
The solubility of $AgCl$ in 0.2M $NaCl$ solution is (${{K}_{sp}}$ of $AgCl$ =$1.20\times {{10}^{-10}}$)
(A)$6.0\times {{10}^{-10}}M$
(B) $1.2\times {{10}^{-10}}M$
(C) $0.2M$
(D) $0.2\times {{10}^{-10}}M$
Answer
362.1k+ views
Hint: Here, a concept of ‘common ion effect’ can be seen. An ionic precipitates solubility is reduced when a soluble compound with the precipitate common ion is added to the solution. This is known as the common ion effect. It prevents a weak acid or base from ionising as much as it would without the presence of a common ion. Here, $C{{l}^{-}}$ions are common in both solute and solvent.
Complete Step by Step Solution:
Let the solubility of $AgCl$ be s.
$AgCl$can be dissociated into ions as
$NaCl$can be dissociated into ions as
Given ${{K}_{sp}}$ of $AgCl$ =$1.20\times {{10}^{-10}}$
As the $C{{l}^{-}}$ions comes from both $AgCl$and $NaCl$, so its solubility is taken as $s+0.2$.
The solubility product ${{K}_{sp}}$ of $AgCl$can be calculated as ${{K}_{sp}}=\left[ A{{g}^{+}} \right]\left[ C{{l}^{-}} \right]$
${{K}_{sp}}=s\times \left( s+0.2 \right)$
${{K}_{sp}}={{s}^{2}}+0.2s$
As ${{s}^{2}}$is very small, therefore it can be neglected.
${{K}_{sp}}$ is thus given as ${{K}_{sp}}=0.2s$
Comparing the calculated value of ${{K}_{sp}}$ with the given value.
$0.2s=1.20\times {{10}^{-10}}$
$s=\frac{1.20\times {{10}^{-10}}}{0.2}$
$s=6\times {{10}^{-10}}M$
Thus, the solubility of $AgCl$ in a 0.2M $NaCl$ solution is $s=6\times {{10}^{-10}}M$.
Correct Option: (A)$6.0\times {{10}^{-10}}M$.
Additional Information: The solubility product is calculated as the product of the concentration of each of the ions, each raised to power to its coefficient in a balanced equation. The solubility product depends upon temperature. With an increase in temperature, the solubility of a substance is increased. Hence, the solubility product increases.
Note: The common ion effect has many applications in the processes involving water treatment; regulation of buffer solutions; production of sodium bicarbonate; salting out of soap; and purification of salts.
Complete Step by Step Solution:
Let the solubility of $AgCl$ be s.
$AgCl$can be dissociated into ions as
$NaCl$can be dissociated into ions as
Given ${{K}_{sp}}$ of $AgCl$ =$1.20\times {{10}^{-10}}$
As the $C{{l}^{-}}$ions comes from both $AgCl$and $NaCl$, so its solubility is taken as $s+0.2$.
The solubility product ${{K}_{sp}}$ of $AgCl$can be calculated as ${{K}_{sp}}=\left[ A{{g}^{+}} \right]\left[ C{{l}^{-}} \right]$
${{K}_{sp}}=s\times \left( s+0.2 \right)$
${{K}_{sp}}={{s}^{2}}+0.2s$
As ${{s}^{2}}$is very small, therefore it can be neglected.
${{K}_{sp}}$ is thus given as ${{K}_{sp}}=0.2s$
Comparing the calculated value of ${{K}_{sp}}$ with the given value.
$0.2s=1.20\times {{10}^{-10}}$
$s=\frac{1.20\times {{10}^{-10}}}{0.2}$
$s=6\times {{10}^{-10}}M$
Thus, the solubility of $AgCl$ in a 0.2M $NaCl$ solution is $s=6\times {{10}^{-10}}M$.
Correct Option: (A)$6.0\times {{10}^{-10}}M$.
Additional Information: The solubility product is calculated as the product of the concentration of each of the ions, each raised to power to its coefficient in a balanced equation. The solubility product depends upon temperature. With an increase in temperature, the solubility of a substance is increased. Hence, the solubility product increases.
Note: The common ion effect has many applications in the processes involving water treatment; regulation of buffer solutions; production of sodium bicarbonate; salting out of soap; and purification of salts.
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