Question

The degree of dissociation of a weak monoprotic acid of concentration $1.2\times 10^{-3}M$ having $K_a=1.0\times 10^{-4}$ is

• A. 1%
• B. 10%
• C. 15%
• D. 25%

Answer 1

Answer: (D)

By $\alpha =\sqrt{\frac{K_a}{C}}=\sqrt{\frac{10^{-4}}{1.2\times 10^{-3}}}=\sqrt{\frac{1}{12}}$
$\%\alpha =28.86\%$ , as value of $\alpha$ is greater than $5\%$ so we can not take assumptions and will have to use the exact formula.
So $\alpha$ will be given by:
$\alpha =\frac{-10^{-4}+\sqrt{10^{-8}+4\times 10^{-3}\times 1.2\times 10^{-4}}}{2\times 1.2\times 10^{-3}}$
So the correct value of $\alpha =25\%$