A solution contains 0.01 M Mn^2+. Given K_sp(Mn(OH)2) = 4.0 x 10^-12 and log 2 = 0.301, the pH at which Mn(OH)2 just begins to precipitate is closest to

JEE Main levelSingle correct optionDifficulty 2 of 3ie188

A solution contains $0.01 M$ $Mn X^{2 +}$ . Given $K_{s p} (Mn (OH) X_{2}) = 4.0 \times 1 0^{- 12}$ and $lo g 2 = 0.301$ , the pH at which $Mn (OH) X_{2}$ just begins to precipitate is closest to

[OH X^{-}]^{2} = K_{s p} / [Mn X^{2 +}] = 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