Question

The $K_a$ of a weak monobasic acid is $1\times 10^{-5}$. The percentage of ionization in a decimolar acid solution is

• A. $0.1$
• B. $10$
• C. $0.01$
• D. $0.5$
• E. $1$

Answer 1

Answer: (E)

Weak monobasic acid ionises in water to give $H_3{O}^{+}$ion according to equation,
$HA+H_{2}O\leftrightharpoons H_3{O}^{+}A$
Let $\alpha$ be the degree of ionisation then the concentration of various species at equilibrium.
$HA+H_{2}O\leftrightharpoons H_3{O}^{+}+A$
$\begin{array}{cccc}\text{ Initialconcentration }& 0.1& 0& 0\\ \text{ Conc.atequilibrium }& 0.1(1-\alpha )& 0.1\alpha & 0.1\alpha \end{array}$
$[\therefore \alpha$ is very small and negligible as compared to $1.]$
Now, $K_a=\frac{0.1\alpha \times 0.1\alpha }{1}$
or $K_a=0.1{\alpha }^2$
But $K_a=1\times 10^{-5}$, (given)
$\therefore 1\times 10^{-5}=0.1{\alpha }^2$
or $\alpha =\sqrt{\frac{10^{-5}}{0.1}}$
$=10^{-2}=0.01$
Hence, degree of ionisation $=0.01$ and percentage of ionisation
$=0.01\times 100=1\%$