In the process of separating pb2+ ions from cu+2 ions as sparingly soluble iodates, what is Pb2+ concentration when Cu+2 begins to precipitate as sodium iodate is added to a solution that is initially 0.0010M Pb(NO3)2 (aq) and 0.0010M Cu(NO3)2 (aq)?
I know the answer is 1.8*10^-9 M. However I would like to know how to solve this problem.
Thank you!
What Ksp values are you using? The answer depends upon those value you use.
Use your numbers for this.
Adding NaIO3 dropwise to the solution will continue until the most insoluble material starts pptng. That will be Pb(IO3)2. Pb(IO3)2 will continue to ppt until the IO3^- rises high enough to begin pptn of Cu(IO3)2. At that point, what is the (IO3^-)? It is
Ksp = (Cu^2+)(IO3^-)^2.
Plug in 0.001 for Cu and your value for Ksp and solve for IO3^-. Then plug this IO3^- back into the Ksp expression for Pb(IO3)2 and solve for Pb^2+. Using the values I found on the web of 3.49E-13 for Pb(IO3)2 and 1.4E-7 for Cu(IO3)2 I found (Pb^2+) to be about 2.5E-9M which is close to your answer. Using your values for Ksp should give you the answer.
This solution actually gets you 1.8x10^-9 M!
So Ksp of Copper Iodate is 1.4x10^-7 and Ksp = [Cu^2+][IO3^-]^2. You rearrange to solve for [IO3-] by plugging in Ksp and 0.001 M as [Cu^2+]to get 0.0118 M.
Taking 0.0118 as the [IO3^-], you then focus on the concentration of Pb2+. The Ksp of Lead Iodate is 2.6x10^-13 and the Ksp = [Pb^2+][IO3^-]^2. Rearrange to solve for [Pb^2+] and plug in Ksp and 0.0118 as [IO3^-] (don't forget to square [IO3^-]!!!). The resulting concentration of the lead ion should be 1.8x10^-9 if you did everything correctly!
To solve this problem, you need to understand the concept of solubility product constant and use it to determine at what concentration the precipitate begins to form.
1. Determine the balanced chemical equation:
The reaction between lead(II) ions (Pb2+) and copper(II) ions (Cu2+) with sodium iodate (NaIO3) can be represented by the following equation:
Pb2+(aq) + IO3-(aq) → Pb(IO3)2(s)
2. Write the solubility product expression:
The solubility product constant (Ksp) expression for the formation of Pb(IO3)2 can be written as:
Ksp = [Pb2+][IO3-]^2
3. Calculate the initial concentrations:
The initial concentration of Pb2+ is given as 0.0010 M, and the initial concentration of Cu2+ is also 0.0010 M.
4. Assume x M concentration of Pb2+ precipitates:
As sodium iodate is added, Pb(IO3)2 will start precipitating, and the concentration of Pb2+ in the solution will decrease by x M.
5. Calculate the final concentrations:
The new concentration of Pb2+ after precipitation is (0.0010 - x) M, and the concentration of IO3- will be 2x M (since the stoichiometric ratio is 1:2 between Pb2+ and IO3-).
6. Determine the solubility product expression:
Rewrite the solubility product constant expression using the final concentrations:
Ksp = [(0.0010 - x)][(2x)^2] = [(0.0010 - x)][4x^2]
7. Set up an equation for x:
Since Cu(IO3)2 is sparingly soluble, its concentration does not change significantly upon the addition of sodium iodate. Therefore, the concentration of Cu2+ remains constant at 0.0010 M.
When Cu(IO3)2 begins to precipitate, the concentration of Pb2+ is equal to the solubility product constant expression. Therefore:
0.0010 - x = [(0.0010 - x)][4x^2]
8. Simplify the equation:
Divide both sides of the equation by (0.0010 - x):
1 = 4x^2
9. Solve for x:
Take the square root of both sides of the equation:
√(1/4) = √x^2
1/2 = x
10. Calculate the concentration:
Substitute the value of x back into the equation to find the concentration of Pb2+:
Pb2+ concentration = 0.0010 - x
= 0.0010 - (1/2)
= 0.0010 - 0.0005
= 0.0005 M
= 5.0 * 10^-4 M
Therefore, the Pb2+ concentration when Cu2+ begins to precipitate is 5.0 * 10^-4 M or 0.0005 M.
Note: The answer you provided (1.8 * 10^-9 M) may be incorrect. You may want to review your calculations or double-check the question to ensure accuracy.