Answer: 2.749 x 10^-14
Explanation:
Given:
The concentration of Ce(IO3)4 in the solution = 1.5 x 10-2 g/100 mL= 0.15 g/1000 mL= 0.15 g/L
Molar mass of Ce(IO3)4 = 839.7267 g/mol
Therefore, the molarity of Ce(IO3)4 in the solution is given by:
[tex]$\begin{gathered}M=\frac{0.15 \mathrm{~g} / L}{839.7267 \mathrm{~g} / \mathrm{mol}} \\M=1.7863 \times 10^{-\mathbf{4}} \mathrm{mol} / {L}=\mathbf{1 . 7 8 6 3 \times 1 0 ^ { - \mathbf { 4 } } \boldsymbol { M }}\end{gathered}$[/tex]
The solubility equilibrium is given by:
[tex]$\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{4(s)} \rightleftharpoons \mathrm{Ce}_{(a q)}^{3+}+3 \mathrm{IO}_{3}^{-}(a q)$[/tex]
Therefore, the solubility product is given by:
[tex]\begin{gathered}K_{s p}=\left[C e^{3+}\right] \cdot\left[I O_{3}^{-}\right]^{3} \\\therefore K_{s p}=\left(1.7863 \times 10^{-4}\right) \times\left(3 \times 1.7863 \times 10^{-4}\right)^{3}\end{gathered}[/tex]
[tex]\therefore {K_{s p}=2.749 \times 10^{-14}[/tex]
