Question

what is the solubility of cd₃(po₄)₂ in water? (ksp 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 $s\ \text{mol/L}$. Then $[\ce{Cd^{2+}}]=3s$ and $[\ce{PO^{3 - }_4}]=2s$.

Step2: Write the solubility - product 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 $K_{sp} = 2.5\times10^{-33}$, we have
$$108s^5=2.5\times 10^{-33}$$
$$s^5=\frac{2.5\times 10^{-33}}{108}$$
$$s^5\approx2.31\times 10^{-35}$$
$$s=\sqrt[5]{2.31\times 10^{-35}}$$
$$s = 1.2\times10^{-7}\ \text{mol/L}$$

Answer:

The solubility of $\ce{Cd3(PO4)2}$ in water is $1.2\times 10^{-7}\ \text{mol/L}$.