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Q.
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
NTA AbhyasNTA Abhyas 2022
Solution:
By $\alpha =\sqrt{\frac{K_{a}}{C}}=\sqrt{\frac{1 0^{- 4}}{1.2 \times 1 0^{- 3}}}=\sqrt{\frac{1}{12}}$
$\% \, \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:
$\alpha =\frac{- 1 0^{- 4} + \sqrt{1 0^{- 8} + 4 \times 1 0^{- 3} \times 1.2 \times 1 0^{- 4}}}{2 \times 1.2 \times 1 0^{- 3}}$
So the correct value of $\alpha =25 \, \%$