Question

By adding which of the following in 1 L 0.1 M solution of HA, $(Ka={10}^{-5})$, the degree of dissociation of HA decreases appreciably?

• A. ${10}^{-3}MHCl,1L$
• B. $0.5MHX(Ka=2\times {10}^{-6}),1L$
• C. $0.1MHN{O}_3,1L$
• D. All of these

Answer 1

The correct option is C $0.1MHN{O}_3,1L$

$HA,\alpha =\sqrt{\frac{{10}^{-5}}{0.1}}=0.01$
(A) On adding $HCl,\left[HA\right]=\frac{0.1}{2},\left[HCl\right]=\frac{{10}^{-3}}{2}$
$HCl\to H^{+}+C{l}^{-}$
$HA\rightleftharpoons H^{+}+A^{-}$
$\frac{0.1}{2}(1-\alpha )\left(\frac{{10}^{-3}}{2}+\frac{0.1\alpha }{2}\right)\frac{0.1\alpha }{2}$
$\Rightarrow {10}^{-5}=\frac{\frac{0.1\alpha }{2}\times \left(\frac{{10}^{-3}}{2}+\frac{0.1\alpha }{2}\right)}{\frac{0.1}{2}(1-\alpha )}$
Neglecting $\alpha$
${10}^{-5}=\alpha \times \left(\frac{{10}^{-3}}{2}+\frac{0.1\alpha }{2}\right)$
$\Rightarrow 0.1{\alpha }^2+{10}^{-3}\alpha -2\times {10}^{-5}=0$
$\Rightarrow {\alpha }^2+{10}^{-2}\alpha -2\times {10}^{-4}=0$
$\Rightarrow \alpha =\frac{-{10}^{-2}+\sqrt{{10}^{-4}+8\times {10}^{-4}}}{2}=\frac{2\times {10}^{-2}}{2}=2$
Which is similiar to initial.
(B) In case of two weak acids
$\left[H^{+}\right]=\sqrt{{10}^{-5}\times \frac{0.1}{2}+2\times {10}^{-6}\times \frac{0.5}{2}}$
$=\sqrt{\frac{{10}^{-6}}{2}+\frac{{10}^{-6}}{2}}={10}^{-3}$
$\alpha =\frac{{10}^{-5}}{{10}^{-3}}={10}^{-2}$
which is same as earlier
(C) $HN{O}_3\to H^{+}+N{O}_3^{-}$
$\frac{0.1}{2}\frac{0.1M}{2}$
$HA\rightleftharpoons H^{+}+A^{-}$
$\frac{0.1M}{2}(1-\alpha )\frac{0.1}{2}+\frac{0.1}{2}\alpha \frac{0.1}{2}\alpha$
$\Rightarrow {10}^{-5}=\frac{\frac{0.1\alpha }{2}\times \frac{0.1}{2}(1+\alpha )}{\frac{0.1}{2}(1-\alpha )}$
Neglecting $\alpha$ due to common ion effect.
$\Rightarrow {10}^{-5}=\frac{0.1\alpha }{2}$
$\Rightarrow \alpha =2\times {10}^{-4}$
Hence, it decreases from $0.01to2\times {10}^{-4}$.