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?
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?
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.
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.
Your calculations are correct. However, there seems to be an error in the given value of Ksp for PbI2. The book states that the Ksp is 9.8E-9, but your calculated Qsp is 2.7E-8. Since Qsp is greater than Ksp, a precipitate will form.
It is possible that there was a typographical error in the book or that you misread or misinterpreted the required values. Double-check the value of Ksp for PbI2 provided in the question or consult a reliable source to confirm the accurate value.