A certain weak acid, HA, has a Ka value of 8.4×10−7. Calculate the percent ionization of HA in a 0.010 M solution.?

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HA(aq) ⇄ H⁺(aq) + A⁻(aq)

equilibrium concentrations:
[ HA ] = 0.010 - x
[ H⁺ ] = x
[ A⁻ ] = x

x² / (0.010 - x) = 8.4 × 10⁻⁷
x = 0.0000912325

%Ionization = (0.0000912325)/(0.010) 100%
%Ionization = 0.91%
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HA(aq) + H₂O ⇌ A⁻(aq) + H₃O⁺(aq) …. Ka = 8.4 × 10⁻⁷

Initial concentrations :
[HA]ₒ = 0.010 M
[A⁻]ₒ = [H₃O⁺]ₒ = 0 M

At equilibrium, assume that α% of HA is ionized.
Equilibrium concentrations :
[HA] = 0.010×(1 - α%) M ≈ 0.010 M (Assume that 1 ≫ α%)
[A⁻] = [H₃O⁺] = α% M

Ka = [A⁻] [H₃O⁺] / [HA]
(α%)² / 0.01 = 8.4 × 10⁻⁷
α% = 9.2 × 10⁻⁵ = 0.0092% (This fulfill the assumption that 1 ≫ α%)

Percent ionization of HA = 0.0092%