The solution is 2 L of 0.02 M NaOH. The so | vedclass

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(A) The solubility product expression for $Zn(OH)_2$ is $K_{sp} = [Zn^{2+}][OH^-]^2$. Given $[OH^-] = 0.02 \ M$ due to the presence of $NaOH$. Substit

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$2.228 	imes 10^{-11} \ g$

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(A) The solubility product expression for $Zn(OH)_2$ is $K_{sp} = [Zn^{2+}][OH^-]^2$. Given $[OH^-] = 0.02 \ M$ due to the presence of $NaOH$. Substituting the values: $4.5 	imes 10^{-17} = [Zn^{2+}](0.02)^2$. $[Zn^{2+}] = \frac{4.5 	imes 10^{-17}}{4 	imes 10^{-4}} = 1.125 	imes 10^{-13} \ M$. The molar mass of $Zn(OH)_2$ is $65.38 + 2 	imes (16 + 1) = 99.38 \ g/mol$. Amount of $Zn(OH)_2$ in $2 \ L$ solution = $Molarity 	imes Volume 	imes Molar \ Mass = (1.125 	imes 10^{-13} \ mol/L) 	imes 2 \ L 	imes 99.38 \ g/mol = 2.236 	imes 10^{-11} \ g$ (approximately $2.228 	imes 10^{-11} \ g$ depending on atomic mass precision).

The solution is $2 \ L$ of $0.02 \ M \ NaOH$. The solubility product of $Zn(OH)_2$ is $4.5 \times 10^{-17}$. Then how many grams of $Zn(OH)_2$ is maximum soluble in the given $NaOH$ solution?

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