose hydrogen ion concentration is 0.005 x 10-

3 kg dm-3?

Solution

In the solution [H+] = 0.005 x 10-3 kg dm-3 = 0.005 x 10-3x 103Lm-3

= 0.005 g dm-3 = 0.005 mol dm-3

11. The pH of blood is maintained at 7.4 due to the presence of HCO-

3 and H2CO3. If Ka of H2CO3 in blood is 8 x 10-7 calculate the ratio [HCO-

3]:[H2CO3] in blood.

Solution

pH = 7.4= -log[H+] log[H+]= -7.4 = 8.6 [H+]= 3.98 x 10-8 4 x 10-8 M


Hydrolysis of salts Dissolution of different salts in water does not always result in neutral solutions. For

example, aqueous solution of copper sulphate is acidic whereas aqueous solution of sodium

acetate is basic and aqueous solution of sodium chloride forms neutral solution. This is due

to the dissociation of the salt in water to form ions. This process of the reaction of anion or

cation of the salt with water to produce an acidic or an alkaline solution is called hydrolysis.

Thus, hydrolysis is reverse of neutralization.

The equilibrium constant of such a reaction is called hydrolysis constant.

Hydrolysis of anion (A-)

Hydrolysis of cation (B+)

The fraction of the total salt that gets hydrolysed at equilibrium is called degree of hydrolysis.

It is written as 'h'.

Relation of Hydrolysis Constant and Ka and Kb

Salt of weak acid and strong base

Here, the anion (A-) is a stronger base than OH-, hence it undergoes hydrolysis to give free

OH- ions. Therefore the resulting solution will be basic in character having pH greater than

7. For example

Other examples of this type of salts are CH3COONa, Na2CO3, Na3PO4, etc.

The aqueous solution of a salt of weak acid and strong base is alkaline. For the general

reaction:

This is called anion hydrolysis.

Hydrolysis constant

The hydrolysis constant may be written as:

Multiplying equation (i) and (ii) and dividing by equation (iii), we get

The hydrolysis constant 'Kh', of the salt is inversely proportional to the dissociation constant,

'Ka' of the weak acid. Therefore, the weaker the acid, the greater is the hydrolysis constant of

the salt.

Degree of hydrolysis

If the original concentration of the salt in the solution is 'c' mol/litre and 'h' is the degree of

hydrolysis at that concentration, then


Intial concentration c 0 0

Concentration at eqilibrium c(1-h) ch ch

If 'h' is very small as compared to 1, we can assume 1 - h 1.

Kh = ch2

pH of the hydrolysed salt solution [OH-] = ch

Now, pH = -log [H+]

Now, -log Kw = pKw - log Ka = pKa

Knowing the molar concentration 'c' of the solution, Ka and Kw, the pH of the solution can be calculated.

Solubility product principle and its application Solubility product is an important concept that is used in explaining phenomena like solubility and precipitation of compounds in analytical chemistry.

Calculation of solubility

Knowing the solubility product of a sparingly soluble salt like AgCl, PbI2, BaSO4 etc. the solubility of the salt can be calculated.

Problem 16. What is the solubility of Ag2CrO4 in water if the value of solubility product, Ksp = 1.7 x 10-11(mol / L) 3.

Solution Ag2CrO4 dissolves in water in accordance with the equilibrium.

If S is the solubility of Ag2CrO4 then in a saturated solution,

Then, Ksp (Ag2CrO4) = [Ag+]2 [CrO42-] = (2S)2 (S)

1.7 x 10-11 (mol / L)3 = 4S3

Thus, the solubility of Ag2CrO4 in water at 298 K is 1.48 x 10-4 mol/ L.


Molar mass of Ag2CrO4 = (2 x 108 + 52 + 4 x 16) g/mol = 332 g/mol So, Solubility of Ag2CrO4 = 1.48 x 10- 4 mol /L

= 1.48 x 10- 4 mol / L x 332 g/mol = 0.049 g /L

In predicting the precipitation in reactions Knowing the solubility product of a salt, it is possible to predict whether on mixing the solution of its ions, a precipitate will be formed or not. For precipitation to occur, its ionic product should exceed solubility product. Therefore, to predict the precipitation reaction, we calculate the ionic product of the ions and find out whether it is greater than Ksp or not. Thus, if

For example in order to precipitate barium sulphate from a solution of barium chloride at a concentration of 0.5 M, the precipitation is done by adding sulphuric acid in small amounts to the solution. Initially no precipitation occurs because the small amount of SO42- is insufficient to make the ionic product, [Ba2+] [SO42-] equal to solubility product of barium sulphate. When we have added sufficient amount of sulphuric acid is added so that the ionic product exceeds solubility product, barium sulphate would get precipitated.

