Find the solubility of Mg(OH)2 in a buffer maintained at pH 9.20 (298 K). K_sp(Mg(OH)2) = 1.2 x 10^-11, K_w = 1.0 x 10^-14, 10^-4.8 = 1.585 x 10^-5. Give the answer in units of 10^-2 mol L^-1 (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3ia081

Find the solubility of $Mg (OH) X_{2}$ in a buffer maintained at pH 9.20 (298 K). $K_{s p} (Mg (OH) X_{2}) = 1.2 \times 1 0^{- 11}$ , $K_{w} = 1.0 \times 1 0^{- 14}$ , $1 0^{- 4.8} = 1.585 \times 1 0^{- 5}$ . Give the answer in units of $1 0^{- 2} mol L^{- 1}$ (two decimals).

Numerical answer type, answer in x 10^-2 mol/L, tolerance 0.01.

pOH = 4.80, so

[OH X^{-}] = 1.585 \times 1 0^{- 5} M

(fixed by the buffer).

s = [Mg X^{2 +}] = K_{s p} / [OH X^{-}]^{2} = 1.2 \times 1 0^{- 11} / (2.51 \times 1 0^{- 10}) = 4.78 \times 1 0^{- 2} M

, i.e. 4.78 in units of

1 0^{- 2} M