Question

The solubility product of $AgCl$in water is $1.5 \times {10^{ - 10}}$. Calculate its solubility in $0.01\,M$ $NaCl$aqueous solution.
(i) $1.5 \times {10^{ - 9}}\,M$
(ii) $1.5 \times {10^{ - 11}}\,M$
(iii) $1.5 \times {10^{ - 8}}\,M$
(iv) $1.5 \times {10^{ - 10}}\,M$

Answer 1

Consider the solubility of $A{g^ + }$ and $C{l^ - }$ ions in water be $S$. Since $NaCl$is a strong electrolyte, it dissociates completely in water and the $C{l^ - }$ ions coming from the dissociation of $NaCl$ will now contribute to the solubility product of $AgCl$ in water. Then by applying the formula of solubility product calculate the solubility.

Complete step-by-step solution: We know that the product of the total molecular concentrations of the ions in a saturated solution of a sparingly soluble salt is termed as the solubility product of the salt.
Suppose we consider a sparingly soluble salt, ${A_x}{B_y}$which ionizes as
${A_x}{B_y}\, \rightleftarrows \,x{A^ + }\, + \,y{B^ - }$
Then the solubility product of the salt is given by,
${K_{sp}}\, = \,{\left[ {{A^ + }} \right]^x}\, \times \,{\left[ {{B^ - }} \right]^y}$
Now, $AgCl$is a sparingly soluble salt which dissociates in water as,
$AgCl\, \rightleftarrows \,A{g^ + }\, + \,C{l^ - }$
So, the solubility product of$AgCl$in water is given as
${K_{sp}}\, = \,\left[ {A{g^ + }} \right]\, \times \,\left[ {C{l^ - }} \right].........(1)$
Let the solubility of $AgCl$be $S\,mol\,{L^{ - 1}}$.
$\therefore$ $\left[ {A{g^ + }} \right]\, = \,S\,mol\,{L^{ - 1}}$ and $\left[ {C{l^ - }} \right]\, = \,S\,mol\,{L^{ - 1}}$
Now, $NaCl$is a strong electrolyte which completely dissociates in water as,
$NaCl \to N{a^ + } + C{l^ - }$
Being a strong electrolyte $\left[ {NaCl} \right]\, \equiv \,\left[ {N{a^ + }} \right]\, \equiv \,\left[ {C{l^ - }} \right]\, = \,0.01\,mol\,{L^{ - 1}}$.
Due to common ion effect $C{l^ - }$ions coming from $NaCl$will now contribute to the solubility product of $AgCl$in water.
Therefore total $\left[ {C{l^ - }} \right]\,$in $0.01\,mol\,{L^{ - 1}}$ aqueous solution of $NaCl\,$ $= \,{\left[ {C{l^ - }} \right]_{AgCl}} + {\left[ {C{l^ - }} \right]_{NaCl\,}}\, = \,\left( {S + 0.01} \right)\,mol\,{L^{ - 1}}$.
Solubility product is only a function of temperature and since no temperature change occur during the entire process, the solubility product remains same.
Therefore from equation $\left( 1 \right)$
${K_{sp}}\, = \,S\left( {S + 0.01} \right)$
$\Rightarrow \,1.5 \times {10^{ - 10}}\, = \,{S^2} + 0.01S$ $(\,\because \,{K_{sp\,}}$of $AgCl$in water $= \,1.5 \times {10^{ - 10}})$
$\Rightarrow \,1.5 \times {10^{ - 10}}\, = \,0.01S$ $($neglecting ${S^2}$ as $S < < < 1)$
$\Rightarrow \,S\, = \,\dfrac{{1.5 \times {{10}^{ - 10}}}}{{0.01}}$
$\therefore \,S\, = \,\,1.5 \times {10^{ - 8}}$
Hence the solubility of $AgCl$ in $0.01\,M\,$aqueous $NaCl$solution is $1.5 \times {10^{ - 8}}\,mol\,{L^{ - 1}}$.

Therefore the correct answer is (iii)$1.5 \times {10^{ - 8}}\,M$.

Note: Make sure to use the idea of common ion effect otherwise the $\left[ {C{l^ - }} \right]\,$will be different and hence the solubility will come out to be different. Also keep in mind $NaCl$is a strong electrolyte and hence undergoes full dissociation so the initial $\left[ {NaCl} \right]$is equal to the $\left[ {N{a^ + }} \right]$and $\left[ {C{l^ - }} \right]$.