Question

what is the solubility of pbf₂ in water? (ksp of pbf₂ is 3.6 × 10⁻⁸)

Explanation:

Step1: Write the dissolution equation

$$\mathrm{PbF}_{2}(s) ightleftharpoons\mathrm{Pb}^{2 +}(aq)+2\mathrm{F}^{-}(aq)$$

Let the solubility of $\mathrm{PbF}_{2}$ be $s$. Then $[\mathrm{Pb}^{2 +}]=s$ and $[\mathrm{F}^{-}]=2s$.

Step2: Write the solubility - product expression

$$K_{sp}=[\mathrm{Pb}^{2 +}][\mathrm{F}^{-}]^{2}$$
Substitute $[\mathrm{Pb}^{2 +}]=s$ and $[\mathrm{F}^{-}]=2s$ into the $K_{sp}$ expression:
$$K_{sp}=s\times(2s)^{2}=4s^{3}$$

Step3: Solve for $s$

Given $K_{sp}=3.6\times 10^{-8}$, then
$$4s^{3}=3.6\times 10^{-8}$$
$$s^{3}=\frac{3.6\times 10^{-8}}{4}=9\times 10^{-9}$$
$$s=\sqrt[3]{9\times 10^{-9}}$$
$$s = 2.1\times10^{-3}\space mol/L$$

Answer:

The solubility of $\mathrm{PbF}_{2}$ in water is $2.1\times 10^{-3}\space mol/L$.