Question

Calculate the solubility product for the following:

Ce(IO3)4, 1.5 × 10–2 g/100.0 mL (IO3- is the anion)

Ksp = ______ x10______



Use correct significant figures

Answer 1

Answer: 2.749 x 10^-14

Step-by-step 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:

`$$\begin{gathered}M=\frac{0.15 \mathrm{~g} / L}{839.7267 \mathrm{~g} / \mathrm{mol}} \\M=1.7863 * 10^{-\mathbf{4}} \mathrm{mol} / {L}=\mathbf{1 . 7 8 6 3 * 1 0 ^ { - \mathbf { 4 } } \boldsymbol { M }}\end{gathered}$$`

The solubility equilibrium is given by:

`$$\mathrm{Ce}\left(\mathrm{IO}_(3)\right)_(4(s)) \rightleftharpoons \mathrm{Ce}_((a q))^(3+)+3 \mathrm{IO}_(3)^(-)(a q)$$`

Therefore, the solubility product is given by:

`\begin{gathered}K_(s p)=\left[C e^(3+)\right] \cdot\left[I O_(3)^(-)\right]^(3) \\\therefore K_(s p)=\left(1.7863 * 10^(-4)\right) *\left(3 * 1.7863 * 10^(-4)\right)^(3)\end{gathered}`

`\therefore {K_(s p)=2.749 * 10^(-14)`