A solution is prepared by mixing 50.0 mL of 0.17 M Pb(NO3)2 with 50.0 mL of 1.6 M KCl. Calculate the concentrations of Pb2+ and Cl - at equilibrium. Ksp for PbCl2(s) is 1.6X 10-5.
Oh, chemistry, my old friend. Let's waltz through this question, shall we?
First, we need to determine the number of moles of Pb(NO3)2 and KCl, as well as the volume of the final solution. Since volume is constant at 100 mL, no worries there. Let's calculate those moles.
For Pb(NO3)2: 0.17 M x 0.05 L = 0.0085 mol
For KCl: 1.6 M x 0.05 L = 0.08 mol
Now, since the stoichiometry of the reaction is 1:2 (one Pb to two Cl), the number of moles of Pb2+ and Cl- ions will differ. And here's where the fun starts.
For every mole of Pb(NO3)2, we get 2 moles of Pb2+ ions.
For every mole of KCl, we get 1 mole of Cl- ions.
So, the final concentration of Pb2+ is 2 x 0.0085 mol / 0.1 L = 0.17 M.
As for Cl-, it's slightly tricky because the reaction reaches equilibrium and some of the Cl- ions combine with the Pb2+ ions to form PbCl2(s). But fear not, Clown Bot is here!
For a moment, imagine Cl- ions are like a crowd at a comedy show. At first, they're all excited, buzzing in the solution. But as the show goes on, some of them leave the solution to go watch the show, leaving fewer Cl- ions in the liquid solution. Thus, the concentration of Cl- will be reduced.
To find the concentration of Cl- ions at equilibrium, we need to figure out how many moles of Cl- ions end up reacting with the Pb2+ ions to form PbCl2(s).
Using the stoichiometry, 1 mole of Pb(NO3)2 reacts with 2 moles of KCl, meaning 2 moles of Cl- are used up for every mole of Pb(NO3)2.
Since we started with 0.0085 moles of Pb(NO3)2, we'll use up 2 x 0.0085 = 0.017 moles of Cl- ions.
Now, considering the total volume remains constant (100 mL), the concentration of Cl- ions at equilibrium will be (0.08 - 0.017) / 0.1 L = 0.063 M.
That's it! We have the concentrations of Pb2+ and Cl- ions at equilibrium.
Remember, chemistry can be confusing, but with a dash of humor, it becomes a little less intimidating. Keep smiling and keep learning!
To calculate the concentrations of Pb2+ and Cl- at equilibrium, we need to consider the reaction between Pb(NO3)2 and KCl:
Pb(NO3)2(aq) + 2KCl(aq) → PbCl2(s) + 2KNO3(aq)
The balanced equation shows that each mole of Pb(NO3)2 produces one mole of PbCl2. Therefore, after mixing the solutions, the total volume of the solution is 100.0 mL (50.0 mL + 50.0 mL).
1. Calculate the moles of Pb(NO3)2:
moles of Pb(NO3)2 = concentration × volume
moles of Pb(NO3)2 = 0.17 M × 0.050 L
moles of Pb(NO3)2 = 0.0085 mol
2. Calculate the moles of KCl:
moles of KCl = concentration × volume
moles of KCl = 1.6 M × 0.050 L
moles of KCl = 0.080 mol
3. From the balanced equation, we know that 1 mole of Pb(NO3)2 produces 1 mole of PbCl2. Therefore, the moles of PbCl2 formed will also be 0.0085 mol.
4. Since PbCl2 is a sparingly soluble salt, it will dissociate partially in water according to its solubility product (Ksp).
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)
The equilibrium expression for Ksp is:
Ksp = [Pb2+][Cl-]^2
5. We can assume that the concentration of Pb2+ ions is equal to the moles of PbCl2 formed divided by the total volume of the solution:
[Pb2+] = moles of PbCl2 / total volume
[Pb2+] = 0.0085 mol / 0.100 L
[Pb2+] = 0.085 M
6. The concentration of Cl- ions is twice the concentration of Pb2+ ions (according to the balanced equation):
[Cl-] = 2[Pb2+]
[Cl-] = 2 × 0.085 M
[Cl-] = 0.17 M
Therefore, the concentrations of Pb2+ and Cl- ions at equilibrium are 0.085 M and 0.17 M, respectively.
To calculate the concentrations of Pb2+ and Cl- at equilibrium, we need to consider the reaction that occurs between Pb(NO3)2 and KCl:
Pb(NO3)2 + 2 KCl -> PbCl2 + 2 KNO3
First, let's determine the moles of Pb(NO3)2 and KCl in the solution. We can use the formula:
moles = (concentration) x (volume in liters)
For Pb(NO3)2:
moles of Pb(NO3)2 = (0.17 M) x (0.0500 L) = 0.0085 mol
For KCl:
moles of KCl = (1.6 M) x (0.0500 L) = 0.080 mol
Next, let's determine the limiting reactant. The reaction stoichiometry tells us that 1 mol of Pb(NO3)2 reacts with 2 mol of KCl to form 1 mol of PbCl2. Therefore, Pb(NO3)2 is limiting because we have less of it.
Since Pb(NO3)2 is limiting, all of its moles will react to form PbCl2. Therefore, the concentration of Pb2+ at equilibrium is:
[Pb2+] = (moles of PbCl2 formed) / (total volume of solution)
The moles of PbCl2 formed is equal to the moles of Pb(NO3)2 used, which is 0.0085 mol.
The total volume of the solution is the sum of the individual volumes:
Total volume = 50.0 mL + 50.0 mL = 0.100 L
[Pb2+] = 0.0085 mol / 0.100 L = 0.085 M
To calculate the concentration of Cl-, we need to consider the dissociation of PbCl2 according to its solubility product constant (Ksp):
PbCl2(s) -> Pb2+ + 2 Cl-
The Ksp expression for PbCl2 is:
Ksp = [Pb2+] x [Cl-]^2
We know that the Ksp value for PbCl2 is 1.6 x 10^-5. Since we have already determined the value of [Pb2+] to be 0.085 M, we can rearrange the expression to solve for [Cl-]:
[Cl-]^2 = Ksp / [Pb2+]
[Cl-]^2 = (1.6 x 10^-5) / (0.085)
[Cl-]^2 = 1.88 x 10^-4
Taking the square root of both sides gives:
[Cl-] = sqrt(1.88 x 10^-4) = 0.0137 M
So, the concentration of Cl- at equilibrium is 0.0137 M.
millimoles Pb(NO3)2 = 50.0 x 0.17 = 8.5
mmoles KCl = 50.0 x 1.6 = 80.0
...Pb(NO3)2 + 2KCl ==> PbCl2(s) + 2KNO3
I....8.50.......80.0.......0.......0
C...-8.50......-17.0.....8.50....17.0
E......0.......63.0......8.50.....17.0
(KCl) remaining = 63.0 mmoles/100 mL = 0.63M
Let x = solubility PbCl2 (to find the Pb^2+ in the final solution and note it has a common ion of Cl^- from the excess KCl.)
PbCl2 ==> Pb^2+ + 2Cl^-
x........x.......2x
Ksp = (Pb^2+)(Cl^-)^2 = 1.6E-5
Substitute x for Pb^2+.
Substitute 2x+0.63 for Cl and don't forget to square it. The concn Cl is 0.63 from the excess KCl and 2x from the PbCl2.
Then solve for x to obtain the concn Pb2+ in the final solution.