20.00mL of a 1.100E-4 mol/L Pb(NO3)2 is mixed with 80.00mL of 4.450E-2 mol/L CaI2. Will a precipitate form?

Question

Here's what I did...
[Pb2+]
c=n/v
= (1.1E-4 mol/L x 0.02L)/(0.02L + 0.08L)
= 2.2E-5 mol/L

[I-]
c=n/v
= (4.45E-2 mol/L x 0.08L)/0.1L
= 3.56E-2 mol/L

Qsp = [Pb2+][I-]
= (2.2E-5)(3.56E-2)^2
= 2.7E-8

Ksp of PbI2 is 9.8E-9

Since Qsp>Ksp, a precipitate will form.

According to the book, Qsp is 3.0E-9; therefore a precipitate will NOT form... What did I do wrong?

Answer 1

I forgot to write ^2 on the [I-] for Qsp. Just a typo though, I did square it when I punched it in my calculator.

Answer 2

Yes you did.
One error.
1. You didn't multiply the final I^- concn by 2 since there are two I^- per molecule CaI2.

Thanks for showing your work.