The solubility product of AgCl is 8 times 10^ | vedclass

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(B) The dissociation of $AgCl$ is given by: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.In the presence of $0.01 \ M \ NaCl$,the concentration of

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$8 	imes 10^{-4}$

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(B) The dissociation of $AgCl$ is given by: $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.In the presence of $0.01 \ M \ NaCl$,the concentration of $Cl^-$ ions increases due to the common ion effect.$NaCl(aq) ightarrow Na^+(aq) + Cl^-(aq)$.Since $NaCl$ is a strong electrolyte,$[Cl^-] = 0.01 \ M$.Let the new solubility of $AgCl$ be $s'$.Then,$[Ag^+] = s'$ and $[Cl^-] = (s' + 0.01) \approx 0.01 \ M$ (since $s'$ is very small).$K_{sp} = [Ag^+][Cl^-] = s' 	imes 0.01$.$8 	imes 10^{-6} = s' 	imes 0.01$.$s' = \frac{8 	imes 10^{-6}}{0.01} = 8 	imes 10^{-4} \ M$.

The solubility product of $AgCl$ is $8 \times 10^{-6}$. Find its new solubility in the presence of $0.01 \ M \ NaCl$.

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