A solution is 2.0×10−2M in both chromate and sulfate anions and . To this solution, 0.55M Pb(NO)3 is slowly added.

What is [Pb2+] at the point at which the second anion begins to precipitate (in M)? The second anion being the, sulfate anion.

We need to think our way through this problem.

............Pb^2+ + CrO42- ==>PbCrO4
initial......0........0.02.........
So when (Pb2+)(CrO42-) = 1.8E-14, PbCrO4 will start to ppt. Therefore, (Pb2+) = 1.8E-14/(0.02) = 9.0E-14 when PbCrO4 FIRST begins to ppt. When all of the PbCrO4 has pptd, the Pb2+ = 1.8E-14. We continue adding Pb(NO3)2 and the PbSO4 begins to ppt when (Pb2+)(SO42-) = 1.6E-8. Since sulfate = 0.02M, then (Pb2+) = (Ksp/0.02) = ?? when the first molecule of PbSO4 forms.
Check my thinking.

Yup that's correct! I was just the wrong Ksp value! Wasted an hour on this question just because of a silly mistake (N)

To determine the concentration of Pb2+ at the point when the second anion (sulfate) begins to precipitate, we need to calculate the solubility product constant (Ksp) for the formation of the PbSO4 precipitate.

The balanced chemical equation for the precipitation of PbSO4 is:

Pb(NO3)2(aq) + Na2SO4(aq) -> PbSO4(s) + 2NaNO3(aq)

The Ksp expression for PbSO4 is:

Ksp = [Pb2+][SO42-]

Given that the initial concentration of Pb2+ is 0 M, and the concentration of the chromate and sulfate anions is 2.0×10^−2 M, we can assume the initial concentration of SO42- is also 2.0×10^−2 M.

Let's assume x M represents the concentration of Pb2+ when the precipitation begins. Since PbSO4 is a 1:1 stoichiometric ratio, the concentration of SO42- will also become x M.

For the balanced equation, [Pb2+] = x M and [SO42-] = x M.

Now we can substitute these values into the Ksp expression and solve for x:

Ksp = [Pb2+][SO42-]
Ksp = (x)(x)
Ksp = x^2

Substituting the given Ksp value for PbSO4 (2.2 x 10^-8), we have:

2.2 x 10^-8 = x^2

Taking the square root of both sides:

x = sqrt(2.2 x 10^-8)
x ≈ 4.69 x 10^-5 M

So the concentration of Pb2+ at the point when the second anion (sulfate) begins to precipitate is approximately 4.69 x 10^-5 M.

To find the concentration of Pb2+ at the point where the second anion (sulfate) begins to precipitate, we need to determine the solubility product constant (Ksp) for the reaction between lead (II) sulfate (PbSO4) and water. The Ksp expression for this reaction is:

Ksp = [Pb2+][SO42-]

Given that the solution initially contains chromate and sulfate anions with a concentration of 2.0×10^−2 M, let's assume that the chromate anion does not precipitate. This means that initially, all the Pb2+ ions react with the sulfate anions to form lead sulfate.

The balanced equation for the reaction is:

Pb(NO3)2 + Na2SO4 → PbSO4 + 2NaNO3

From the balanced equation, we can see that the ratio between Pb(NO3)2 and PbSO4 is 1:1. Therefore, as the sulfate anion begins to precipitate, half of the Pb2+ ions would have reacted and precipitated as lead sulfate.

Using the initial concentration of Pb(NO3)2 (0.55 M) and the stoichiometry of the reaction, we can calculate the concentration of Pb2+ when half of it has reacted:

[Pb2+] = (Initial concentration of Pb2+) × (Fraction reacted)
= 0.55 M × 0.5
= 0.275 M

Therefore, the concentration of Pb2+ at the point where the sulfate anion starts to precipitate is 0.275 M.