posted by Cody .
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=7.1e-9.
Answer=9 drops. Please show steps.
I calculate 4 drops, so I think I may be doing something wrong.
(Pb^2+)(I^-)^2 = 7.1E-9
(0.01)(I^-)^2 = 7.1E-9
(I^-) = about 8.4E-4M
mols in 100 mL = M x L = about 8E-5 mols.
How many mols do you have in a drop. That's M x L = 0.2M x 5E-5 = 1E-5
Then (1E-5 mols/drop) x #drops = 8.4E-5
# 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
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 10-7)=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 = 4.21E-4M you must evaluate 2x. So 4.21E-4 x 2 = 8.42E-4 whereas I had 8.43E-4 with a lot less algebra. :-).
Wow, I can't believe I did that.