The solubility of CaCl_2 is 1.11 times 10^{-8 | vedclass

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(B) The solubility $s$ in $g/100 \ mL$ is $1.11 	imes 10^{-8}$.To convert this to $mol/L$ (molarity):$s = \frac{1.11 	imes 10^{-8} \ g}{100 \ mL}

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$4 	imes 10^{-27}$

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(B) The solubility $s$ in $g/100 \ mL$ is $1.11 	imes 10^{-8}$.To convert this to $mol/L$ (molarity):$s = \frac{1.11 	imes 10^{-8} \ g}{100 \ mL} 	imes \frac{1000 \ mL}{1 \ L} 	imes \frac{1 \ mol}{111 \ g} = 10^{-9} \ mol/L$.For the dissociation $CaCl_2 ightleftharpoons Ca^{2+} + 2Cl^-$,the solubility product expression is $K_{sp} = [Ca^{2+}][Cl^-]^2 = (s)(2s)^2 = 4s^3$.Substituting $s = 10^{-9}$:$K_{sp} = 4 	imes (10^{-9})^3 = 4 	imes 10^{-27}$.

The solubility of $CaCl_2$ is $1.11 \times 10^{-8} \ g/100 \ mL$. If the molar mass of $CaCl_2$ is $111 \ g/mol$,find the $K_{sp}$.

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