The concentration of KI and KCl in a solution | vedclass

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(B) The concentration of $I^-$ and $Cl^-$ ions in the final mixture becomes $0.0005 \ M$ each due to dilution ($20 \ mL$ added to $20 \ mL$ makes $40

Vedclass · Accepted Answer

$AgI$ will be precipitated

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(B) The concentration of $I^-$ and $Cl^-$ ions in the final mixture becomes $0.0005 \ M$ each due to dilution ($20 \ mL$ added to $20 \ mL$ makes $40 \ mL$ total volume).In a saturated solution of $AgI$,the concentration of $Ag^+$ ions is $[Ag^+] = K_{sp}(AgI) / [I^-]_{sat}$.When the solutions are mixed,the ionic product of $AgI$ is $Q_{sp} = [Ag^+][I^-]_{mix}$.Since $[I^-]_{mix} > [I^-]_{sat}$,the ionic product $Q_{sp}$ exceeds the solubility product $K_{sp}$ of $AgI$.Therefore,$AgI$ will be precipitated.

The concentration of $KI$ and $KCl$ in a solution containing both is $0.001 \ M$ each. If $20 \ mL$ of this solution is added to $20 \ mL$ of a saturated solution of $AgI$ in water,what will happen?

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