Question

The solubility of $PbS{O}_4$ water is 0.038 g/L. Calculate the solubility product constant of $PbS{O}_4$. Molar mass $PbS{O}_4=304g/mole\frac{380}{304}=1.25(1.25{)}^2=1.56$

Answer 1

Molar mass of $PbS{O}_4=304g/mol$

Number of moles= $\frac{0.038}{304}$

$=0.000125$ moles

$PbS{O}_4\rightleftharpoons P{b}^{2+}+S{O}_4^{2-}$

$\Rightarrow K_{sp}=\left[P{b}^{2+}\right][S{O}_4{]}^{2-}$

$=(0.000125{)}^2$

$=1.56\times {10}^{-8}$