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 acid. (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 $AgI$
$$ \begin{aligned} = & \frac{8.5 \times 10^{-17}}{0.10}=8.5 \times 10^{-16} M \ K _{\text {sp }}\left(HgI _2\right) & =2.5 \times 10^{-26}=\left[Hg^{2+}\right]\left[I^{-}\right]^{2} \end{aligned} $$
$\left[I^{-}\right]$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 to $5 \times 10^{-13}$, AgI gets precipitated almost completely.
When $HgI _2$ starts precipitating,
$$ \begin{aligned} {\left[Ag^{+}\right] } & =\frac{8.5 \times 10^{-17}}{5 \times 10^{-13}}=1.70 \times 10^{-4} M \ % Ag^{+} \text {remaining } & =\frac{1.70 \times 10^{-4} \times 100}{0.10}=0.17 \ % Ag^{+} \text {precipitated } & =100-0.17=99.83 \end{aligned} $$
