Question

The ksp for cobalt (ii) phosphate (mm = 366.73 g/mol) is 2.05 * 10-35. what is the solubility of this salt (ng/l) in a 0.029 m sodium phosphate solution?

Answer 1

`\boxed{\text{3.54 ng/L}}`

Step-by-step explanation:

At equilibrium we have

`\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}`

`K_(sp) = [\text{Co}^(2+)]^(3)[\text{PO}_(4)^(3-)]^(2)= 2.05*10^(-35)\\\\(3s)^(3)* (0.029 + 2s)^(2) = 2.05*10^(-35)`

Assume that s ≪ 0.029. Then

`27s^(3)* (0.029)^(2) = 2.05*10^(-35)\\\\2.27 * 10^(-2)s^(3) = 2.05*10^(-35)\\\\s^(3)= 9.028*10^(-34)\\\\s = \sqrt[3]{9.028*10^(-34)}= 9.665*10^(-12)`

`s = \frac{9.665*10^(-12)\text{ mol}}{\text{1 L}}* \frac{ \text{366.73 g} }{\text{1 mol}}\\\\ = 3.54* 10^(-9) \text{g/L} = \text{ 3.54 ng/L}`

The solubility of cobalt(II) phosphate is
`\boxed{\textbf {3.54 ng/L}}`.