Question

For propanoic acid $K_a=1.34\times {10}^{-5}\text{M}$ at ${25}^{\circ }\text{C}$. Then what will be the degree of dissociation for a $0.01\text{M}$ solution of the acid ?

• A. $3.66\times {10}^{-4}$
• B. $3.66\times {10}^{-2}$
• C. $5\times {10}^{-4}$
• D. $5\times {10}^{-2}$

Answer 1

The correct option is B $3.66\times {10}^{-2}$

If $\alpha$ is the degree of dissociation of propanoic acid of concentration c, then the concentrations of various species in solution are
$C{H}_3\underset{c(1-\alpha )}{C{H}_2}COOH+H_{2}O\rightleftharpoons C{H}_3\underset{c\alpha }{C{H}_2}CO{O}^{-}+\underset{c\alpha }{H_3{O}^{+}}$
With these concentrations, the equilibrium constant becomes
$K_a=\frac{\left[C{H}_{3}C{H}_{2}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C{H}_{3}C{H}_{2}COOH\right]}=\frac{\left(c\alpha \right)\left(c\alpha \right)}{c(1-\alpha )}=\frac{c{\alpha }^2}{1-\alpha }\approx c{\alpha }^2$
or $\alpha =\sqrt{\frac{K_a}{c}}=\sqrt{\frac{(1.34\times {10}^{-5}\text{M})}{\left(0.01\text{M}\right)}}=\sqrt{1.34\times {10}^{-3}}$
Thus, $\alpha =3.66\times {10}^{-2}$