The solubility product of AgCl is K_sp. Then the solubility of AgCl in xM KCl is

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$\frac{K_{sp}}{x}$

$K_{sp} \times x^{2}$

$\frac{K_{sp}}{x^{2}}$

$\frac{X}{K_{sp}}$

answer is D.

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Detailed Solution

$\begin{array}{l} ∴ & S (S + x) = K_{sp} \\ or, & S^{2} + xS = K_{sp} \end{array} ∵ S^{2} is very small, it is neglected \begin{aligned} \Rightarrow XS = K_{sp} \\ \Rightarrow S = \frac{K_{sp}}{X} \end{aligned}$

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