Zinc hydroxide dissolves as Zn^2+ (with K_sp = 3.0x10^-17) and, in strong base, as zincate: Zn(OH)2(s) + 2OH- [Zn(OH)4]^2- with K = 1.2x10^-2. The total dissolved zinc is smallest at a pH of (use log 5 = 0.699)

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia213

Zinc hydroxide dissolves as $Zn X^{2 +}$ (with $K_{s p} = 3.0 \times 1 0^{- 17}$ ) and, in strong base, as zincate: $Zn (OH) X_{2} (s) + 2 OH X^{-} [Zn (OH) X_{4}] X^{2 -}$ with $K = 1.2 \times 1 0^{- 2}$ . The total dissolved zinc is smallest at a pH of (use $lo g 5 = 0.699$ )

s = [Zn X^{2 +}] + [Zn (OH) X_{4} X^{2 -}] = K_{s p} / [OH X^{-}]^{2} + K [OH X^{-}]^{2}

. This is minimum when

[OH X^{-}]^{2} = K_{s p} / K = 2.5 \times 1 0^{- 15} = 5.0 \times 1 0^{- 8}

, so

[OH X^{-}] = 2.2 \times 1 0^{- 4} M

and

pOH = 3.651

. Hence

pH = 10.35

and

s_{min} = 2 K_{s p} K = 1.2 \times 1 0^{- 9} M