If the solubility products of AgCl and AgBr a | vedclass

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Question

(A) For a salt of type $AB$,the solubility product $K_{sp}$ is related to solubility $S$ by the formula $K_{sp} = S^2$,which implies $S = \sqrt{K_{sp}

Vedclass · Accepted Answer

$S$ of $AgBr$ is less than that of $AgCl$

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(A) For a salt of type $AB$,the solubility product $K_{sp}$ is related to solubility $S$ by the formula $K_{sp} = S^2$,which implies $S = \sqrt{K_{sp}}$.For $AgCl$: $K_{sp} = 1.0 	imes 10^{-10}$.$S_{AgCl} = \sqrt{1.0 	imes 10^{-10}} = 1.0 	imes 10^{-5} \ M$.For $AgBr$: $K_{sp} = 3.5 	imes 10^{-13}$.$S_{AgBr} = \sqrt{3.5 	imes 10^{-13}} = \sqrt{35 	imes 10^{-14}} \approx 5.91 	imes 10^{-7} \ M$.Comparing the two,$5.91 	imes 10^{-7} Therefore,the solubility of $AgBr$ is less than that of $AgCl$.

If the solubility products of $AgCl$ and $AgBr$ are $1.0 \times 10^{-10}$ and $3.5 \times 10^{-13}$ respectively,then the relation between the solubilities (denoted by the symbol $S$) of these salts can correctly be represented as:

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