A solution is 0.10M Pb(NO3)2 and 0.10M AgNO3. If solid NaCl is added to the solution what is [Ag+] when PbCl2 begins to precipitate? (Ksp PbCl2 = 1.7 x 10^-5; AgCl = 1.8 x 10^-10)
The answers is 1.4 x 10^-8
How do I find the answer
......PbCl2(s) ==> Pb^2+ + 2Cl^-
I......solid.......0.1.......0
C......solid.........x.......2x
E......solid.....0.1+x.......2x
Ksp = (Pb^2+)(Cl^-)^2 = 1.7E-5
.........AgCl(s)==> Ag^+ + Cl^-
I........solid......0.1.....0
C........solid......+x.......x
E.........solid...0.1+x......x
Ksp = (Ag^+)(Cl^-) =1.8E-10
These are done this way.
When NaCl is added, AgCl will ppt first because it has the smaller Ksp. It will continue pptng until the Ksp for PbCl2 is exceeded. What is the (Cl^-) when PbCl2 first ppts? That is
1.7E-5 = (Pb^2+)(Cl^-)^2
Substitute and solve for Cl^-. Remember Pb^2+ is 0.1M. I get something like 0.013 M but you need to confirm that.
Then plug this Cl^- into Ksp for AgCl to find Ag^+ when PbCl2 just reaches that point.
Ksp AgCl = (Ag^+)(Cl^-)
1.8E-10 = (Ag^+)(0.013)
Ag^+ = 1.8E-10/0.013 = 1.38E-8 M which rounds to 1.4E-8M
To find the concentration of Ag+ when PbCl2 begins to precipitate, you need to determine how much of the Ag+ ions will be consumed to form PbCl2, based on the stoichiometry of the reaction.
First, let's write the balanced equations for the reactions that occur:
Pb(NO3)2 (aq) + 2NaCl (aq) ⇌ PbCl2 (s) + 2NaNO3 (aq)
AgNO3 (aq) + NaCl (aq) ⇌ AgCl (s) + NaNO3 (aq)
The solubility product constant (Ksp) for PbCl2 is given as 1.7 x 10^-5, which means the equilibrium expression is:
Ksp = [Pb2+][Cl-]^2
Since we are given the concentration of Pb(NO3)2 as 0.10 M, the initial concentration of Pb2+ is also 0.10 M.
For AgCl, the Ksp is given as 1.8 x 10^-10, which means the equilibrium expression is:
Ksp = [Ag+][Cl-]
Since we are given the concentration of AgNO3 as 0.10 M, the initial concentration of Ag+ is also 0.10 M.
Now, let's assume that x moles of PbCl2 and x moles of NaNO3 are formed.
Using the stoichiometry of the reaction, x moles of Pb2+ will react with 2x moles of Cl- ions to form x moles of PbCl2.
Therefore, the equilibrium concentrations of Pb2+ and Cl- ions are:
[Pb2+] = 0.10 - x M
[Cl-] = 2x M
Using the solubility product constant expression for PbCl2, we have:
Ksp = [Pb2+][Cl-]^2
1.7 x 10^-5 = (0.10 - x)(2x)^2
1.7 x 10^-5 = 4x^3 - 0.4x^2
Now, let's solve this equation to find the value of x:
4x^3 - 0.4x^2 - 1.7 x 10^-5 = 0
To solve this equation you would need to use a numerical method such as a calculator or a solver tool. In this case, the value of x is approximately 0.0102.
Now, let's calculate the concentration of Ag+ ions remaining in the solution when PbCl2 begins to precipitate.
Since 2x moles of Cl- ions are required to react with 1 mole of Ag+ ions to form AgCl, the concentration of Ag+ remaining in the solution is:
[Ag+] = (0.10 - x)/2
[Ag+] = (0.10 - 0.0102)/2
[Ag+] ≈ 0.0449 M or 4.49 x 10^-2 M
However, the question asks for the concentration of [Ag+] when PbCl2 begins to precipitate, which means that all the Pb2+ ions have reacted. So, the concentration of [Ag+] when PbCl2 begins to precipitate is actually zero.
Therefore, the answer provided (1.4 x 10^-8) does not seem to be correct based on the given information and calculations. The correct answer should be zero (0 M).
To find the answer, you need to determine the point at which PbCl2 begins to precipitate and calculate the concentration of Ag+ at that point. Here are the steps to follow:
Step 1: Write the balanced equation for the precipitation reaction that occurs when PbCl2 begins to precipitate:
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)
Step 2: Identify the stoichiometry of the reaction. From the equation, you can see that 1 mol of PbCl2 produces 1 mol of Pb2+ ions and 2 mol of Cl- ions.
Step 3: Calculate the initial concentration of Cl- ions. Since you have added solid NaCl to the solution, we can assume it fully dissolves, providing an additional source of Cl- ions. The initial concentration of Cl- ions is therefore 0.10M (from AgNO3) + X (from NaCl), where X represents the concentration of Cl- ions from NaCl.
Step 4: Use the Ksp expression for PbCl2 to determine the concentration of Pb2+ ions at the point of precipitation:
Ksp = [Pb2+][Cl-]^2
From the equation, you can see that the concentration of Pb2+ ions is equal to the concentration of Cl- ions. Therefore, at the point of precipitation:
Ksp = [Pb2+][Cl-]^2
1.7 x 10^-5 = x(X)^2
Step 5: Solve the equation for X to obtain the concentration of Cl- ions at the point of precipitation (X = [Cl-]).
Step 6: Calculate the concentration of Ag+ ions at the point of precipitation using the Ksp expression for AgCl:
Ksp = [Ag+][Cl-]
1.8 x 10^-10 = (0.10 - X)(X)
Step 7: Solve the equation for X to obtain the concentration of Ag+ ions at the point of precipitation (X = [Ag+]).
The answer you provided, 1.4 x 10^-8, is correct and represents the concentration of Ag+ ions when PbCl2 begins to precipitate.