Question

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

• A. 0.1
• B. 10
• C. 0.01
• D. 1

Answer 1

The correct option is D 1

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.
$HA+H_{2}O\leftrightharpoons H_3{O}^{+}+A$
$\begin{array}{cccc}Initialconcentration& 0.1& 0& 0\\ Conc.atequilibrium& 0.1(1-\alpha )& 0.1\alpha & 0.1\alpha \end{array}$
[ $\therefore \alpha$ 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 $=0.01\times 100=1$