Answer:
[tex]\boxed{\text{3.54 ng/L}}[/tex]
Explanation:
At equilibrium we have
[tex]\begin{array}{cccccc} &\text{Co}_{3}\text{(PO}_{4})_{2} & \rightleftharpoons &3\text{Co}^{2+}&+ & 2\text{PO}_{4}^{3-}\\\text{I:}& & & 0 & & 0.029\\\text{C:}& & & +3s & & 0.029 + 2s \\\text{E:}& & & 3s & &0.029+2s\\\end{array}[/tex]
[tex]K_{sp} = [\text{Co}^{2+}]^{3}[\text{PO}_{4}^{3-}]^{2}= 2.05\times10^{-35}\\\\(3s)^{3}\times (0.029 + 2s)^{2} = 2.05\times10^{-35}[/tex]
Assume that s ≪ 0.029. Then
[tex]27s^{3}\times (0.029)^{2} = 2.05\times10^{-35}\\\\2.27 \times 10^{-2}s^{3} = 2.05\times10^{-35}\\\\s^{3}= 9.028\times10^{-34}\\\\s = \sqrt[3]{9.028\times10^{-34}}= 9.665\times10^{-12}[/tex]
[tex]s = \dfrac{9.665\times10^{-12}\text{ mol}}{\text{1 L}}\times \dfrac{ \text{366.73 g} }{\text{1 mol}}\\\\ = 3.54\times 10^{-9} \text{g/L} = \text{ 3.54 ng/L}[/tex]
The solubility of cobalt(II) phosphate is [tex]\boxed{\textbf {3.54 ng/L}}[/tex].