Problems 17. The concentration of Ca++ in blood is 0.0025M. If an oxalate solution with oxalate ion concentration 1 x 10-7M is added to it, state whether precipitation occurs or not. Solubility product of calcium oxalate = 2.3 x 10-9.

Solution In the resulting solution, [Ca++ ] = 0.0025M. [C2O42-] = 1 x 10-7 M. [Ca++][C2O4

2-] = 0.0025 x 10-7 = 2.5 x 10-10

This is less than the solubility product of calcium oxalate. Thus, precipitation of calcium oxalate does not occur. 18. A solution is prepared by mixing equal volumes of 0.01M MgCl2, and 0.02M MgC2O4 at 18°C. Would MgC2O4 precipitate out? Ksp of MgC2O4 at 18°C = 8.57 x 10-5.

Solution

When mixed, the total volume gets doubled and hence the effective concentrations of the ions would be half of the initial concentration, i.e., in solution

[Mg2+] =(0.01/2)=0.005 mol/L

[C2O42-] = (0.02/2) = 0.01 mol/L

These ions would react to form sparingly soluble salt MgC2O4 in accordance with the reaction,

Then, the ionic product function in the solution is given by,

[Mg2+] [C2O42-] = 0.005 x 0.01 = 5 x 10-5

the Ksp value for MgC2O4 at 18°C is 8.57 x 10-5. Since, the ionic product function in the solution is less than the Ksp value, precipitation does not take place.


In inorganic qualitative analysis

The concepts of solubility product and common ion effect play an important role in qualitative analysis for the separation of basic radicals (cations) into different groups. Weak acids and weak bases ionise in water slightly and an equilibrium is established in their solutions. For example, in the ionization of a weak base NH4OH as:

The ionization constant for the base,

If solid NH4Cl is added to the solution, the concentration of NH4+ ions increases. According

to Le Chatelier's principle, the equilibrium shifts to the left. As a result, the concentration of OH- is considerably decreased and the weak base NH4OH becomes even weaker in the presence of its salt.

This is common ion effect and may be defined as the suppression of the degree of dissociation of a weak acid or a weak base by the addition of a strong electrolyte containing a common ion.

Qualitative analysis

The common ion effect is generally employed in qualitative analysis. The cations of group II (Hg2+, Pb2+, Bi3+, Cu2+, As3+, Sb3+, Sn2+) are precipitated as their sulphides (such as CuS, PbS) by passing H2S gas in the presence hydrochloric acid (Common H+ ions).

The cations of group III are precipitated as their hydroxides by NH4OH in the presence of NH4Cl. The cations of group V are precipitated as their carbonates by the addition (NH4)2CO3, in the presence of HCl.

Purification of sodium chloride

Sodium chloride obtained from sea-water or lakes is always impure. It can be purified on the basis of common ion effect as described below:

The saturated solution of impure sodium chloride is prepared by dissolving in minimum quantity of water. HCl gas is then passed through this solution. The following equilibria are set up:

Due to the presence of common chloride ions, the dissociation of sodium chloride is suppressed. This is known as common ion effect. The dissociation of sodium chloride is


decreased to such an extent that the ionic product of NaCl exceeds its solubility product and it is thrown down as a precipitate.

Salting out of soap

Soap is a sodium salt of higher fatty acids e.g. sodium stearate, sodium oleate etc. When soap is prepared it floats over spent lye (the residual aqueous solution containing unused alkali, glycerol etc.). A significant amount of soap remains dissolved in this solution. To recover this soap, sodium chloride is added to the boiling soap solution. The recovered soap separates out due to the common ion effect of Na+, in accordance with the reactions.

