Given:

Ionization constant of chloroacetic acid is 1.35 × 10^–3

C = 0.1M

Ionization of chloroacetic acid:

ClCH₂COOH ↔ ClCH₂COO^- + H⁺

Using the formula,

K_a =

⇒K_a =

As it is completely ionized,

∴ [ClCH₂COO^-] = [H⁺]

As [ClCH₂COOH] = 0.02M (given)

As K_a = 1.35× 10^-3 (given)

∴ 1.35× 10^-3=

^⇒[H⁺] = √0.02× 1.35× 10^-3

⇒[H⁺] = 1.16 × 10^-2

To calculate pH of the solution, we apply the formula,

pH= -log[H⁺]

⇒pH= -log (1.16 × 10^-2)

⇒pH = -log 1.16 – (-2) log 10

⇒pH= 1.94

Thus, the pH of 0.1 M acid is 1.94

Now, To calculate the pH of 0.1M sodium salt solution:

As we know that ClCH₂COONa is salt of weak acid(ClCH₂COOH) and a stong base(NaOH)

Hence, by applying the formula:

pH = -1/2 [logK_w + log K_a – log c]

Where K_w is the ionic product of water

K_a is the ionization constant

C is the concentration

As we know the value of K_w is 10^-14

K_a = 1.35 × 10^–3 (given)

C = 0.1 M (given)

∴ pH = -1/2 [log10^-14 + log (1.35 × 10^–3) + log 0.1]

⇒pH = -1/2 [-14 + (-3 + 0.1303) – (-1)]

⇒pH = 7+ 1.44+ 0.5

⇒pH = 7.94

Thus, the pH of 0.1 M sodium salt solution is 7.94