Two 50mL beaker. One contains 50mL of 1.0M PbSO4 and the other contains 10mL of 0.1M Pb(NO3)2 added with 40mL of .10M KI. What is the Ksp of PbI2?

Ksp=[Pb][I]^2

[Pb]=.10M*(10ml/(40mL+10mL+50mL)=.01M
[I]=.10M*(40/(40mL+10mL+50mL))=.04M
Ksp=1.6E-05
This is what i got

PbSO4 is insoluble. How do you make 1M PbSO4?

there should be an E(V) too. let suppose your E(V)=.128.. your calculation should somewhat look like this if I'm not wrong,

Ecell=E^o cell - (RT/nF)lnQ
.125=0-(.0592/2)lnQ
e^(-.128/.0296)=[I]/.1M
[I]=.001324
[Pb]= [I]x(1/2)=.000662

ksp=[Pb][I^2]=1.16E-09

To determine the Ksp (solubility product constant) of PbI2, we need to understand the reaction that occurs between Pb(NO3)2 and KI.

The balanced equation for the reaction is as follows:

Pb(NO3)2 + 2KI -> PbI2 + 2KNO3

The reaction involves the formation of PbI2, which is the substance of interest. It is important to note that PbI2 is an ionic compound that dissociates into its constituent ions when dissolved in water.

Given that one beaker contains 50 mL of 1.0M PbSO4 and the other contains 10 mL of 0.1M Pb(NO3)2 mixed with 40 mL of 0.10M KI, we need to determine the concentration of Pb2+ and I- ions in the solution.

For the first beaker containing 50 mL of 1.0M PbSO4, we can assume that all the PbSO4 dissociates to release one mole of Pb2+ ions. Therefore, the concentration of Pb2+ ions in this beaker is 1.0M.

For the second beaker, we need to determine the concentrations of Pb2+ and I- ions after the reaction.

The initial concentration of Pb2+ ions from the 10 mL of 0.1M Pb(NO3)2 is (0.1M)(10mL / 50mL) = 0.02M.

The initial concentration of I- ions from the 40 mL of 0.10M KI is (0.10M)(40mL / 50mL) = 0.08M.

According to the balanced equation, the stoichiometry of the reaction is 1:1 between Pb(NO3)2 and PbI2. This means that for each mole of Pb(NO3)2, one mole of PbI2 is formed. Therefore, the concentration of Pb2+ and I- ions will be the same after the reaction.

Hence, the final concentration of Pb2+ and I- ions in the second beaker is 0.02M and 0.08M, respectively.

Now, we can calculate the Ksp of PbI2 using the equation:

Ksp = [Pb2+][I-]^2

Substituting the concentrations of Pb2+ and I- ions:

Ksp = (0.02M)(0.08M)^2 = 0.000032M^3

Therefore, the Ksp of PbI2 is 0.000032M^3.