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A certain weak acid, ha, has a ka value of 3.6×10−7. part a calculate the percent ionization of ha in a 0.10 m solution. express your answer to two significant figures and include the appropriate units. view available hint(s)

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Answer is: the percent ionization is 0,19%.
Chemical reaction: HA(aq) ⇄ H⁺(aq) + A⁻(aq).
Ka(HA) = 3,6·10⁻⁷.
c(HA) = 0,1 M.
[H⁺] = [A⁻] = x; equilibrium concentration.
[HA] = 0,1 M - x.
Ka = [H⁺] · [A⁻] / [HA].
0,00000036 = x² / 0,1 M - x.
Solve quadratic equation: x = 0,00019 M.
α = 0,00019 M ÷ 0,1 M · 100% = 0,19%.

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