Chemical and Ionic Equilibrium - Result Question 133
76. The dissociation constant of a weak acid $H A$ is $4.9 \times 10^{-8}$. After making the necessary approximations, calculate
(i) $pH$
(ii) $OH^{-}$ concentration in a decimolar solution of the base. (Water has a $pH$ of 7).
(1983, 2 M)
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Solution:
- $K _{\text {sp }}(AgI)=8.5 \times 10^{-17}=\left[Ag^{+}\right]\left[I^{-}\right]$
$\left[I^{-}\right]$required to start precipitation of $AgI$
$ \begin{alignedat} = & \frac{8.5 \times 10^{-17}}{0.10}=8.5 \times 10^{-16} M \\ K _{\text {sp }}\left(HgI _2\right) & =2.5 \times 10^{-26}=\left[Hg^{2+}\right]\left[I^{-}\right]^{2} \end{aligned} $
$\left[I^{-}\right]$ required to start precipitation of $HgI_2$
$ =\sqrt{\frac{2.5 \times 10^{-26}}{0.10}}=5 \times 10^{-13} M $
The above calculation indicates that lower $\left[I^{-}\right]$ is required for precipitation of AgI. When $\left[I^{-}\right]$ reaches $5 \times 10^{-13}$, AgI gets precipitated almost completely.
When $HgI _2$ starts precipitating,
$ \begin{alignedat} {\left[Ag^{+}\right] } & =\frac{8.5 \times 10^{-17}}{5 \times 10^{-13}}=1.70 \times 10^{-4} M \\ % Ag^{+} \text {remaining } & =\frac{1.70 \times 10^{-4} \times 100}{0.10}=0.17 \\ % Ag^{+} \text {precipitated } & =100-0.17=99.83 \end{aligned} $
BETA
