Question

The ph of 0.01m lactic acid is 2.43. Calculate ka and pka.

Answer 1

Given,

Concentration of acetic acid, c = 0.01 M

pH of lactic acid = 2.43

As lactic acid is a weak acid (HA), so it is dissociated into:

                               HA ⇔ H⁺ + A⁻

∴ Equilibrium acid dissociation constant, Kₐ = $\frac{[H⁺][A⁻]}{[HA]}$

We know,

pH = -log[H⁺]

⇒ [H⁺] = 10-pH

⇒ [H⁺] = 10-2.43

⇒ [H⁺] = 3.71 × 10⁻³ M

From the dissociation equation, [H⁺] and [A⁻] are in 1:1 molar ratio.

∴ [A⁻] = 3.71 × 10⁻³ M

Also, [HA] = 0.01 M

∴ Kₐ of lactic acid = $\frac{(3.71 × 10⁻³)(3.71 × 10⁻³)}{0.01}$

       = 1.37 × 10⁻³ M

Again,

pKₐ = -log(Kₐ)

∴ pKₐ of lactic acid = -log(1.37 × 10⁻³)

                                = - (-2.86)

                                = 2.86