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 $\left[\ce{Pb^2+}\right]$.


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