JEE Main 2025 Jan 24 Shift 1, Chemistry Q7: Solubility Equilibria Of Sparingly Soluble Salts

JEE Main2025Jan 24, Shift 1Chemistry

$K_{s p}$ for $Cr (OH)_{3}$ is $1.6 \times 1 0^{- 30}$ . What is the molar solubility of this salt in water?

Solution

Let the molar solubility be $s$ . The dissolution gives:

$Cr (OH)_{3} (s) ⇌ C r^{3 +} (aq) + 3 O H^{-} (aq)$

Equilibrium concentrations: $[C r^{3 +}] = s$ , $[O H^{-}] = 3 s$ .

$K_{s p} = s \cdot (3 s)^{3} = 27 s^{4}$

$27 s^{4} = 1.6 \times 1 0^{- 30}$

$s = (\frac{1.6 \times 1 0 ^{- 30}}{27})^{1/4} = 4 \frac{1.6 \times 1 0 ^{- 30}}{27}$

$∴$ Option (1).

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