The increased concentration of Na+ in the solution due to the dissociation of NaCl, shifts the equilibrium towards left and thus soap is precipitated. The recovery of a dissolved salt by adding another salt to the solution is termed salting out.

Comparison of solubility product and ionic product

Solubility product Ionic product

It is the product of the concentration of ions

of the electrolyte each raised to the power of

their coefficients in the balanced chemical

equation in a saturated solution

It is the product of the concentration of ions of

the electrolyte each raised to the power of their

coefficients in the balanced chemical equation

in a solution at any concentration

It is applicable to only saturated solutions It is applicable to all types of solutions of any

concentration

It has a constant value for an electrolyte at a

constant temperature

Its value is not constant and varies with

change in concentration

***

Common ion effects and its application The degree of ionization of an electrolyte is suppressed by the addition of a strong electrolyte

containing common ion. This effect is known common ion effect.

In other words: The phenomenon of lowering the degree of ionization of a weak electrolyte by

adding a solution of a strong electrolyte having a common ion is called common ion effect.

Application of common ion effect

Knowledge of common ion effect is very useful in analytical chemistry. It is frequently applied

in qualitative analysis.

An electrolyte is precipitated only when the concentration of its ions exceeds the solubility

product (KSP). The precipitation is obtained only when the concentration of any one ion is

increased. Thus by adding a common ion, the solubility product can be increased.


PRECIPITATION OF THE CATIONS OF GROUP IV

Cations of groups IV are precipitated as sulphides by passing H2S gas through the solution in

the presence of NH4OH.

Ionization of NH4OH:

NH4OH NH4+ + OH-

In this analysis NH4OH provides OH- ions which combines with H+ ions of H2S to form H2O.

H2S 2H+ + S-2 : H+ + OH- H2O

Removal of H+ ions from product side shifts the equilibrium to rightand the concentration of

S-2 increases which is enough to exceed theKSP of the sulphides of group IV. In this way CoS, NiS or

ZnS can easily be precipitated.

PRECIPITATION OF THE CATIONS OF GROUP II

Sulphides of basic radicals of groups II are precipitated by passing H2S gas through the

acidified solution by HCl.

Ionization of H2S:

H2S 2H+ + S-2

Here HCl provides common ion H+ which shifts the above equilibrium to the left as given

by

Le-Chatelier's principle.

HCl H+ + Cl-

Addition of HCl suppresses the ionization of H2S and lowers the concentration of S-2 ions,

just enough to exceeds the KSP of II group sulphides. In this way only cations of group II are

precipitated as CuS, PbS, CdS etc. but precipitation of the sulphides of group IV is

prevented because they have high KSP values as compared to the sulphides of group II.

PRECIPITATION OF THE CATIONS OF GROUP III

Cations of groups III are precipitated as hydroxides by passing NH4OH in the presence of

NH4Cl. Here

NH4Cl provides common ion NH4+ which suppresses the ionization of NH4OH.

NH4OH NH4+ + OH- : NH4Cl NH4

+ + Cl-

Common ion NH4+ shifts the equilibrium to left side and the concentration of OH- ions

decreases. Under these circumstances, theKSP of the hydroxides of Al, Fe and Cr is only

exceeded and they are precipitated as Al (OH)3, Fe (OH)3 and Cr (OH)3 but the hydroxides of

Zn, Ni and Co are not precipitated as they have high values of KSP.

PRECIPITATION OF THE CATIONS OF GROUP IV

Cations of groups IV are precipitated as sulphides by passing H2S gas through the solution in

the presence of NH4OH.

Ionization of NH4OH:

NH4OH NH4

+ + OH-


In this analysis NH4OH provides OH- ions which combines with H+ions of H2S to form

H2O.

H2S 2H+ + S-2 : H+ + OH- H2O

Removal of H+ ions from product side shifts the equilibrium to rightand the concentration

of S-2 increases which is enough to exceed theKSP of the sulphides of group IV. In this way CoS, NiS

or ZnS can easily be precipitated.

Application of solubility product principle in qualitative analysis

The concepts of solubility product and common ion effect play an important role in qualitative analysis for the separation of basic radicals (cations) into different groups. Weak acids and weak bases ionise in water slightly and an equilibrium is established in their solutions. For example, in the ionization of a weak base NH4OH as:

The ionization constant for the base,

If solid NH4Cl is added to the solution, the concentration of NH4+ ions increases. According

to Le Chatelier's principle, the equilibrium shifts to the left. As a result, the concentration of OH- is considerably decreased and the weak base NH4OH becomes even weaker in the presence of its salt.

