Question

What is the pH of a 1.15 M solution of methanoic acid? Ka = 1.8 x 10^-4?

Answer 1

$HCOOH_{(aq)}+ H_2O_{(l)} ⇌ HCOO^-_{(aq)} + H_3O^+_{(aq)}$

Using the ICE table;

Initial: 1.15M + --- ⇌ 0M + 0M

Change:. -xM + --- ⇌ +xM + +xM

Equilibrium: (1.15-x)M + --- ⇌ xM + xM

$K_a = \dfrac{[HCOO^−][H_3O^+]}{[HCOOH]}$ $= \dfrac{x^2}{[1.15-x]}$

Since K_a is small compared with the initial concentration of the acid, you can approximate (1.15−x) with 1.15. This will give;

$1.8×10^{-4} = \dfrac{x^2}{1.15}$

$x = \sqrt{1.8×10^{-4}×1.15} = 0.0144$

since $[x] = [H_3O] = 0.0144$

pH = $-log[H_3O] = 1.84$

Therefore, the pH of the 1.15 M solution of methanoic acid is 1.84