In a hypothetical case, $Pb (OH) X_{2}$ ( $K_{s p} = 1.2 \times 1 0^{- 15}$ ) dissolves in alkali as $Pb (OH) X_{2} + OH X^{-} [Pb (OH) X_{3}] X^{-}$ because $Pb X^{2 +} + 3 OH X^{-} [Pb (OH) X_{3}] X^{-}$ has $K_{f} = 3.8 \times 1 0^{14}$ . What total concentration of NaOH is needed to dissolve exactly $0.020 mol$ of $Pb (OH) X_{2}$ in $1.0 L$ (no volume change; neglect free $Pb X^{2 +}$ )? Give the answer in units of $1 0^{- 2} M$ to two decimals.

Numerical answer type, answer in x 10^-2 M, tolerance 0.02.

Answer: 6.385965 x 10^-2 M

Equilibrium constant

K = K_{s p} K_{f} = 0.456

= [[Pb (OH) X_{3}] X^{-}] / [OH X^{-}]

. At the point of complete dissolution

[[Pb (OH) X_{3}] X^{-}] = 0.020 M

, so free

[OH X^{-}] = 0.020/0.456 = 0.0439 M

. Total NaOH

= 0.0439 + 0.020 = 0.0639 M

, i.e.

6.39 \times 1 0^{- 2} M