For a sparingly soluble salt $\mathrm{AB}_{2}$, the equilibrium concentrations of $\mathrm{A}^{2+}$ ions and $\mathrm{B}^{-}$ions are $1.2 \times 10^{-4} \mathrm{M}$ and $0.24 \times 10^{-3} \mathrm{M}$, respectively. The solubility product of $\mathrm{AB}_{2}$ is :

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Equilibrium

For a sparingly soluble salt

AB_{2}

, the equilibrium concentrations of

A^{2 +}

ions and

B^{-}

ions are

1.2 \times 1 0^{- 4} M

and

0.24 \times 1 0^{- 3} M

, respectively. The solubility product of

AB_{2}

is :

Answer: (B)

For

AB_{2} (s) ⇌ A^{2 +} (a q) + 2 B^{-} (a q)

K_{s p} = [A^{2 +}] [B^{-}]^{2}

With

[A^{2 +}] = 1.2 \times 1 0^{- 4}

M and

[B^{-}] = 2.4 \times 1 0^{- 4}

K_{s p} = (1.2 \times 1 0^{- 4}) (2.4 \times 1 0^{- 4})^{2} = (1.2 \times 1 0^{- 4}) (5.76 \times 1 0^{- 8}) = 6.91 \times 1 0^{- 12}

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option