Answer:
B) 0.32 %
Explanation:
Given that:
[tex]K_{a}=1.8\times 10^{-5}[/tex]
Concentration = 1.8 M
Considering the ICE table for the dissociation of acid as:-
[tex]\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}[/tex]
The expression for dissociation constant of acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}[/tex]
[tex]1.8\times 10^{-5}=\frac{x^2}{1.8-x}[/tex]
[tex]1.8\left(1.8-x\right)=100000x^2[/tex]
Solving for x, we get:
x = 0.00568 M
Percentage ionization = [tex]\frac{0.00568}{1.8}\times 100=0.32 \%[/tex]
Option B is correct.
