Chemistry
posted by Cody on .
How many drops (1 drop=0.05mL) of 0.20M KI must we add to 100.0 mL of 0.010M Pb(NO3)2 to get a precipitate of lead iodide to start. Ksp of PbI2=7.1e9.
Answer=9 drops. Please show steps.

I calculate 4 drops, so I think I may be doing something wrong.

(Pb^2+)(I^)^2 = 7.1E9
(0.01)(I^)^2 = 7.1E9
(I^) = about 8.4E4M
mols in 100 mL = M x L = about 8E5 mols.
How many mols do you have in a drop. That's M x L = 0.2M x 5E5 = 1E5
Then (1E5 mols/drop) x #drops = 8.4E5
# drops = 8.5; therefore, you mut round that to 9 drops. 
PbI2> Pb + 2I
*****We differ at the setup, why did you not include the 2?
Ksp=7.1 x 10^9=products/reactants=[0.01M][2x]^2
Equation becomes,
7.1 x 10^7=4x^2
Solving for x,
7.1 x 10^7/4=x^2
1.78 x 10^7=x^2
sqrt*(1.78 x 107)=x
x= 4.21 x 10^4 M=I concentration
4.21 x 10^4 M *0.1 L=4.21 x10^5 moles of I
4.21 x10^5 moles of I/0.2 M=2.11 x 10^4 L
2.11 x 10^4 L*(10^3 mL/1L)=0.211 mL
0.211 mL/0.05 mL/drop=about 4 drops 
I chose to call I^ simply x because that simplifies a lot of stuff. However, you may call it anything you wish BUT you must follow through with it. If you choose to call it 2x, then when you solve for x = 4.21E4M you must evaluate 2x. So 4.21E4 x 2 = 8.42E4 whereas I had 8.43E4 with a lot less algebra. :).

Wow, I can't believe I did that.