Answer:
Ka = 4.70x10⁻⁴M
Explanation:
The general dissociation of a weak acid, HX, is:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
And Ka is written as:
Ka = [H⁺] [X⁻] / [HX]
Where [] represents the molar concentration in equilibrium of each specie.
The equilibrium is reached when X of HX is dissociate in X H⁺ and X X⁻, that is:
[HX] = 0.0129M - X
[H⁺] = X
[X⁻] = X
As pH = -log [H⁺]:
10^-pH = [H⁺] = X = 2.239x10⁻³M
Solving:
[HX] = 0.0129M - 2.239x10⁻³M = 0.01066M
[H⁺] = 2.239x10⁻³M
[X⁻] = 2.239x10⁻³M
Ka = [H⁺] [X⁻] / [HX]
Ka = [2.239x10⁻³M] [2.239x10⁻³M] / [0.01066M]
Ka = 4.70x10⁻⁴M
