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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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).

Metal sulfide	Solubility product
$A$. $PbS$	$I$. $4.0 \times 10^{-53}$
$B$. $HgS$	$II$. $8.0 \times 10^{-28}$
$C$. $MnS$	$III$. $1.6 \times 10^{-24}$
$D$. $ZnS$	$IV$. $2.5 \times 10^{-13}$

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?

Similar Questions

Which of the following has maximum solubility at low $pH$?

Match the following metal sulfides with their respective solubility product $(K_{sp})$ values:

Metal sulfide Solubility product

$A$. $PbS$ $I$. $4.0 \times 10^{-53}$

$B$. $HgS$ $II$. $8.0 \times 10^{-28}$

$C$. $MnS$ $III$. $1.6 \times 10^{-24}$

$D$. $ZnS$ $IV$. $2.5 \times 10^{-13}$

For the precipitation reaction of $Ag^{\oplus}$ ions with $NaCl,$ which of the following statements is true?

The solubility product constant $K_{sp}$ of $Mg(OH)_2$ is $9.0 \times 10^{-12}.$ If a solution is $0.010 \ M$ with respect to $Mg^{2+}$ ion,what is the maximum hydroxide ion concentration which could be present without causing the precipitation of $Mg(OH)_2$?

The solubility product $(K_{sp})$ for sodium chloride is $36 \text{ mol}^2/\text{L}^2$. The solubility of sodium chloride is:

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