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