Percent Ionization
Percent ionization for a weak acid (HA) is determined by the following formula:

Percent ionization=[HA] ionized[HA] initial×100%

For strong acids, ionization is nearly complete (100%) at most concentrations. However, for weak acids, the percent ionization changes significantly with concentration. The more diluted the acid is, the greater percent ionization.
A certain weak acid, HA, has a Ka value of 7.6×10−7.
Part A
Calculate the percent ionization of HA in a 0.10 M solution.
Express your answer as a percent using two significant figures.

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Part B
Calculate the percent ionization of HA in a 0.010 M solution.
Express your answer as a percent using two significant figures.

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[tex]\begin{matrix}&HA&\rightleftharpoons &A^-&&H^+\\ At\ time, t = 0 &0.10&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&0.10-x&&x&&x\end{matrix}[/tex]

The expression for dissociation constant of acid is:

[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {A}^- \right ]}{[HA]}[/tex]

[tex]7.2\times 10^{-7}=\frac{x^2}{0.10-x}[/tex]

[tex]7.2\left(0.10-x\right)=10000000x^2[/tex]

Solving for x, we get:

x = 0.00026 M

Percentage ionization = [tex]\frac{0.00026}{0.10}\times 100=0.26 \%[/tex]

(b)

Concentration = 0.010 M

Considering the ICE table for the dissociation of acid as:-

[tex]\begin{matrix}&HA&\rightleftharpoons &A^-&&H^+\\ At\ time, t = 0 &0.010&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&0.010-x&&x&&x\end{matrix}[/tex]

The expression for dissociation constant of acid is:

[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {A}^- \right ]}{[HA]}[/tex]

[tex]7.2\times 10^{-7}=\frac{x^2}{0.010-x}[/tex]

[tex]7.2\left(0.010-x\right)=10000000x^2[/tex]

Solving for x, we get:

x = 0.00008 M

Percentage ionization = [tex]\frac{0.00008}{0.010}\times 100=0.80 \%[/tex]