# Concentration of dissolved lead(II) bromide

Calculate the concentration of $\ce{Pb^2+}$ and $\ce{Br-}$ when $\ce{PbBr2}$ dissolves in water. $K = 7.9 \times 10^{-5}$

$$\ce{PbBr2 <=> Pb^2+ + 2Br-}$$

In most cases we have the concentration of the solute, but here we don’t. How should I be thinking now?

if it’s $1~\mathrm{M}$ for example

$$K = \frac{[x][x]^2}{[1-x]}$$

The given value $K=7.9\times10^{-5}$ actually is the value of the solubility product constant $K_\mathrm{sp}$ of $\ce{PbBr2}$

$$K_\mathrm{sp}=a_{\ce{Pb^2+}}\cdot a_{\ce{Br-}}^2=7.9\times10^{-5}$$

which approximately is (when incorrectly neglecting activity coefficients $\gamma$, the difference of molality $b$ and concentration $c$, and the correct units)

$$K_\mathrm{sp}\approx \left[\ce{Pb^2+}\right] \cdot \left[\ce{Br-}\right]^2$$

When $\ce{PbBr2}$ dissolves in water according to the given chemical equation, the resulting concentrations of $\ce{Pb^2+}$ and $\ce{Br-}$ correspond to

$$\left[\ce{Br-}\right]=2\cdot\left[\ce{Pb^2+}\right]$$

These equations can be used for calculating a value for $\left[\ce{Pb^2+}\right]$.