The molar solubility of Zn(OH)_{2} in 0.1 , M | vedclass

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(C) The dissolution equilibrium of $Zn(OH)_{2}$ is given by:$Zn(OH)_{2(s)} \rightleftharpoons Zn^{2+}_{(aq)} + 2OH^{-}_{(aq)}$Let the molar solubility

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$2$

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(C) The dissolution equilibrium of $Zn(OH)_{2}$ is given by:$Zn(OH)_{2(s)} ightleftharpoons Zn^{2+}_{(aq)} + 2OH^{-}_{(aq)}$Let the molar solubility be $S$. The concentration of $OH^{-}$ ions from $NaOH$ is $0.1 \, M$. Since $S$ is very small,the total concentration of $OH^{-}$ is approximately $0.1 \, M$.$K_{sp} = [Zn^{2+}][OH^{-}]^{2}$$2 	imes 10^{-20} = S 	imes (0.1)^{2}$$2 	imes 10^{-20} = S 	imes 10^{-2}$$S = \frac{2 	imes 10^{-20}}{10^{-2}} = 2 	imes 10^{-18} \, M$Comparing this with $x 	imes 10^{-18} \, M$,we get $x = 2$.

The molar solubility of $Zn(OH)_{2}$ in $0.1 \, M \, NaOH$ solution is $x \times 10^{-18} \, M$. The value of $x$ is ...... . (Nearest integer)
(Given : The solubility product of $Zn(OH)_{2}$ is $2 \times 10^{-20}$)

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The molar solubility of $Zn(OH)_{2}$ in $0.1 \, M \, NaOH$ solution is $x \times 10^{-18} \, M$. The value of $x$ is ...... . (Nearest integer)(Given : The solubility product of $Zn(OH)_{2}$ is $2 \times 10^{-20}$)