Salt Hydrolysis

Salt hydrolysis

A salt is a compound that contains a metallic or positive radical rather than hydrogen (H+) and acidic or negative radical rather than hydroxyl ion (OH⁻).

Classification of salts

Salts are categorized into four major classes:

1. Normal salts: Salts with strong cations and strong anions.
2. Salts with strong cation and weak anion: Also known as basic salts, often organic salts or salts of carboxylic acid.
3. Salts with weak cation and strong anion: These are acidic salts.
4. Salts with weak cation and weak anion: Undergo both cationic and anionic salt hydrolysis.

Definition of salt hydrolysis

Salt hydrolysis is the reaction between salt and water to produce acid and base. It is the reversible reaction of a neutralization reaction. In salt hydrolysis, only weak ions react with water to give respective products.

Hydrolysis constant ($K_h$)

Definition: $K_h$ is the ratio of the product of molar concentrations of the products to the reactant concentration, raised to the power equal to the balancing number in the hydrolysis equation.

Hydrolysis equation

For anionic salt hydrolysis:

$CH_3COONa + H_2O \rightleftharpoons CH_3COOH + OH^-$

$K_h = \frac{K_w}{K_a}$

For cationic salt hydrolysis:

$NH_4Cl + H_2O \rightleftharpoons NH_4OH + H^+$

$K_h = \frac{K_w}{K_b}$

Formulas

Degree of Hydrolysis ($h$): $h = \sqrt{K_h / c}$, where $c$ = concentration of the salt.

pH of Anionic Hydrolysis: $pH = 7 + 0.5(\log_{10}(K_w/K_a) + \log_{10} C)$

pH of Cationic Hydrolysis: $pH = 7 - 0.5(\log_{10}(K_w/K_b) + \log_{10} C)$

Examples

1. Example 1: Calculate the hydrolysis constant for 0.05 M $NH_4NO_3$ if $K_b = 6.67 \times 10^{-6}$.

   Solution:

   $K_h = \frac{K_w}{K_b} = \frac{10^{-14}}{6.67 \times 10^{-6}} = 1.5 \times 10^{-9}$

2. Example 2: Calculate the pH of a 0.001M $NH_4NO_3$ solution if $K_b(NH_4OH) = 1.8 \times 10^{-5}$.

   Solution:

   $K_h = \frac{K_w}{K_b} = \frac{10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$

   $h = \sqrt{K_h / c} = \sqrt{5.56 \times 10^{-10} / 0.001} = 2.36 \times 10^{-4}$

   $[H^+] = c \times h = 0.001 \times 2.36 \times 10^{-4} = 2.36 \times 10^{-7}$

   $pH = -\log_{10}([H^+]) = -\log_{10}(2.36 \times 10^{-7}) = 6.63$

3. Example 3: Derive the relationship between $K_a$, $K_w$, and $K_h$ for a salt with weak cation and weak anion (e.g., ammonium acetate).

   Solution:

   $K_a \times K_b = K_w$

   $K_h = \frac{K_w}{K_a \times K_b}$

Conceptual questions

1. Why is the aqueous solution of sodium carbonate alkaline?

   Sodium carbonate hydrolyzes to form OH⁻ ions:

   $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$

2. Calculate the pH of 0.2M NaCN if $K_a$ for HCN $= 4 \times 10^{-10}$.

   Solution:

   $K_w = K_a \times K_b$

   $K_b = \frac{K_w}{K_a} = \frac{10^{-14}}{4 \times 10^{-10}} = 2.5 \times 10^{-5}$

   pH = 7 + 0.5(\log_{10}(K_b \times c)) = 7 + 0.5(\log_{10}(2.5 \times 10^{-5} \times 0.2)) = 11.03}