Question

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

$$\ce{Cd3(PO4)2(s) <=> 3Cd^{2 + }(aq) + 2PO^{3 - }_4(aq)}$$
Let the solubility of $\ce{Cd3(PO4)2}$ be $s$ (in mol/L). Then $[\ce{Cd^{2+}}]=3s$ and $[\ce{PO^{3 - }_4}]=2s$.

Step2: Write the $K_{sp}$ expression

$$K_{sp}=[\ce{Cd^{2+}}]^3[\ce{PO^{3 - }_4}]^2$$
Substitute $[\ce{Cd^{2+}}]=3s$ and $[\ce{PO^{3 - }_4}]=2s$ into the $K_{sp}$ expression:
$$K_{sp}=(3s)^3\times(2s)^2$$
$$K_{sp}=27s^3\times4s^2=108s^5$$

Step3: Solve for $s$

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

Answer:

The solubility of $\ce{Cd3(PO4)2}$ in water is approximately $1.6\times 10^{-7}\text{ mol/L}$.