Excess PbBr2(s) is placed in water at 25 C. At equilibrium, the solution contains 0.012 M Pb 2+(aq). What is the equilibrium constant for the following reaction? PbBr2(s) --> Pb2+(aq) + 2Br-(aq)

Nov 24, 2015

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

${K}_{s p} = \left[P {b}^{2 +}\right] {\left[B {r}^{-}\right]}^{2}$ $=$ ??

Explanation:

This is just another equilibrium reaction in which the amount of undissolved lead bromide is irrelevant, and, therefore, does not appear in the equilibrium expression.

We are given that at equilibrium, $\left[P {b}^{2 +}\right] = 0.012$ $m o l \cdot {L}^{- 1}$. By the stoichiometry of the reaction, $\left[B {r}^{-}\right] = 0.024$ $m o l \cdot {L}^{- 1}$, because for each lead iron in solution, two bromide ions go up.

So, ${K}_{s p} = \left[P {b}^{2 +}\right] {\left[B {r}^{-}\right]}^{2}$ $=$ [0.012][0.024]^2 = ??

Would you expect ${K}_{s p}$ to increase or decrease at higher temperature? Why? Also, if the solution already had bromide ion present, would you expect solubility to go up or down?