The solubility product of AgI at 25,text{°C} | vedclass

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(C) The dissociation of $AgI$ is given by: $AgI(s) ightleftharpoons Ag^+(aq) + I^-(aq)$.Given $K_{sp} = 1.0 	imes 10^{-16}$.In a $10^{-4}\,	ext{N}

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$1.0 	imes 10^{-12}$

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(C) The dissociation of $AgI$ is given by: $AgI(s) ightleftharpoons Ag^+(aq) + I^-(aq)$.Given $K_{sp} = 1.0 	imes 10^{-16}$.In a $10^{-4}\,	ext{N}$ solution of $KI$,the concentration of $I^-$ ions is $[I^-] = 10^{-4}\,	ext{M}$ (since $KI$ is a strong electrolyte).Let $S$ be the solubility of $AgI$ in the presence of $KI$.Then $[Ag^+] = S$ and $[I^-] = 10^{-4} + S \approx 10^{-4}$ (since $S$ is very small).$K_{sp} = [Ag^+][I^-] = S 	imes 10^{-4} = 1.0 	imes 10^{-16}$.$S = \frac{1.0 	imes 10^{-16}}{10^{-4}} = 1.0 	imes 10^{-12}\,	ext{mol L}^{-1}$.

The solubility product of $AgI$ at $25\,\text{°C}$ is $1.0 \times 10^{-16}\,\text{mol}^2\,\text{L}^{-2}$. The solubility of $AgI$ in $10^{-4}\,\text{N}$ solution of $KI$ at $25\,\text{°C}$ is approximately (in $\text{mol L}^{-1}$)

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