The solubility product of AgI at 25^{circ} C | vedclass

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(A) The solubility product constant $(K_{sp})$ for $AgI$ is given by: $K_{sp} = [Ag^+][I^-] = 1.0 	imes 10^{-16}$.In the presence of $10^{-4} \ N \ K

Vedclass · Accepted Answer

$1.0 	imes 10^{-12} \ mol \ L^{-1}$

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(A) The solubility product constant $(K_{sp})$ for $AgI$ is given by: $K_{sp} = [Ag^+][I^-] = 1.0 	imes 10^{-16}$.In the presence of $10^{-4} \ N \ KI$,the concentration of $I^-$ ions is $[I^-] = 10^{-4} \ M$.Let the solubility of $AgI$ be $S$. Then $[Ag^+] = S$.Substituting these values into the $K_{sp}$ expression: $S 	imes 10^{-4} = 1.0 	imes 10^{-16}$.Solving for $S$: $S = \frac{1.0 	imes 10^{-16}}{10^{-4}} = 1.0 	imes 10^{-12} \ mol \ L^{-1}$.

The solubility product of $AgI$ at $25^{\circ} C$ is $1.0 \times 10^{-16} \ mol^2 \ L^{-2}$. What is the solubility of $AgI$ in $10^{-4} \ N \ KI$ at $25^{\circ} C$?

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