Answer
a) PbCO3 will precipitate first.
b) $[CO_3^{2-}]=7.55\times10^{-6}\ M$
Work Step by Step
a)
$Ksp(PbI_2)=[Pb^{2+}][I^-]^2$
$9.8\times10^{-11}=[Pb^{2+}]\times0.10^2$
$[Pb^{2+}]=9.8\times10^{-9}\ M$
$Ksp(PbCO_3)=[Pb^{2+}][CO_3^{2-}]$
$7.4\times10^{-14}=[Pb^{2+}]\times0.10$
$[Pb^{2+}]=7.4\times10^{-15}\ M$
PbCO3 will precipitate first.
b) At the saturation of PbI2,
$Ksp(PbCO_3)=[Pb^{2+}][CO_3^{2-}]$
$7.4\times10^{-14}=9.8\times10^{-9}\times[CO_3^{2-}]$
$[CO_3^{2-}]=7.55\times10^{-6}\ M$
