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-?

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

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.

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.

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?

Oh, the student must be having a "mixing" time in the lab! To determine the final concentration of Pb 2+ and I-, we first need to take a look at what's happening in the reaction. The equation shows that one Pb 2+ ion reacts with two I- ions to form one molecule of PbI2.

Now, since the student mixed equal volumes of the two solutions, we can assume that the initial concentrations of the Pb 2+ ions and I- ions are equal. So, we'll use the initial concentration of 0.0020 M (let's call it x) for both Pb 2+ and I-.

During the reaction, one Pb 2+ ion reacts with two I- ions, which means the concentration of Pb 2+ will decrease by x, and the concentration of I- ions will decrease by 2x.

Now, if we want to find the trial ion product, we multiply the final concentrations of Pb 2+ and I-. Since the concentration of Pb 2+ decreases by x and I- decreases by 2x, the final concentration of Pb 2+ would be x - x = 0, and the final concentration of I- would be x - 2x = -x.

But hey, negative concentrations don't exist in the real world! So, we can conclude that the trial ion product is just 0. Ha! No need to worry about juggling negative concentrations here!

To determine the final concentration of Pb2+ and I-, we need to understand the concept of stoichiometry and how it relates to chemical reactions. In this case, the reaction is PbI2 <-> Pb2+ + 2I-.

The initial concentrations of Pb2+ and I- are given as 0.0020 M and 0.0040 M, respectively. Since equal volumes of solutions are mixed, the two concentrations remain the same after mixing.

To find the final concentration of Pb2+ and I-, we need to consider the reaction stoichiometry. According to the stoichiometry, for every 1 mole of Pb2+ ion, 1 mole of PbI2 is formed, and for every 2 moles of I- ions, 1 mole of PbI2 is formed.

Since the initial concentrations of Pb2+ and I- are equal, we can assume there is initially 1 mole of each ion. When they react, 1 mole of Pb2+ combines with 2 moles of I- to form 1 mole of PbI2. This means that the available number of moles of each ion will be halved.

Therefore, the final concentration of Pb2+ will be 0.0020 M / 2 = 0.0010 M, and the final concentration of I- will be 0.0040 M / 2 = 0.0020 M.

Now, to find the trial ion product, we multiply the final concentrations of Pb2+ and I- together, taking into account their stoichiometric coefficients.

The trial ion product is given by the expression [Pb2+] * [I-]^2, so substituting the values we calculated:

Trial ion product = (0.0010 M) * (0.0020 M)^2
Trial ion product = 4.0 x 10^-9 M^3