Salt Hydrolysis

Salt Hydrolysis

When an acid reacts with a base, a salt and water are formed and the reaction is called neutralization. Salts completely dissociate in aqueous solutions to give their constituent ions. The ions so produced are hydrated in water. In certain cases, the cation, anion or both react with water and the reaction is called salt hydrolysis. Hence, salt hydrolysis is the reverse of neutralization reaction.

Salts of strong acid and a strong base

Let us consider the reaction between NaOH and nitric acid to give sodium nitrate and water.

NaOH(aq)+HNO₃ (aq) → NaNO₃ (aq) + H₂O(l)

The salt NaNO₃ completely dissociates in water to produce Na⁺ and NO₃^− ions.

NaNO₃ (aq) → Na⁺ (aq) + NO₃^- (aq)

Water dissociates to a small extent as

H₂O(l) ↔ H⁺ (aq) + OH^- (aq)

Since [H + ]=[OH ^- ], water is neutral

NO₃^- ion is the conjugate base of the strong acid HNO₃ and hence it has no tendency to react with H⁺ .

Similarly, Na⁺ is the conjugate acid of the strong base NaOH and it has no tendency to react with OH^- .

It means that there is no hydrolysis. In such cases [H⁺] = [OH^-] pH is maintained and, therefore, the solution is neutral.

Hydrolysis of Salt of strong base and weak acid (Anionic Hydrolysis).

Let us consider the reactions between sodium hydroxide and acetic acid to give sodium acetate and water.

NaOH (aq) + CH₃ COOH(aq) ↔ CH₃COONa(aq) + H₂O(l)

In aqueous solution, CH₃COONa is completely dissociated as below

CH₃ COONa (aq) → CH₃COO^- (aq)+Na⁺ (aq)

CH₃COO^- is a conjugate base of the weak acid CH₃COOH and it has a tendency to react with H⁺ from water to produce unionised acid .

There is no such tendency for Na+ to react with OH- .

 CH₃COO^- (aq) + H ₂ O(l) ↔ CH₃COOH (aq) + OH^- (aq) and therefore [OH^- ] > [H⁺ ] , in such cases, the solution is basic due to hydrolysis and the pH is greater than 7.

Let us find a relation between the equilibrium constant for the hydrolysis reaction (hydrolysis constant) and the dissociation constant of the acid.

(1) × (2)

⇒ K_h .K_a =[H⁺ ][OH^- ]

we know that [H⁺ ][OH^- ]=K_w

K_h .K_a =K_w

K_h value in terms of degree of hydrolysis (h) and the concentration of salt (C) for the equilibrium can be obtained as in the case of ostwald's dilution law. K_h = h²C. and i.e [OH ^- ] = √ (K_h .C)

pH of salt solution in terms of K_a and the concentration of the electrolyte.

pH + pOH = 14

pH = 14 - p OH = 14 - {-log [OH^- ]}

= 14 + log [OH^- ]

∴pH = 14 + log (K_hC)^1/2

Hydrolysis of salt of strong acid and weak base (Cationic Hydrolysis)

Let us consider the reactions between a strong acid, HCl, and a weak base, NH₄OH, to produce a salt, NH₄Cl , and water

HCl (aq) + NH₄OH (aq) ↔ NH₄Cl(aq) + H₂O(l)

NH₄Cl(aq) → NH₄⁺ +Cl^- (aq)

NH₄⁺ is a strong conjugate acid of the weak base NH₄OHand it has a tendency to react with OH^- from water to produce unionised NH₄OH shown below.

NH₄⁺ (aq) + H₂O(l) ↔ NH₄OH (aq) + H⁺ (aq)

There is no such tendency shown by Cl^- and therefore [H⁺] > [OH^- ]; the solution is acidic and the pH is less than 7.

As discussed in the salt hydrolysis of strong base and weak acid. In this case also, we can establish a relationship between the K_h and K_b as

K_h .K_b = K_w

Let us calculate the K_h value in terms of degree of hydrolysis (h) and the concentration of salt

 = 1/2 log K_w – 1/2 log C + 1/2 log K_b

pH = 7 – ^1/₂ pK_b – ^1/₂ log C.

Hydrolysis of Salt of weak acid and weak base (Anionic & Cationic Hydrolysis).

Let us consider the hydrolysis of ammonium acetate.

CH₃ COONH ₄ (aq) → CH ₃ COO ^- (aq) + NH₄⁺ (aq)

In this case, both the cation (NH₄⁺ ) and anion (CH₃COO^−) have the tendency to react with water

CH₃COO^− + H₂O ↔ CH₃COOH + OH^−

NH₄⁺ + H₂O ↔ NH₄OH + H⁺

The nature of the solution depends on the strength of acid (or) base i.e, if K_a >K_b ; then the solution is acidic and pH < 7, if K_a < K_b ; then the solution is basic and pH > 7, if K_a =K_b ; then the solution is neutral.

The relation between the dissociation constant (K_a ,K_b ) and the hydrolysis constant is given by the following expression.

K_a.K_b.K_h = K_w

pH of the solution

pH of the solution can be calculated using the following

expression, pH = 7 + ^1/₂ pK_a − ^1/₂ pK_{b .}

Example 8.8

Calculate i) degree of hydrolysis, ii) the hydrolysis constant and iii) pH of 0.1M CH₃COONa solution ( pK_a for CH₃COOH is 4.74).

Solution (a) CH₃COONa is a salt of weak acid (CH ₃COOH) and a strong base (NaOH).

Hence, the solutions is alkaline due to hydrolysis.

CH₃COO ^- (aq) + H₂O(aq) ↔ CH₃COOH (aq) + OH^-(aq)

Evaluate yourself – 10

Calculate the i) hydrolysis constant, ii) degree of hydrolysis and iii) pH of 0.05M sodium carbonate solution ( pK_a for HCO₃^− is 10.26).

Answer:

Sodium carbonate is a salt of weak acid, H₂CO₃ and a strong base, NaOH, and hence the solution is alkaline due to hydrolysis.

Na₂CO₃ (aq) → 2Na⁺ (aq) + CO₃^2- (aq)

CO₃^2- (aq)+ H₂O (l) ↔ HCO₃^- +OH^-

i) h=  √{ (K_w) / (K_a × C) }

= √ ( 1 × 10^-14 /  5.5 × 10^-11 × 0.05 )

h = 6.03 × 10^-2

Given that pK_a =10.26

pK_a = -log K_a

i.e., K_a =antilog of (-pK_a )

= antilog of (-10.26)

= antilog of (-11 + 0.74)

= 10^-11 × 5.5

[antilog of 0.74 = 5.49 ≈ 5.5]

ii) K_h = K_w/K_a = ( 1×10^-14 /  5.5 ×10^-11 )

 = 1.8 ×10^-4

iii) pH = 7 + pK_a/2 + logC/2

 = 7 + (10.26/2) + (log 0.05)/2 = 7 + 5.13 −0.65

 = 11.48