Chemical and Ionic Equilibrium 2 Question 78
76. The dissociation constant of a weak acid $\mathrm{H} A$ is $4.9 \times 10^{-8}$. After making the necessary approximations, calculate (i) $\mathrm{pH}$
(ii) $\mathrm{OH}^{-}$concentration in a decimolar solution of the acid. (Water has a $\mathrm{pH}$ of 7).
$(1983,2 \mathrm{M})$
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Solution:
- $K_{\text {sp }}(\mathrm{AgI})=8.5 \times 10^{-17}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{I}^{-}\right]$
$\left[\mathrm{I}^{-}\right]$required to start precipitation of $\mathrm{AgI}$
$$ \begin{aligned} = & \frac{8.5 \times 10^{-17}}{0.10}=8.5 \times 10^{-16} \mathrm{M} \ K_{\text {sp }}\left(\mathrm{HgI}_{2}\right) & =2.5 \times 10^{-26}=\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2} \end{aligned} $$
$\left[\mathrm{I}^{-}\right]$required to start precipitation of $\mathrm{HgI}_{2}$
$$ =\sqrt{\frac{2.5 \times 10^{-26}}{0.10}}=5 \times 10^{-13} \mathrm{M} $$
The above calculation indicates that lower $\left[\mathrm{I}^{-}\right]$is required for precipitation of AgI. When $\left[\mathrm{I}^{-}\right]$reaches to $5 \times 10^{-13}$, AgI gets precipitated almost completely.
When $\mathrm{HgI}_{2}$ starts precipitating,
$$ \begin{aligned} {\left[\mathrm{Ag}^{+}\right] } & =\frac{8.5 \times 10^{-17}}{5 \times 10^{-13}}=1.70 \times 10^{-4} \mathrm{M} \ % \mathrm{Ag}^{+} \text {remaining } & =\frac{1.70 \times 10^{-4} \times 100}{0.10}=0.17 \ % \mathrm{Ag}^{+} \text {precipitated } & =100-0.17=99.83 \end{aligned} $$