This is common ion effect and may be defined as the suppression of the degree of dissociation of a weak acid or a weak base by the addition of a strong electrolyte containing a common ion.

Qualitative analysis

The common ion effect is generally employed in qualitative analysis. The cations of group II (Hg2+, Pb2+, Bi3+, Cu2+, As3+, Sb3+, Sn2+) are precipitated as their sulphides (such as CuS, PbS) by passing H2S gas in the presence hydrochloric acid (Common H+ ions).

The cations of group III are precipitated as their hydroxides by NH4OH in the presence of NH4Cl. The cations of group V are precipitated as their carbonates by the addition (NH4)2CO3, in the presence of HCl.

Purification of sodium chloride

Sodium chloride obtained from sea-water or lakes is always impure. It can be purified on the basis of common ion effect as described below:


The saturated solution of impure sodium chloride is prepared by dissolving in minimum quantity of water. HCl gas is then passed through this solution. The following equilibria are set up:

Due to the presence of common chloride ions, the dissociation of sodium chloride is suppressed. This is known as common ion effect. The dissociation of sodium chloride is decreased to such an extent that the ionic product of NaCl exceeds its solubility product and it is thrown down as a precipitate.

Salting out of soap

Soap is a sodium salt of higher fatty acids e.g. sodium stearate, sodium oleate etc. When soap is prepared it floats over spent lye (the residual aqueous solution containing unused alkali, glycerol etc.). A significant amount of soap remains dissolved in this solution. To recover this soap, sodium chloride is added to the boiling soap solution. The recovered soap separates out due to the common ion effect of Na+, in accordance with the reactions.

The increased concentration of Na+ in the solution due to the dissociation of NaCl, shifts the equilibrium towards left and thus soap is precipitated. The recovery of a dissolved salt by adding another salt to the solution is termed salting out.

Buffer solution Certain solutions, such as that of ammonium acetate, have a tendency to resist any change in its hydronium ion concentration or pH, whenever a small amount of a strong acid or a strong base is added to it. This property of a solution is known as buffer action.

Buffer Solution

A solution which resists any change of pH when a small amount of a strong acid or a strong base is added to it, is called a buffer solution or simply as a buffer. Alternatively, a buffer solution may be defined as a solution whose pH value does not change appreciably upon the addition of small amounts of a strong acid, base and/or water from outside.

Thus, buffers have reserve acidity and reserve alkalinity. Buffer solutions usually consist of a mixture of a weak acid and its salt with a strong base e.g., CH3COOH and CH3COONa, or that of a weak base and its salt with a strong acid e.g., NH4OH and NH4Cl. The solution of any salt of a weak acid and a weak base e.g., ammonium acetate, also shows buffering property.

Types of Buffers

There are two types of buffers, acid buffer and basic buffer.


Acid buffer A buffer solution containing a large amounts of a weak acid, and its

salt with a strong base, is termed as an acid buffer. Such buffer solutions have pH on the acidic side i.e., pH is less than 7 at 298 K. The pH of an acid buffer is given by the equation.

where Ka is the acid dissociation constant of the weak acid.

Basic buffer

A buffer solution containing relatively large amounts of a weak base and its salt with a strong acid, is termed as a basic buffer. Such buffers have pH on the alkaline side i.e., pH is higher than 7 at 298 K. The pH of a basic buffer is given by the equation:

where Kb is the base dissociation constant of the weak base.

These equations are called Henderson equation.

Buffer-capacity and Buffer-range The effectiveness of any buffer is described in terms of its buffer capacity. It is defined as, 'the number of equivalents of a strong acid (or a strong base) required to change the pH of one litre of a buffer solution by one unit, keeping the total amount of the acid and the salt in the buffer constant'. The buffer capacity of a buffer is maximum when acid to salt or base to salt ratio is equal to 1 i.e., it contains equal number of moles of acid (or base) and the salt. All buffer solutions remain effective over a small pH range: this pH-range is characteristic of the buffer and is termed as the buffer-range.

