A solution contains 0.0330 M Pb2+and 0.0210 M Ag+. Can 99% of Pb2+be precipitated by chromate(CrO42-), without precipitating Ag+? What will the concentration of Pb2+be, when Ag2CrO4begins to precipitate?
Remember that I'm using values for Ksp that I obtained from the web. Be sure to use the values in your text or notes from your class.
Two ions in solution. 0.0330 M Pb2+and 0.0210 M Ag+. Write the reaction with CrO4^2- and solve for (CrO4^2-) at equilibrium; i.e., just when one more drop of CrO4^- will ppt PbCrO4 or Ag2CrO4.
....................Pb^2+ + CrO4^2- ==> PbCrO4
Ksp = 3E-13 = (Pb^2+)(CrO4^2-)
(CrO4^2-) = 3E-13/0.0330 = 9.09E-12 M
Same for Ag2CrO4
Ksp = 1.12E-12 = (Ag^+)^2(CrO4^2-)
(CrO4^2-) = 1.12E-12/(0.0210)^2 = 2.54E-9 M
These two values tell us that if CrO4^2- is dripped slowly into the solution that PbCrO4 will start pptng first. It will continue a bit before Ag2CrO4 starts to ppt. Can 99% of the Pb^2+ be pptd before Ag2CrO4 starts? We can calculate that. Pb^2+ = 0.0330 x 0.01 = 0.000330 Pb^+ remaining if 99% has been ppt. What will the CrO4^2- be at that point?
(CrO4^2-) = 3E-13/0.000330 = 9.09E-10 so PbCrO4 will start pptng first and it will continue pptng until (CrO4^2-) reaches 2,54E-9 M (calculated above) Since 9.09E-10 is smaller than 2.54E-9 then yes, we can ppt 99% of the PbCrO4 before the Ag2CrO4 starts contaminating it.
Next question. What will be (Pb^2+) when Ag2CrO4 first starts to ppt?
First, remember that the (CrO4^2-) = 5.4E-9 M when Ag2CrO4 first starts. Plug this into the Ksp for PbCrO4 and solve for Pb^2+.
(Pb^2+) = Ksp/(CrO4^-) = 3E-13/2.54E-9 = 1.18E-4 M which is rather small. Calculations like this tell us that a mixture of Pb^2+ and Ag^+ can be completely separated with Na2CrO4 and determined independently of each other. Post your work if you get stuck.
Thank you so much for your help.
To determine if 99% of Pb2+ can be precipitated by chromate (CrO42-) without precipitating Ag+, we need to compare the solubility product constants (Ksp) of the respective precipitates.
The solubility product constant for lead chromate (PbCrO4) is given by:
Ksp (PbCrO4) = [Pb2+][CrO42-]
The solubility product constant for silver chromate (Ag2CrO4) is given by:
Ksp (Ag2CrO4) = [Ag+]^2[CrO42-]
Given that the concentration of Pb2+ is 0.0330 M and the concentration of Ag+ is 0.0210 M, we can substitute these values into the solubility product equations.
For PbCrO4 precipitation:
Ksp (PbCrO4) = (0.0330)([CrO42-]) [1]
For Ag2CrO4 precipitation:
Ksp (Ag2CrO4) = (0.0210)^2([CrO42-]) [2]
To determine the concentration of Pb2+ when Ag2CrO4 begins to precipitate, we can set the Ksp values [1] and [2] equal to each other since neither compound should be fully precipitated:
(0.0330)([CrO42-]) = (0.0210)^2([CrO42-])
Simplifying the equation:
0.0330 = (0.0210)^2
Taking the square root of both sides:
0.182 = 0.0210
Therefore, when Ag2CrO4 begins to precipitate, the concentration of Pb2+ will be 0.182 M.
To determine if 99% of Pb2+ can be precipitated by chromate (CrO42-) without precipitating Ag+, we need to compare the solubility products (Ksp) of the two compounds involved.
First, we'll find the Ksp of PbCrO4 (lead chromate) and Ag2CrO4 (silver chromate):
1. PbCrO4:
- The balanced chemical equation for the precipitation reaction is:
Pb2+(aq) + CrO42-(aq) -> PbCrO4(s)
- The solubility product expression for PbCrO4 is:
Ksp(PbCrO4) = [Pb2+][CrO42-]
- Assuming the solubility in pure water is negligible, Ksp is the equilibrium constant.
2. Ag2CrO4:
- The balanced chemical equation for the precipitation reaction is:
2Ag+(aq) + CrO42-(aq) -> Ag2CrO4(s)
- The solubility product expression for Ag2CrO4 is:
Ksp(Ag2CrO4) = [Ag+]^2[CrO42-]
- Assuming the solubility in pure water is negligible, Ksp is the equilibrium constant.
Now, let's compare the Ksp values:
Ksp(PbCrO4) = [Pb2+][CrO42-]
Ksp(Ag2CrO4) = [Ag+]^2[CrO42-]
Given in the problem:
[Pb2+] = 0.0330 M
[Ag+] = 0.0210 M
To determine the concentration of Pb2+ when Ag2CrO4 begins to precipitate, we need to find the point where the concentration of Pb2+ exceeds the solubility product (Ksp) of Ag2CrO4. At this point, the Ag2CrO4 will start precipitating.
Let's assume x M of Pb2+ forms PbCrO4 and precipitates. Thus, the concentration of Pb2+ will decrease by x M.
The concentration of CrO42- will increase by x M (assuming complete reaction).
Substituting the given values and the assumed values into the Ksp expressions:
For PbCrO4:
Ksp(PbCrO4) = (0.0330 - x)(x) = 1.7 x 10^-12 (the Ksp value for PbCrO4)
For Ag2CrO4:
Ksp(Ag2CrO4) = (0.0210)^2(0.0330 - x) = 9 x 10^-12 (the Ksp value for Ag2CrO4)
To solve for x and find the concentration of Pb2+ when Ag2CrO4 begins to precipitate, we will equate the two Ksp expressions:
(0.0330 - x)(x) = (0.0210)^2(0.0330 - x)
Solving this equation will give us the concentration of Pb2+ when Ag2CrO4 begins to precipitate.