Question

Calculate the Ka for a 0.3 M solution of HA (unknown weak acid) if the pH = 3.65. The reaction can be modelled as HA (aq) + H2O (l) ←→ A- (aq) + H3O+ (aq).

Answer 1

The Ka : 1.671 x 10⁻⁷

Further explanation

Given

Reaction

HA (aq) + H2O (l) ←→ A- (aq) + H3O+ (aq).

0.3 M HA

pH = 3.65

Required

Ka

Solution

pH = - log [H3O+]

`\tt [H_3O^+]=10^(-3.65)=2.239* 10^(-4)`

ICE method :

HA (aq) ←→ A- (aq) + H3O+ (aq).

0.3 0 0

2.239.10⁻⁴ 2.239.10⁻⁴ 2.239.10⁻⁴

0.3-2.239.10⁻⁴ 2.239.10⁻⁴ 2.239.10⁻⁴

`\tt Ka=([H_3O^+][A^-])/([HA])\\\\Ka=\frac{(2.239.10^(-4)){^2}}{0.3-2.239.10^(-4)}\\\\Ka=1.671* 10^(-7)`