Rebecca S. answered • 07/18/23
20+ years teaching AP, Honors, and General Chemistry
Answer to first question:
The 𝐾sp of PbBr2 is 6.60×10−6. What is the molar solubility of PbBr2 in pure water?
Follow these steps to solve the problem.
1) Write the equilibrium constant expression for the reaction of PbBr2 dissolving in water.
PbBr2 --> Pb+2 + 2 Br-1
Ksp = [Pb+2+][Br-1]2
NOTE: since PbBr2 is a solid it does not get written in the Ksp expression.
2) Set up an ICE chart to help organize your work.
PbBr2 --> Pb+2 + 2 Br-1
I --- 0 0
C --- +x +2x
E --- x 2x
3) Plug the equiibrium concentrations (E on the ICE chart) into the expression written in step 1.
Ksp = [Pb+2+][Br-1]2
6.60x10-6 = (x)(2x)2
4) Solve for x. "x" is the molar solubility of PbBr2.
6.60x10-6 = 4x3
÷ ÷
4 4
1.65x10-6 = x3
3√1.65x10-6 = x
x = 0.0118 M
Answer to second question:
What is the molar solubility of PbBr2 in 0.500 M KBr solution?
To answer this question, you will use the same steps laid out above except the concentration of the Br- is different in the ICE chart. Adding KBr to the solution increases the amount of Br- ion. The addition of the Br- will cause the equilibrium to shift to the reactants which decreases the conctration of Pb+2. Therefore, the concentration of Pb+2 ,x, is less than the answer to the first question.
Set up the ICE chart:
PbBr2 --> Pb+2 + 2 Br-1
I --- 0 0.500
C --- +x +2x
E --- x 0.500 NOTE: the amount of Br- contributed from the PbBr2 when it dissolves is going to be much smaller than the 0.500 M added due to the addition of KBr. Therefore, the 2x from the C in the ICE chart can be considered negligable and not included in the final amount!!!!! This makes our math much easier.
Ksp = [Pb+2+][Br-1]2
6.60x10-6 = (x)(0.500)2
x = 2.64x10-5 M
Answer to third question:
What is the molar solubility of PbBr2 in a 0.500 M Pb(NO3)2 solution?
This question is done the same way as the second question except in the ICE chart 0.500 will be substituted for the "x" for Pb+2. Again the x can be considered negligable.
