A solution is $0.0100 M$ in $Pb X^{2 +}$ . Dilute $HCl$ is added so that the equilibrium chloride concentration is $0.10 M$ (assume it is unchanged by precipitation). With $K_{s p} (PbCl X_{2}) = 1.6 \times 1 0^{- 5}$ , the percentage of lead that continues into the group II filtrate is

Answer: A

At equilibrium

[Pb X^{2 +}] = K_{s p} / [Cl X^{-}] X^{2} = 1.6 \times 1 0^{- 5} / (0.10)^{2} = 1.6 e - 03 M

. Fraction remaining

= 1.6 e - 03/0.0100 = 16%

, so 84% is precipitated as

PbCl X_{2}

. The remaining 16% is precipitated as

PbS

in group II; this explains the dual appearance of lead.