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Q.
The solubility product of AgCl is $4.0 \times 10^{-10}$ at 298 K. The solubility of AgCl in 0.04 M $CaCl_2$ will be
Equilibrium
Solution:
$
\begin{array}{l}
x =[ Ag ]=\left[ A ^{-}\right]+[ HA ]---(1) \\
\text { from } \frac{[4 A ]}{[ A ]}=10 \\
x =\left[ Ag ^{+}\right]=\left[ A ^{-}\right]+10\left[ A ^{-}\right]=11\left[ A ^{-}\right] \\
x =\left[ Ag ^{+}\right]=11 \times\left[ A ^{-}\right]---(2)
\end{array}
$
equation (2) can be written as
$
\begin{array}{l}
A ^{-}+ H _{2} O \rightleftharpoons HA + OH \\
kw =\left[ OH ^{-}\right]\left[ H ^{+}\right] \\
{\left[ OH ^{-}\right]=\frac{ kw }{\left[ H ^{+}\right]}=\frac{10^{-14}}{10^{-4}}=10^{-5}}
\end{array}
$
At eq $\left[ A ^{+}\right]=\left[ OH ^{-}\right]=10^{-5}$
from eq (1) $\left[ A ^{+}\right]=\left[ A ^{-}\right]=10^{-5}$
$
\begin{array}{l}
\operatorname{krp}=\left[ Ag ^{+}\right]\left[\left( A ^{-}\right)+( HA )\right] \\
=10^{-5} \times 11 \times\left[ A ^{-}\right] \\
=10^{-5} \times 11 \times\left[10^{-5}\right] \\
=10^{-5} \times 11=\frac{11}{10} \times 10^{-10} \times 10 \\
=1.1 \times 10^{-9}
\end{array}
$