# What is the molar solubility of PbBr_2 in pure water?

Dec 10, 2015

You need a value for the solubility product, ${K}_{s p}$ at a given temperature.

#### Explanation:

The interwebs give a value for the solubility product of lead bromide as ${K}_{s p}$ $=$ $1.86 \times {10}^{- 5}$ at 20""^@C. This is an equilibrium constant for the reaction:

$P b B {r}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s P {b}^{2 +} + 2 B {r}^{-}$

And so ${K}_{s p}$ $=$ $1.86 \times {10}^{- 5}$ $=$ $\left[P {b}^{2 +}\right] {\left[B {r}^{-}\right]}^{2}$. We cannot speak of the concentration of a solid, which is why $P b B {r}_{2}$ does not appear in the expression. If we were to call the solubility of lead bromide $S$, then ${K}_{s p} = S \times {\left(2 S\right)}^{2}$ $=$ $4 {S}^{3}$.

So $S$ $=$ $\sqrt[3]{\frac{1.86 \times {10}^{- 5}}{4}}$. The answer has units $m o l \cdot {L}^{-} 1$ as required. Please do not trust my arithmetic.

Would you expect ${K}_{s p}$ to increase or decrease at elevated temperature? Why or why not??

If we were to make $\left[B {r}^{-}\right]$ artificially high (say by adding quantities of $N a B r$), would the solubility of the lead salt decrease or increase; why or why not?