The solubility of AgCl is 4 times 10^{-6} M. | vedclass

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(D) For $AgCl$ in pure water: $AgCl(s) ⇌ Ag^+(aq) + Cl^-(aq)$.$K_{sp} = [Ag^+][Cl^-] = s^2 = (4 	imes 10^{-6})^2 = 16 	imes 10^{-12}$.In the presenc

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$4 	imes 10^{-9} \ M$

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(D) For $AgCl$ in pure water: $AgCl(s) ⇌ Ag^+(aq) + Cl^-(aq)$.$K_{sp} = [Ag^+][Cl^-] = s^2 = (4 	imes 10^{-6})^2 = 16 	imes 10^{-12}$.In the presence of $2 	imes 10^{-3} \ M \ CaCl_2$,the concentration of $Cl^-$ ions from $CaCl_2$ is $[Cl^-] = 2 	imes (2 	imes 10^{-3}) = 4 	imes 10^{-3} \ M$.Let the new solubility be $s'$. The total concentration of $Cl^-$ is $(s' + 4 	imes 10^{-3}) \approx 4 	imes 10^{-3} \ M$ (since $s'$ is very small).$K_{sp} = [Ag^+][Cl^-] = s' 	imes (4 	imes 10^{-3}) = 16 	imes 10^{-12}$.$s' = \frac{16 	imes 10^{-12}}{4 	imes 10^{-3}} = 4 	imes 10^{-9} \ M$.

The solubility of $AgCl$ is $4 \times 10^{-6} \ M$. Find its $K_{sp}$. Also,find its new solubility in the presence of $2 \times 10^{-3} \ M \ CaCl_2$.

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