Question

The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?

Answer 1

Given,

Ionization constant of chloroacetic acid (Cl₂CHCOOH), Kₐ = 1.35 × 10⁻³

Chloroacetic acid dissociates as: -

Cl₂CHCOOH ⇔ Cl₂CHCOO⁻ + H⁺

We know,

[H⁺] = $\sqrt{Kₐ × c}$, c is the concentration of the acid

      = $\sqrt{1.35 × 10⁻³ × 0.1}$

      = 1.16 × 10⁻²

∴ pH of 0.1 M Cl₂CHCOOH = -log[H⁺]

                                            = -log(1.16 × 10⁻²)

                                            = - (-1.93)

                                            = 1.93

Sodium salt of the acid is a mixture of Cl₂CHCOOH (weak acid) and NaOH (strong base).

∴ pH of 0.1 M Cl₂CHCOO⁻ = -$\frac{1}{2}$[log Kw + log Kₐ - log c]

                                           = $\frac{1}{2}$[-log Kw - log Kₐ + log c]

                                         = $\frac{1}{2}$[ +14 - log(1.35 × 10⁻³) + log(0.1)]

                                          = $\frac{1}{2}$[14 + 2.87 - 1]

                                          = 7.94