If S_1, S_2, S_3 and S_4 are the solubilities | vedclass

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(B) The solubility of a sparingly soluble salt like $AgCl$ decreases in the presence of a common ion due to the common ion effect.$S_1$ is the solubil

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$S_1 > S_3 > S_2 > S_4$

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(B) The solubility of a sparingly soluble salt like $AgCl$ decreases in the presence of a common ion due to the common ion effect.$S_1$ is the solubility in water (no common ion).$S_2$ is the solubility in $0.01 \ M \ CaCl_2$. Since $CaCl_2 ightarrow Ca^{2+} + 2Cl^-$,the concentration of $Cl^-$ is $2 	imes 0.01 = 0.02 \ M$.$S_3$ is the solubility in $0.01 \ M \ NaCl$. Since $NaCl ightarrow Na^+ + Cl^-$,the concentration of $Cl^-$ is $0.01 \ M$.$S_4$ is the solubility in $0.05 \ M \ AgNO_3$. Since $AgNO_3 ightarrow Ag^+ + NO_3^-$,the concentration of $Ag^+$ is $0.05 \ M$.Comparing the common ion concentrations: $[Cl^-]$ in $S_2$ is $0.02 \ M$,$[Cl^-]$ in $S_3$ is $0.01 \ M$,and $[Ag^+]$ in $S_4$ is $0.05 \ M$.Since $S = K_{sp} / [	ext{common ion}]$,higher common ion concentration leads to lower solubility.$S_4$ has the highest common ion concentration $(0.05 \ M)$,so it is the lowest.Comparing $S_2$ and $S_3$,$S_2$ has a higher $[Cl^-]$ $(0.02 \ M)$ than $S_3$ $(0.01 \ M)$,so $S_2 Thus,the order is $S_1 > S_3 > S_2 > S_4$.

If $S_1, S_2, S_3$ and $S_4$ are the solubilities of $AgCl$ in water,in $0.01 \ M \ CaCl_2$,in $0.01 \ M \ NaCl$ and in $0.05 \ M \ AgNO_3$ respectively at a certain temperature,the correct order of solubilities is

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