What will be the solubility of AlCl_3 in a so | vedclass

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(B) The dissociation of $AlCl_3$ is given by: $AlCl_3(s) \rightleftharpoons Al^{3+}(aq) + 3Cl^-(aq)$.Let the solubility of $AlCl_3$ be $S$.Then,$[Al^{

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$\frac{K_{sp}}{8C^3}$

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(B) The dissociation of $AlCl_3$ is given by: $AlCl_3(s) ightleftharpoons Al^{3+}(aq) + 3Cl^-(aq)$.Let the solubility of $AlCl_3$ be $S$.Then,$[Al^{3+}] = S$ and $[Cl^-] = 3S$.In a solution of $CaCl_2$ with concentration $C$,the dissociation is $CaCl_2 ightarrow Ca^{2+}(aq) + 2Cl^-(aq)$.Thus,the concentration of $Cl^-$ ions from $CaCl_2$ is $2C$.The total concentration of $Cl^-$ ions is $[Cl^-] = (3S + 2C)$.Since $S$ is very small compared to $C$,we can approximate $[Cl^-] \approx 2C$.The solubility product expression is $K_{sp} = [Al^{3+}][Cl^-]^3$.Substituting the values: $K_{sp} = (S)(2C)^3$.$K_{sp} = S 	imes 8C^3$.Therefore,$S = \frac{K_{sp}}{8C^3}$.

What will be the solubility of $AlCl_3$ in a solution of $CaCl_2$ with concentration $C$?

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