The molar solubility of Cd(OH)_2 is 1.84 time | vedclass

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Question

(D) The solubility product constant $K_{sp}$ for $Cd(OH)_2$ is given by $K_{sp} = 4s^3$,where $s$ is the molar solubility in water.$K_{sp} = 4 	imes

Vedclass · Accepted Answer

$2.49 	imes 10^{-10} \ M$

Vedclass · Answer

(D) The solubility product constant $K_{sp}$ for $Cd(OH)_2$ is given by $K_{sp} = 4s^3$,where $s$ is the molar solubility in water.$K_{sp} = 4 	imes (1.84 	imes 10^{-5})^3$.Given $pH = 12$,we have $pOH = 14 - 12 = 2$.Therefore,$[OH^-] = 10^{-2} \ M$.Using the expression $K_{sp} = [Cd^{2+}][OH^-]^2$,we substitute the values:$4 	imes (1.84 	imes 10^{-5})^3 = [Cd^{2+}] 	imes (10^{-2})^2$.$[Cd^{2+}] = \frac{4 	imes (1.84)^3 	imes 10^{-15}}{10^{-4}}$.$[Cd^{2+}] = 4 	imes 6.2295 	imes 10^{-11} \approx 2.49 	imes 10^{-10} \ M$.

The molar solubility of $Cd(OH)_2$ is $1.84 \times 10^{-5} \ M$ in water. The expected solubility of $Cd(OH)_2$ in a buffer solution of $pH = 12$ is

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