K_sp of Mg(OH)2 is 4.0x10^-12. At what pH (two decimals) does Mg(OH)2 just begin to precipitate from 0.010 M Mg^2+? Use log 2 = 0.301 and K_w = 10^-14.

JEE Main levelNumerical answerDifficulty 2 of 3ia186

$K_{s p}$ of $Mg (OH) X_{2}$ is $4.0 \times 1 0^{- 12}$ . At what pH (two decimals) does $Mg (OH) X_{2}$ just begin to precipitate from $0.010 M$ $Mg X^{2 +}$ ? Use $lo g 2 = 0.301$ and $K_{w} = 1 0^{- 14}$ .

Numerical answer type, tolerance 0.01.

[OH X^{-}]^{2} = K_{s p} /0.010 = 4.0 \times 1 0^{- 10}

, so

[OH X^{-}] = 2 \times 1 0^{- 5} M

pOH = 5 - 0.301 = 4.699

and

pH = 9.30