Q19 / 75 · JEE Main 2025 (4 Apr Shift 2) · Chemistry
x 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 $\mathrm{H}_2 \mathrm{O}$)
(Given : $\mathrm{Mg}(\mathrm{OH})_2$ is assumed to dissociate completely in $\mathrm{H}_2 \mathrm{O}$)
Correct answer: 3
Solution
$\begin{aligned}<br />& \mathrm{pH}=10 \\<br>& \mathrm{pOH}=4 \\<br>& {\left[\mathrm{OH}^{-}\right]=10^{-4}}<br />\end{aligned}$
no. of moles of $\mathrm{OH}^{-}=10^{-4}$
no. of moles of $\operatorname{Mg}(\mathrm{OH})_2=\frac{10^{-4}}{2}=5 \times 10^{-5}$
$\text { mass of } \begin{aligned}<br />\mathrm{Mg}(\mathrm{OH})_2 & =5 \times 10^{-5} \times 58 \times 10^3 \mathrm{mg} \\<br>& =2.9<br />\end{aligned}$
no. of moles of $\mathrm{OH}^{-}=10^{-4}$
no. of moles of $\operatorname{Mg}(\mathrm{OH})_2=\frac{10^{-4}}{2}=5 \times 10^{-5}$
$\text { mass of } \begin{aligned}<br />\mathrm{Mg}(\mathrm{OH})_2 & =5 \times 10^{-5} \times 58 \times 10^3 \mathrm{mg} \\<br>& =2.9<br />\end{aligned}$
Hence, the correct answer is 3.