Question

The percentage degree of dissociation of $0.05M$ acetic acid $\left(C{H}_{3}COOH\right)$ solution with $K_a=1.8\times {10}^{-5}$ is :

• A. 0.55
• B. 5.5
• C. 1.9
• D. 19

Answer 1

The correct option is C 1.9

$K_a=1.8\times {10}^{-5}$, degree of dissociation $\left(\alpha \right)$
The dissociation of acetic acid is given by
$C{H}_{3}COOH(aq.)\rightleftharpoons H^{+}(aq.)+C{H}_{3}CO{O}^{-}(aq.)Initial:C00Atequilibrium:C(1-\alpha )C\alpha C\alpha$

$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}=\frac{(C\alpha {)}^2}{C(1-\alpha )}{K}_a=\frac{C{\alpha }^2}{1-\alpha }1.8\times {10}^{-5}=\frac{0.05\times {\alpha }^2}{1-\alpha }$
Since $C{H}_{3}COOH$ is a weak acid,
So, $\alpha <<11-\alpha \approx 1$

$1.8\times {10}^{-5}=0.05\times {\alpha }^2\alpha =\sqrt{\frac{1.8\times {10}^{-5}}{0.05}}=\sqrt{3.6\times {10}^{-4}}\alpha =1.9\times {10}^{-2}=0.019$
%$\alpha =0.019\times 100=1.9$%