For the two types of buffers, it is given by

Buffer range in pH units

Acid buffer: pKa -1 to pKa +1

Basic buffer: (pKw - pKb) -1 to (pKw - pKb) +1

Consider acetic acid - sodium acetate buffer, an acid buffer. The acid dissociation constant (Ka) of acetic acid is 1.84 x 10-5. Therefore, pKa for acetic acid is 4.74. Then, the buffer range of an acetic acid - sodium acetate buffer is, pH = (pKa) - 1 to (pKa) + 1

= 4.74 1 to 4.74 + 1

= 3.74 to 5.74

Thus, the acetic acid - sodium acetate buffer will act as an effective buffer over the pH range 3.74 to 5.74. The pH of a buffer solution depends only on the ratio of the concentrations of the salt and the acid, or salt and the base. It does not depend on the individual concentration. Since, the ratio remains the same even when the solution is diluted. However, at higher dilutions, buffers do not remain effective as, they are not able to resist a change in the pH value due to the addition of a strong acid or a strong base.


Mechanism of Buffer Action

The buffering action of buffer solutions can be explained in terms of the Bronsted-Lowry concept of acids and bases as follows:

Mechanisms

Action of an acid buffer

An acid buffer contains relatively large amounts of a weak acid (HA) and its salt with a strong base (say NaA). The buffer solution thus contains large concentration of HA and A- (due to the dissociation of the salt), apart from H3O+ and OH-. An addition of small amount of a strong acid causes the reaction,

to proceed in such a direction that an equivalent amount of A- combines with H3O+ to give the same amount of undissociated weak acid, HA. Thus, the added acid is picked up by the anions (from the salt) present in large concentrations in the buffer. As long as the added strong acid is in smaller amounts, the changes in the concentrations of salt and that of the weak acid, (HA) are small. Therefore, the acid to salt ratio does not change appreciably by the addition of strong acid to the buffer solution. As a result no noticeable change is seen in the pH value of the buffer. Addition of a strong base to an acid buffer on the other hand causes the reaction

to proceed in the forward direction, resulting in the formation of an equivalent amount of the salt at the cost of the buffer acid. As long as the added base is in small amounts, the ratio of weak acid to salt remains virtually unchanged. As a result, no observable change in the pH value is seen.

Action of basic buffer

A basic buffer contains a weak base (BOH), and its salt with strong acid (BX). The buffer solution thus contains large amounts of the weak base BOH, and the cation B+ (coming from the dissociation of the salt BX), in addition to H3O+ and OH-. The addition of an acid or a base to the basic buffer causes the following reactions:

proceeds in the forward direction. It is clear that the addition of an acid or a base to any buffer solution does cause a change in the concentrations of the buffer acid (or base) and the salt. But, because of the relatively much larger concentrations of these in the buffer solution, for all practical purposes, the ratio, [Salt] / [Acid] or [Salt] / [Base] remains constant. Hence, the pH does not change.

Problem 12. Calculate the pH of a buffer solution containing 0.2 mole of NH4Cl and 0.1 mole of NH4OH per litre. Kb for NH4OH = 1.85 x 10-5.


Solution According to Henderson's equation:

Applications of Buffers

Buffers find extensive applications in a variety of fields.

In biochemical systems

pH plays a very significant role in biochemical reactions. For example, the blood in our bodies is buffered at a pH value of 7.36-7.42 due to bicarbonate - carbonic acid buffer. A mere change of 0.2 pH units can cause death. Certain enzymes get activated only at certain definite pH values.

Agriculture

The pH of the soil is very important for having proper crop yield. The soils get buffered due to the presence of salts such as carbonates, bicarbonates, phosphates and organic acids. The choice of fertilizers depends upon pH of the soil.

Industry

Practically all industries use buffers in one process or the other. Major industries, which employ buffers are paper, dyes, ink, paints and drugs industries.

Analytical chemistry

Buffers find extensive use in analytical chemistry, viz., both in qualitative and quantitative analysis. For example, qualitative analysis of Group III and Group IV is done in solutions buffered by NH4Cl + NH4OH. Buffers are used in the removal of interfering radicals such as phosphate, oxalate, borate and fluoride etc. The control of pH is very important in the field of food preservation

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