The molar solubility of $\mathrm{Zn}(\mathrm{OH})_{2}$ in $1 \mathrm{M}$ ammonia solution at room temperature is $\left(K_{\mathrm{sp}}\right.$ of $\mathrm{Zn}(\mathrm{OH})_{2}=1.6 \times 10^{-17} ; K_{\text {stab }}$ of
$\left.\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}=1.6 \times 10^{10}\right)$
(a) $4 \times 10^{-3} \mathrm{M}$
(b) $1.58 \times 10^{-6} \mathrm{M}$
(c) $4 \times 10^{-9} \mathrm{M}$
(d) $2.56 \times 10^{-7} \mathrm{M}$