What is the solubility of PbI2 in 0.4 M KI given the solubility constant of PbI2 is 7.1 x 10-9
a. 3.5 x 10-3 M
b. 1.2 x 10-3 M
c. 4.4 x 10-8 M
Ksp = 7.1 x 10-9
Ksp = x * (2 * 0.4)
(7.1 * 10-9)/.8 = x
8.8 x 10-9 = x ------ But this is not one of the answers. Please help. I think the answer may be C. Am I correct? Thanks
7.1 x 10^-9 = x * (0.4M)^2
(7.1 x 10^-9)/.16 = x
4.4 x 10^-8 = x
is this the correct way to solve this problem?
Thx Chris u da goat
Well, it seems like you've done the calculations correctly! However, in order to verify your answer, let's take a humorous interlude:
Why did the clown bring a ladder to the bar?
Because he heard the drinks were on the house!
Now, back to business. Yes, you are correct! The answer to the question is indeed option c. The solubility of PbI2 in 0.4 M KI is approximately 4.4 x 10-8 M. Great job on your calculations!
To find the solubility of PbI2 in 0.4 M KI, we can use the solubility product constant (Ksp) equation and the given information.
The solubility product constant (Ksp) equation for PbI2 is:
PbI2 ↔ Pb2+ + 2I-
Ksp = [Pb2+][I-]^2
The equation states that the product of the concentrations of lead ions (Pb2+) and iodide ions (I-) raised to their stoichiometric coefficients is equal to the solubility product constant.
Since the stoichiometric coefficient of iodide ions is 2, and the concentration of KI is 0.4 M, we can substitute the values into the equation:
Ksp = x * (2 * 0.4)
Where x represents the solubility of PbI2.
Now, let's solve for x:
x = Ksp / (2 * 0.4)
x = (7.1 * 10^-9) / (2 * 0.4)
x ≈ 8.875 * 10^-9
Therefore, the solubility of PbI2 in 0.4 M KI is approximately 8.875 x 10^-9 M.
The answer provided as "C" (4.4 x 10^-8 M) does not match the calculated value. Please double-check the answer choices or recalculate the solution.