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)

Answer 1

#PbBr_2(s)rightleftharpoons Pb^(2+) + 2Br^-#

#K_(sp) = [Pb^(2+)][Br^-]^2# #=# #??#

The quantity of undissolved lead bromide is insignificant in this equilibrium reaction, and as a result, it is not included in the equilibrium expression.

We are given that at equilibrium, #[Pb^(2+)] = 0.012# #mol*L^(-1)#. By the stoichiometry of the reaction, #[Br^(-)] = 0.024# #mol*L^(-1)#, because for each lead iron in solution, two bromide ions go up.

So, #K_(sp) = [Pb^(2+)][Br^-]^2# #=# #[0.012][0.024]^2 = ??#

Would you expect #K_(sp)# 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?

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Answer 2

Olivia Anton

The equilibrium constant ( K_{\text{eq}} ) for the reaction is ( 2.5 \times 10^{-6} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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) --&gt; Pb2+(aq) + 2Br-(aq)

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)