JEE Main 2025 Apr 4 Shift 2, Chemistry Q25: Ionic Equilibrium In Solution

JEE Main2025Apr 4, Shift 2Chemistry

$x$ mg of $M g (O H)_{2}$ (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 : $M g (O H)_{2}$ is assumed to dissociate completely in $H_{2} O$ )

Your answer

Solution

From $pH = 10$ , $pOH = 4$ , so $[O H^{-}] = 1 0^{- 4}$ mol L $^{- 1}$ .

In 1.0 L, moles of $O H^{-} = 1 0^{- 4}$ mol.

$M g (O H)_{2}$ dissociates as $M g (O H)_{2} \to M g^{2 +} + 2 O H^{-}$ , so

moles of $M g (O H)_{2} = \frac{1 0 ^{- 4}}{2} = 5 \times 1 0^{- 5}$ mol.

Mass $= 5 \times 1 0^{- 5} \times 58$ g $= 2.9 \times 1 0^{- 3}$ g $= 2.9$ mg $\approx 3$ mg (nearest integer).

$∴ x \approx 3$ mg.

Concept behind this question

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