what is the solubility of cd₃(po₄)₂ in water? (k_sp of cd₃(po₄)₂ is 2.5 × 10⁻³³)

Explanation:

Step1: Write the dissolution equation

$$\mathrm{Cd}{3}(\mathrm{PO}{4}){2}(s)\rightleftharpoons 3\mathrm{Cd}^{2 +}(aq)+2\mathrm{PO}{4}^{3 -}(aq)$$ Let the solubility of $\mathrm{Cd}{3}(\mathrm{PO}{4}){2}$ be $s$. Then $[\mathrm{Cd}^{2 +}]=3s$ and $[\mathrm{PO}{4}^{3 -}]=2s$.

Step2: Write the $K_{sp}$ expression

$$K_{sp}=[\mathrm{Cd}^{2 +}]^{3}[\mathrm{PO}{4}^{3 -}]^{2}$$ Substitute $[\mathrm{Cd}^{2 +}]=3s$ and $[\mathrm{PO}{4}^{3 -}]=2s$ into the $K_{sp}$ expression: $$K_{sp}=(3s)^{3}(2s)^{2}$$ $$K_{sp}=27s^{3}\times4s^{2}=108s^{5}$$

Step3: Solve for $s$

Given $K_{sp}=2.5\times 10^{-33}$, then $$108s^{5}=2.5\times 10^{-33}$$ $$s^{5}=\frac{2.5\times 10^{-33}}{108}$$ $$s=\sqrt[5]{\frac{2.5\times 10^{-33}}{108}}$$ $$s\approx 1.6\times 10^{-7}\space mol/L$$

Answer:

The solubility of $\mathrm{Cd}{3}(\mathrm{PO}{4})_{2}$ in water is approximately $1.6\times 10^{-7}\space mol/L$.