I look at it this way.
Assume we add 100 mL of each.
So we add 0.2 millimole Pb^+ to 0.4 millimole I^- which is the exact ratios for PbI2 to be pptd with no excess of either reagent. Therefore, the final concn of Pb^+ and I^- depends upon the Ksp.
Set up Ksp and solve for solubility PbI2.
In an experiment, a student mixes equal volumes of 0.0020M Pb 2+ ions with 0.0040M I- ions. The trial ion product is?
PbI2 <-> Pb 2+ + 2I-
How do you determine the final concentration of Pb 2+ and I-?
4 answers
The question asks for the ksp, so don't I try to find the final concentration first? How do you take 100mL of each molar concentration?
Is it,
Final [Pb 2+]=0.0020M x (100ml/200ml)=0.0010M Pb 2+
Final [I-]=0.0040M x (100ml/200mL)=0.0020M I-
Ksp=(0.0010)(0.0020)^2
Ksp=4.0x10^-9
Final [Pb 2+]=0.0020M x (100ml/200ml)=0.0010M Pb 2+
Final [I-]=0.0040M x (100ml/200mL)=0.0020M I-
Ksp=(0.0010)(0.0020)^2
Ksp=4.0x10^-9
No. Excuse me. I misread the question. The question asks for the trial ion product. That is NOT Ksp, at least not in my book. Some books refer to this as Qsp. That is (Pb^+2)(I^-)^2 = ??
(0.002)(0.004)^2 = ??
The idea here is to see if Qsp exceeds Ksp. If it does, then PbI2 will ppt, it not, it will not ppt.
(0.002)(0.004)^2 = ??
The idea here is to see if Qsp exceeds Ksp. If it does, then PbI2 will ppt, it not, it will not ppt.
5 answers