$K_{s p}$ of $PbI X_{2}$ is $6.4 \times 1 0^{- 9}$ . Calculate its solubility in $0.02 M$ $KI$ solution, in units of $1 0^{- 6} mol L^{- 1}$ (nearest integer; take $[I X^{-}] = 0.02 M$ ).

Numerical answer type, answer in x 10^-6 mol/L, tolerance 0.5.

Answer: 16 x 10^-6 mol/L

K_{s p} = [Pb X^{2 +}] [I X^{-}]^{2}

, so

s = [Pb X^{2 +}] = 6.4 \times 1 0^{- 9} / (0.02)^{2} = 1.6 \times 1 0^{- 5} M

, i.e. 16 in units of

1 0^{- 6} M