Will a precipitate form if I do this

[Dixie]

I'm working on a chem problem for my exam tommorow. Can anyone help me?
Q1: Will a precipitate form if 45cm3 of 0.0035M NaI is added to 95cm3 of 0.0013M Pb(NO3)2? The Ksp of PbI2 is 7.08 x 10^-9.

My teacher says the answer's no precipitate and the trial Ksp is 8.8 x10^-10. But I got 1.1 x 10^-9. Can someone show me how to do this?

 

[GCT]

try the chemistry sub forum located here at Physicsforum

 

[thepatientmental]

A: Based on the given information, it is possible to calculate the expected concentration of PbI2 in solution and compare it to the Ksp value to determine if a precipitate will form. The equation for Ksp is [Pb2+][I-]^2, where [Pb2+] and [I-] represent the concentrations of Pb2+ and I- ions in solution, respectively.

First, we need to calculate the moles of each ion present in the solution. For NaI, we have 0.0035 mol/L x 0.045 L = 0.0001575 mol of I- ions. For Pb(NO3)2, we have 0.0013 mol/L x 0.095 L = 0.0001235 mol of Pb2+ ions.

Next, we need to determine the expected concentration of PbI2 in solution. Since the reaction is 1:2 (1 mol of Pb2+ reacts with 2 mol of I- to form 1 mol of PbI2), the concentration of PbI2 will be half of the concentration of Pb2+ ions. Therefore, the expected concentration of PbI2 is 0.0001235 mol/L.

Comparing this to the given Ksp value of 7.08 x 10^-9, we can see that the expected concentration of PbI2 is higher than the Ksp value. This means that a precipitate will form, and the teacher's answer of no precipitate is incorrect.

To calculate the trial Ksp value, we can use the expected concentration of PbI2 (0.0001235 mol/L) and the actual volume of the solution (0.045 L + 0.095 L = 0.14 L). This gives us a trial Ksp value of 8.8 x 10^-9, which is close to your calculated value of 1.1 x 10^-9.

In conclusion, a precipitate will form in this solution and your calculated value for Ksp is correct. It is important to double check your calculations and make sure you are using the correct values for the volume and concentrations of the ions in solution. Good luck on your exam!