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A weak monobasic acid is $10 \%$ dissociated in $0.05 \mathrm{M}$ solution. What is its percentage dissociation in $0.10 \mathrm{M}$ solution?
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The correct answer is:
$7.17 \%$
Weak monobasic acid (HA)
$$
\begin{aligned}
& \underset{\mathrm{C}}{\mathrm{HA}_{(\mathrm{aq})}} \rightleftharpoons \mathrm{H}_{\mathrm{aq}}^{+}+\mathrm{A}_{\mathrm{aq}}^{-} \\
& \text {C-C } \quad \mathrm{C} \propto \quad \mathrm{C} \propto \\
& \mathrm{Ka}=\frac{\mathrm{C} \propto^2}{1-\propto} \\
& 1-\propto \simeq 1 \\
& \left(\begin{array}{l}
\mathrm{C}_1=0.05, \propto=10 \%=0.1 \\
\mathrm{C}_2=0.1, \propto_2=?
\end{array}\right. \\
& \mathrm{K}_{\mathrm{a}}=\mathrm{C}^2 \\
& \mathrm{C}_1 \propto_1^2=\mathrm{C}_2 \propto \boldsymbol{c}_2^2 \\
& 0.05 \times 0.1 \times 0.1=0.1 \times \propto_2^2 \\
& \alpha_2^2=5 \times 10^{-3} \\
& =\frac{5}{1000}=\frac{1}{200} \\
& \propto_2=\sqrt{\frac{1}{200}}=0.717 \\
& \propto_2=7.17 \% \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \underset{\mathrm{C}}{\mathrm{HA}_{(\mathrm{aq})}} \rightleftharpoons \mathrm{H}_{\mathrm{aq}}^{+}+\mathrm{A}_{\mathrm{aq}}^{-} \\
& \text {C-C } \quad \mathrm{C} \propto \quad \mathrm{C} \propto \\
& \mathrm{Ka}=\frac{\mathrm{C} \propto^2}{1-\propto} \\
& 1-\propto \simeq 1 \\
& \left(\begin{array}{l}
\mathrm{C}_1=0.05, \propto=10 \%=0.1 \\
\mathrm{C}_2=0.1, \propto_2=?
\end{array}\right. \\
& \mathrm{K}_{\mathrm{a}}=\mathrm{C}^2 \\
& \mathrm{C}_1 \propto_1^2=\mathrm{C}_2 \propto \boldsymbol{c}_2^2 \\
& 0.05 \times 0.1 \times 0.1=0.1 \times \propto_2^2 \\
& \alpha_2^2=5 \times 10^{-3} \\
& =\frac{5}{1000}=\frac{1}{200} \\
& \propto_2=\sqrt{\frac{1}{200}}=0.717 \\
& \propto_2=7.17 \% \\
&
\end{aligned}
$$
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