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+]

$\tt [H_3O^+]=10^{-3.65}=2.239\times 10^{-4}$

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

$\tt Ka=\dfrac{[H_3O^+][A^-]}{[HA]}\\\\Ka=\dfrac{(2.239.10^{-4}){^2}}{0.3-2.239.10^{-4}}\\\\Ka=1.671\times 10^{-7}$