Can someone show and explain to me step by step how to do the problems?
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..
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.
Let's go through each problem step by step and explain the calculations you've made.
a) The solubility of PbI2 is given as 3.67 mg per mL. You correctly converted this to molarity by dividing by the molar mass of PbI2, which is 461.01 g/mol. So, you have calculated the molar solubility (concentration) of PbI2 to be 7.96x10^-6 M.
The Ksp expression for PbI2 is [Pb^2+][I^-]^2. Plugging in the molar solubility value, you correctly calculated the Ksp as (3.98x10^-6)(7.96x10^-6)^2, which equals 2.52x10^-16.
Your calculations for part a are correct.
b) For part b, you correctly converted the number of drops (3 drops) to volume by multiplying it by the conversion factor (0.05 mL/1 drop), resulting in 0.15 mL.
Next, you calculated the concentration of KI (potassium iodide) in the solution by dividing the moles of KI (0.15 mL x 0.20 M KI) by the total volume of the solution (1.6 L). However, it seems there is a mistake in your calculation. It should be:
(0.15 L x 0.20 mol/L KI)/1.6 L = 0.01875 M KI
Similarly, you calculated the concentration of Pb(NO3)2 (lead nitrate) in the solution by dividing the moles of Pb(NO3)2 (0.1 L x 0.010 M Pb(NO3)2) by the total volume of the solution (1.6 L). However, there is a mistake in this calculation as well. It should be:
(0.1 L x 0.010 mol/L Pb(NO3)2)/1.6 L = 6.25x10^-4 M Pb(NO3)2
Therefore, the corrected concentrations are 0.01875 M KI and 6.25x10^-4 M Pb(NO3)2.
Now, to determine if a precipitate will form, you need to compare the reaction quotient (Q) with the Ksp. The Q expression for this reaction is [K^+][I^-]2 / [Pb^2+].
So, the corrected Q value is (0.01875)^2 / (6.25x10^-4). Simplifying this expression gives:
Q = 5.625 / 6.25x10^-4 = 9x10^3
To determine if a precipitate will form, compare the Q value to the Ksp value you calculated in part a. If Q > Ksp, a precipitate will form. If Q < Ksp, no precipitate will form.
In this case, Q (9x10^3) is much larger than Ksp (2.52x10^-16), which means a precipitate will form.
Let's go through each problem step by step and make sure the calculations are correct.
a) To find the Ksp, we need to convert the solubility of PbI2 from mg/mL to moles per liter (M). The molar mass of PbI2 is 461.01 g/mol.
Step 1: Convert the solubility from mg/mL to moles/L.
3.67 mg/mL x (1 g/1,000 mg) x (1 mol/461.01 g PbI2) = 7.96x10^-6 mol/L
Step 2: Calculate the Ksp using the formula: Ksp = [Pb^2+][I^-]^2
Ksp = (7.96x10^-6)^3 = 4.67 x 10^-16 (rounded to nearest significant figure)
So, the Ksp is approximately 4.67 x 10^-16.
b) To determine if a precipitate will form, we need to calculate the ion concentrations after the reaction.
Step 1: Convert the volume of potassium iodide from drops to liters.
3 drops x (0.05 mL/1 drop) = 0.15 mL = 0.150 L
Step 2: Calculate the ion concentrations.
KI: (0.150 L x 0.20 M)/0.100 L = 0.300 M
Pb(NO3)2: (0.100 L x 0.010 M) / 0.100 L = 0.010 M
Step 3: Calculate the ion product Q using the formula: Q = [KNO3]^2 / ([Pb(NO3)]^2[KI]^2)
Q = (0.300)^2 / (0.010^2) = 900
Since the value of Q is greater than the Ksp value calculated in part a (4.67 x 10^-16), a precipitate is expected to form.
Please let me know if there is anything else I can help you with.