$x \mathrm{mg}$ of $\mathrm{Mg}(\mathrm{OH})_{2}$ (molar mass $=58$ ) is required to be dissolved in

Question

$x \mathrm{mg}$ of $\mathrm{Mg}(\mathrm{OH})_{2}$ (molar mass $=58$ ) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K . The value of $x$ is $\_\_\_\_$ mg. (Nearest integer) [Given: $\mathrm{Mg}(\mathrm{OH})_{2}$ is assumed to dissociate completely in $\mathrm{H}_{2} \mathrm{O}$ ]\

Professor PC Thomas Classes · Answer

(3) : Given : Molar mass of $\mathrm{Mg}(\mathrm{OH})_{2}=58 \mathrm{~g} / \mathrm{mol}$\ $V_{\left(\mathrm{H}_{2} \mathrm{O}ight)}=1 \mathrm{~L} ; \mathrm{pH}=10 ; T=298 \mathrm{~K}$\ $\mathrm{pH}=10$ means $\left[\mathrm{H}^{+}ight]=10^{-10}$;\ hence $\left[\mathrm{OH}^{-}ight]=10^{-4}$ $$ \underset{0.5 	imes 10^{-4} \mathrm{M}}{\mathrm{Mg}(\mathrm{OH})_{2}} \longrightarrow \mathrm{Mg}^{2+}+\underset{1 	imes 10^{-4} \mathrm{M}}{2 \mathrm{OH}^{-}} $$ $$ \begin{aligned} 	ext { Mass of } \mathrm{Mg}(\mathrm{OH})_{2} & =0.5 	imes 10^{-4} 	imes 58 	imes 1 \ & =2.9 \mathrm{mg}=3 \mathrm{mg} \end{aligned} $$