The ionization constant of nitrous acid is 4.5 × 10 - 4 . Calculate the p H of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

N a N O 2 is the salt of a strong base N a O H and a weak acid H N O 2 . N O 2 - + H 2 O ⇋ HN O 2 + O H - K h = HN O 2 O H - N O 2 - = HNO 2 OH - [ H + ] NO 2 - [ H + ] ∴ k w = OH - H + = 10 - 14 ; K a = NO 2 - H + HNO 2 ⇒ K h = K w K a = 1 0 - 14 4.5 × 1 0 - 4 = 0.22 × 1 0 - 10 Now, If x moles of the salt undergoes hydrolysis, then the concentration of various species present in the solution will be: N O 2 - = 0.04 - x ≈ ⁡ 0 . 04 HN O 2 = x O H - = x K h = x 2 0.04 = 0.22 × 1 0 - 10 x 2 = 0.0088 × 1 0 - 10 x = 0.093 × 1 0 - 5 ∴ O H - = 0.093 × 1 0 - 5 M H 3 O + = K w H O - = 1 0 - 14 0.093 × 1 0 - 5 = 10.75 × 1 0 - 9 M ⇒ pH = - log ⁡ 10.75 × 1 0 - 9 pH = - [ log ⁡ 10.75 ) + l o g ⁡ ( 1 0 - 9 ] pH = - [ log ⁡ 10.75 ) - 9 l o g 1 0 = 7.96 Therefore, the degree of hydrolysis (h) = x 0.04 = 0 . 093 × 1 0 - 5 0.04 = 2.325 × 1 0 - 5 .

The ionization constant of chloroacetic acid is 1.35 × 10 - 3 . What will be the p H of 0. 1 M acid and its 0.1 M sodium salt solution?

K a for C l C H 2 C O O H = 1.35 × 10 – 3 . ⇒ K a = c α 2 ∴ α = K α c = 1.35 × 1 0 - 3 0.1 ∵ c o n c e n t r a t i o n o f a c i d = 0.1 M α = 1.35 × 1 0 - 2 = 0.116 ∴ H + = c α = 0.1 × 0.116 = 0.0116 pH = - log ⁡ H + = - log ⁡ 0 .0116 ≈ 1 .93 ClCH 2 COONa is the salt of a weak acid ( C l C H 2 C O O H ) and a strong base ( N a O H ) . ClC H 2 CO O - + H 2 O ⇌ ClC H 2 COOH + O H - 0 . 1 x x K h = ClCH 2 COOH OH - ClCH 2 COO - Multiplying both the sides with [ H + ] K h = ClCH 2 COOH OH - H + ClCH 2 COO - H + ( ∴ OH - H + = K w ; K a = ClCH 2 COO - H + ClCH 2 COOH ) K h = K w K a K h = 1 0 - 14 1.35 × 1 0 - 3 = 0.740 × 1 0 - 11 Also, K h = x 2 0.1 (where x is the concentration of O H - and C l C H 2 C O O H ) 0.740 × 1 0 - 11 = x 2 0.1 0.074 × 1 0 - 11 = x 2 ⇒ x 2 = 0.74 × 1 0 - 12 x = 0.86 × 1 0 - 6 O H - = 0.86 × 1 0 - 6, ∴ H + = K w 0.86 × 1 0 - 6 = 1 0 - 14 0.86 × 1 0 - 6 H + = 1.162 × 1 0 - 8 pH = - log ⁡ H + = - log ( 1.162 × 1 0 - 8 ) = 7.94 .