Ksp for $\mathrm{Cr}(\mathrm{OH})_{3}$ is $1.6 \times 10^{-30}$. What is the molar solubility of this salt in water?

(A) $\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}$

(B) $\frac{1.8 \times 10^{-30}}{27}$

(C) $\sqrt[5]{1.8 \times 10^{-30}}$

(D) $\sqrt[2]{1.6 \times 10^{-30}}$

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Equilibrium

Ksp for

Cr (OH)_{3}

1.6 \times 1 0^{- 30}

. What is the molar solubility of this salt in water?

Answer: (A)

Dissolution:

C r (O H)_{3} (s) ⇌ C r^{3 +} (a q) + 3 O H^{-} (a q)

If the molar solubility is

s

, then

[C r^{3 +}] = s

and

[O H^{-}] = 3 s

, giving

K_{s p} = [C r^{3 +}] [O H^{-}]^{3} = s (3 s)^{3} = 27 s^{4}

Setting

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

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

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option