Chemistry · Ionic Equilibrium

JEE Main 2025 — 4 April, Evening Shift — Question 15

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 $\mathrm{H}_{2} \mathrm{O}$ ]

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

For $\mathrm{pH}=10,\left[\mathrm{OH}^{-}\right]=10^{-4}$

\because \quad \mathrm{Mg}(\mathrm{OH})_{2} \longrightarrow \mathrm{Mg}^{2+}+2 \mathrm{OH}^{-}

$\left[\mathrm{Mg}(\mathrm{OH})_{2}\right]=0.5 \times 10^{-4} \mathrm{M}$

Mass of $\mathrm{Mg}(\mathrm{OH})_{2}=5 \times 10^{-5} \times 1 \times 58=2.9 \mathrm{mg}$

$\approx 3 \mathrm{mg}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Evening Shift
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: Solutions containing one Acid or Base