Answer
$Kps = 6.423 \times 10^{-9}$
Work Step by Step
1. Convert the mass to nº of moles:
$mm(PbI_2) = 207.2 + 126.9*2 = 461 g/mol$
$n(moles) = \frac{mass(g)}{mm}$
$n(moles) = \frac{0.54}{461}$
$n(moles) = 1.171 \times 10^{-3}$
2. Find the concentration of $PbI_2$:
$C = \frac{n(moles)}{V(L)}$
$C = \frac{1.171 \times 10^{-3}}{1}$
$C = 1.171 \times 10^{-3} M$
3. Write the Kps Formula:
$Kps = [Pb^{2+}]*[I^-]^2$
$[PbI_2] = 1.171 \times 10^{-3}M$
$[Pb^{2+}] = 1.171 \times 10^{-3}M$
$[I^-] = 1.171 \times 10^{-3} * 2 = 2.342 \times 10^{-3}M$
4. Calculate the Kps:
$Kps = 1.171 \times 10^{-3} * (2.342 \times 10^{-3})^2$
$Kps = 1.171 \times 10^{-3} * 5.485 \times 10^{-6}$
$Kps = 6.423 \times 10^{-9}$
