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.
(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.
I calculate 4 drops, so I think I may be doing something wrong.
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 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.
To find the number of drops of 0.20M KI needed to form a precipitate of lead iodide, we need to calculate the maximum amount of PbI2 that can be formed and compare it to the amount of PbI2 that can be formed from the given concentration of Pb(NO3)2.
Let's start by calculating the maximum amount of PbI2 that can be formed. The Ksp expression for PbI2 is:
Ksp = [Pb2+][I-]^2
Substituting the concentrations:
7.1e-9 = [Pb2+][I-]^2
Since the concentration of Pb2+ is given and is 0.010M, we can solve for the concentration of I-:
[I-]^2 = Ksp / [Pb2+]
[I-]^2 = (7.1e-9) / (0.010M)
Taking the square root:
[I-] = √[(7.1e-9) / (0.010M)]
[I-] = 8.43e-6 M
Now, let's calculate the number of moles of I- needed to form PbI2 in the given volume of Pb(NO3)2 solution:
moles of I- = concentration of I- x volume of solution
moles of I- = (8.43e-6 M) x (100.0 mL / 1000 mL/mL)
moles of I- = 8.43e-7 moles
Since 1 drop is equal to 0.05 mL, we can calculate the number of drops of KI needed:
number of drops = (moles of I-) / (moles of KI in 1 drop)
number of drops = (8.43e-7 moles) / (0.20 M x 0.05 mL)
number of drops = 8.43e-7 / 0.01
number of drops = 8.43e-5 drops
Rounding to the nearest whole number, the answer is approximately 9 drops.