The molar solubility of a salt AB2 is 1.5 x 10^-4 mol L^-1. Its K_sp is

NEET levelSingle correct optionDifficulty 1 of 3ie172

The molar solubility of a salt $AB X_{2}$ is $1.5 \times 1 0^{- 4} mol L^{- 1}$ . Its $K_{s p}$ is

AB X_{2} A X^{2 +} + 2 B X^{-}

gives

[A] = s

and

[B] = 2 s

, so

K_{s p} = s (2 s)^{2} = 4 s^{3} = 4 (1.5 \times 1 0^{- 4})^{3} = 1.3 \times 1 0^{- 11}