Question

1 Write down the Hydrolysis reaction for acetic acid (CH3COOH)
2 Write the equilibrium expression of these reaction(Ka)
3 Calculate the pH of 0.1M solution of acetic acid

Answer 1

1. The hydrolysis reaction equation for the acid is;

$CH_3COOH+H_2O\leftrightharpoons H^+_{(aq)}+CH_3COO^-_{(aq)}$

$H^+_{(aq)}+H_2O_{(l)}\to H_3O^+_{(aq)}$ .

The overall hydrolysis reaction is;

$CH_3OOH+H_2O\leftrightharpoons CH_3COO^-_{(aq)}+H_3O^+_{(aq)}$ 2. The equilibrium expression for the reaction is;

$K_a =$ ${[CH_3COO^-][H_3O^+]}\over {[CH_3COOH]}$

3. For acetic acid$pK_a=4.76$ and $-log _{10} [H^+]=pH$

$4.76=log_ {10}(0.1)+2.pH$

$2.pH=4.76+1$

$2.pH= 5.76$

$pH=2.88$

The pH for $0.1 M$ acetic acid$=2.88$