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
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.
Looks like you're having a hard time getting the answer, but don't worry, I'm here to help you out! It seems like you're on the right track with your ICE diagram and the Ksp expression. However, I have to point out a small error in your calculation. When you substitute the values into the Ksp expression, it should be:
1.72 x 10^-5 = (x) * (0.023)^2
This will give you the correct value for x, which represents the Pb2+ concentration. Give it another try using the correct equation, and I'm sure you'll get the right answer! And remember, if you still need help, I'm always here to lend a funny hand.
To solve this problem, you correctly set up the initial, change, and equilibrium (ICE) table. However, your calculations are slightly incorrect.
The equilibrium expression for the dissolution of PbCl2 is given by:
Ksp = [Pb2+][Cl-]^2
Since the concentration of Cl- is given as 0.023 M, you can substitute that value into the equation:
1.72x10^-5 = [Pb2+](0.023)^2
Rearranging the equation, you get:
[Pb2+] = (1.72x10^-5) / (0.023)^2
Calculating this expression, you find:
[Pb2+] = 0.026 M
So, the concentration of Pb2+ in a saturated solution of PbCl2 is 0.026 M.
To solve this problem, you correctly set up the ICE table based on the balanced equation for the dissociation of PbCl2. However, at the end of your calculation, you made a small mistake in substituting the concentration of chloride ion back into the equation.
Let's correct the solution step-by-step:
1. Write down the balanced chemical equation for the dissociation of PbCl2:
PbCl2 (s) ⇌ Pb2+ (aq) + 2 Cl- (aq)
2. Set up an ICE table:
PbCl2 (s) ⇌ Pb2+ (aq) + 2 Cl- (aq)
I: 0 M 0 M 0.023 M
C: +x M +x M +2x M
E: x M x M 0.023 M
3. Substitute the equilibrium concentrations into the Ksp expression:
Ksp = [Pb2+][Cl-]²
Ksp = (x)(0.023)²
4. Solve for x:
(1.72 × 10^-5) = x(0.023)²
x = (1.72 × 10^-5) / (0.023)²
x ≈ 3.28 × 10^-4 M
So, the concentration of Pb2+ in the saturated solution of PbCl2 is approximately 3.28 × 10^-4 M.