The concentration of $500 \ mL$ $NaOH$ solution is $0.02 \ M$. How many grams of $FeSO_4$ must be added to this solution to initiate the precipitation of $Fe(OH)_2$? The $K_{sp}$ of $Fe(OH)_2$ is $1.5 \times 10^{-15}$. The molecular mass of $FeSO_4$ is $152 \ g \ mol^{-1}$.

(N/A) $1$. The concentration of $OH^-$ ions in the $NaOH$ solution is $[OH^-] = 0.02 \ M$.
$2$. The solubility product expression for $Fe(OH)_2$ is $K_{sp} = [Fe^{2+}][OH^-]^2$.
$3$. To initiate precipitation,the ionic product must exceed the $K_{sp}$. Thus,$[Fe^{2+}][OH^-]^2 > 1.5 \times 10^{-15}$.
$4$. Substituting $[OH^-] = 0.02$,we get $[Fe^{2+}](0.02)^2 > 1.5 \times 10^{-15}$.
$5$. $[Fe^{2+}](4 \times 10^{-4}) > 1.5 \times 10^{-15} \implies [Fe^{2+}] > 3.75 \times 10^{-12} \ M$.
$6$. The number of moles of $FeSO_4$ required in $500 \ mL$ $(0.5 \ L)$ is $n = M \times V = 3.75 \times 10^{-12} \times 0.5 = 1.875 \times 10^{-12} \ mol$.
$7$. The mass of $FeSO_4$ required is $mass = n \times \text{molar mass} = 1.875 \times 10^{-12} \ mol \times 152 \ g \ mol^{-1} = 2.85 \times 10^{-10} \ g$.
$8$. Therefore,more than $2.85 \times 10^{-10} \ g$ of $FeSO_4$ must be added.