Can someone show and explain to me step by step how to do the problems?

Question

I feel like I am doing it incorrectly.

a) The solubility of PbI2 is 3.67 mg per mL. What is the Ksp?

b) Three drops of 0.20 M potassium iodide are added to 100.0 mL of 0.010 M lead nitrate. Will a precipitate form? (Assume 1 drop = 0.05 mL)

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For part a, I converted..
3.67 mg x (1 g/1,000 mg) x (1 mol/461.01 g PbI2) = 7.96x10^-6 M PbI2
Then the Ksp = [Pb^2+][I^-]^2
= (3.98x10^-6)(7.96x10^-6)^2
= 2.52x10^-16

For part b, I did...
3 drops x (0.05 mL/1 drop) = 0.15 mL

(1.5x10^-4 L x 0.20 M KI2)/1.6 L = 0.1875 M KI2
(0.1 L x 0.010 M Pb(NO3)2)/1.6 L = 6.2x10^-4 M Pb(NO3)2

Q = [KNO3]^2 / ([Pb(NO3)]^2[Ki]^2)

But I may not be going the right way in solving it..

Answer 1

a. 3.67 mg/mL = 3.67 g/L
3.67 x (1 mol/461.01) = 7.96E-3 = (PbI2)
..........PbI2 --> Pb^2+ + 2I^-
I.........solid....0........0
C.........solid....x........2x
E.........solid....x........x

Ksp PbI2 = (Pb^2+)(I^-)^2
Ksp = (7.96E-3)(2*7.96E-3)^2 = ?
NOTE: (I^-) = twice (Pb^2+)

b)
3 x 0.05 = 0.15 mL = 1.5E-4 L and that + 0.1 L (100 ml) is still just 0.1L.
So the original KI that was 0.2M is not 0.2 x (1.5E-4/0.1) = 03.00E-4M
Then Q = (Pb^2+)(I^-)^2
(Pb^2+) = 0.01M
(I^-) = 3E-4M
Q = (0.01)(3E-4)^2 = 9E-10 which is smaller than the Ksp calculated in part a.