Asked by Lynn
What is the Pb2+ concentration in a saturated solution of PbCl2 if the concentration of chloride ion is 0.023 M? Units: M ksp = 1.72x10^-5
i can't seem to get the answer. i have tried and ICE diagram and going from there:
PbCl2 (s) <--> Pb ^2+ = 2Cl-
I: 0.023 M = Cl-
C: Pb2+ = +X
E: Pb2+ = x, 2Cl- = 0.023
k = [Pb2+]*[cl-]^2
1.72E-5 = x * 0.023
Where x = (1.72E-5/0.023) = 0.0325
Please help
i can't seem to get the answer. i have tried and ICE diagram and going from there:
PbCl2 (s) <--> Pb ^2+ = 2Cl-
I: 0.023 M = Cl-
C: Pb2+ = +X
E: Pb2+ = x, 2Cl- = 0.023
k = [Pb2+]*[cl-]^2
1.72E-5 = x * 0.023
Where x = (1.72E-5/0.023) = 0.0325
Please help
Answers
Answered by
DrBob222
From your answer above, you have
k = [Pb2+]*[cl-]^2
k you have.
(Pb^2+) is x as you have it.
(Cl^-) is 0.023 from the problem BUT you didn't square it. Do that and solve for x.
Note: I don't think it will make much difference BUT since the Ksp for PbCl2 is so large, a somewhat more accurate solution (also it gives you a quadratic equation which is a little harder to solve)
Ksp = (x)(0.023+x)^2 and solve for x. Many calculators can solve a quadratic; if you don't have one of those you can find sites on the web that will do it for you. Post your work if you get stuck.
k = [Pb2+]*[cl-]^2
k you have.
(Pb^2+) is x as you have it.
(Cl^-) is 0.023 from the problem BUT you didn't square it. Do that and solve for x.
Note: I don't think it will make much difference BUT since the Ksp for PbCl2 is so large, a somewhat more accurate solution (also it gives you a quadratic equation which is a little harder to solve)
Ksp = (x)(0.023+x)^2 and solve for x. Many calculators can solve a quadratic; if you don't have one of those you can find sites on the web that will do it for you. Post your work if you get stuck.
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