The solubility product of AgCl is 4.0 times 1 | vedclass

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(C) The dissociation of $AgCl$ is given by: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.$CaCl_2$ is a strong electrolyte and dissociates complete

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$5.0 	imes 10^{-9} \ m$

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(C) The dissociation of $AgCl$ is given by: $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.$CaCl_2$ is a strong electrolyte and dissociates completely as: $CaCl_2(aq) ightarrow Ca^{2+}(aq) + 2Cl^-(aq)$.Since the concentration of $CaCl_2$ is $0.04 \ m$,the concentration of $Cl^-$ ions from $CaCl_2$ is $2 	imes 0.04 \ m = 0.08 \ m$.Let the solubility of $AgCl$ be $s$. Then $[Ag^+] = s$ and $[Cl^-] = (0.08 + s) \approx 0.08 \ m$ (since $s$ is very small).The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.$4.0 	imes 10^{-10} = s 	imes 0.08$.$s = \frac{4.0 	imes 10^{-10}}{0.08} = 5.0 	imes 10^{-9} \ m$.

The solubility product of $AgCl$ is $4.0 \times 10^{-10}$ at $298 \ K$. The solubility of $AgCl$ in $0.04 \ m \ CaCl_2$ will be

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