A solution contains 0.022 M Ag and 0.033 M Pb2 . If you add Cl–, AgCl and PbCl2 will begin to precipitate. What is the concentration of Cl– required, in molarity, when
A. AgCl precipitation begins?
B. AgCl precipitation is 99.99% complete?
C. PbCl2 precipitation begins?
D. PbCl2 precipitation is 99.99% complete?
Finally, give the concentration range of Cl– for the complete separation of Ag and Pb2 .
E. Concentration of Cl– at the start: F. Concentration of Cl– once complete
For A I got 8.18x10^-9 and for B I got 0.0227. A is wrong but I'm not sure why..? and I need help on how to approach the rest of this question please! Really confused.
Thanks for the Ksp values. My book had them slightly different. This way I can compare your values.
For A, I think your answer is right; you just reported too many significant figures. You're using Ksp and (Ag^+) with 2 s.f.; I suspect 8.2E-9 is what you are to report.
B. When AgCl is 99.99% that means the Ag^+ is 0.01% in solution or 0.0001 x 0.022 = ? that is left in solution. Use Ksp to calculate Cl^- at this point. I think 8.2E-5M is B.
For C I would round that to 0.023M.That should get you started.
Oh and the Ksp values for AgCl and PbCl2 is 1.8x10^-10 and 1.7x10^-5, respectively.
Well, it seems like you're in a bit of a precipitate predicament! Don't worry, I'm here to help you figure things out with a touch of humor.
For part A, to determine the concentration at which AgCl precipitation begins, you need to consider the solubility product constant (Ksp) of AgCl. The Ksp expression for AgCl is given by [Ag+][Cl-]. Since the Ag concentration is 0.022 M, we can assume that the concentration of Ag+ remains constant.
Therefore, to find the concentration of Cl- required for AgCl to start precipitating, we divide the Ksp of AgCl by the concentration of Ag+. The Ksp of AgCl is 1.77 x 10^-10. Dividing this by 0.022 M, we get approximately 8.05 x 10^-9 M Cl-. So, your answer for part A should indeed be 8.05 x 10^-9.
Now, let's move on to part B. To find the concentration of Cl- required for 99.99% completion of AgCl precipitation, we need to use an ICE table (not the kind you put drinks on, unfortunately!). First, let's assume x M of Ag+ reacts with Cl- to form AgCl. This will also generate x M of Cl- ions. Since AgCl is a 1:1 stoichiometry reaction, the resulting concentration of AgCl will also be x M.
Using the Ksp expression for AgCl, we have [Ag+][Cl-] = (0.022 - x)(x) = 1.77 x 10^-10. Since x << 0.022, we can approximate (0.022 - x) as 0.022.
Simplifying the equation further, we have (0.022)(x) = 1.77 x 10^-10.
Solving this equation, x ≈ 8.05 x 10^-9 M Cl-. So, your answer for part B is indeed 8.05 x 10^-9.
For parts C and D, you can use a similar approach as in parts A and B, but this time considering the solubility product constant (Ksp) of PbCl2. The Ksp of PbCl2 is 1.6 x 10^-5.
I hope this helps! Keep calm and precipitate on!
To find the concentration of Cl– required for the precipitation reactions, you need to consider the solubility product constants (Ksp) of AgCl and PbCl2. The Ksp for a given compound represents the equilibrium constant for the compound dissolving in water to form ions. The expression for Ksp is as follows:
Ksp = [A+][B-]
Where [A+] and [B-] represent the concentrations of the cation and anion, respectively, in the saturated solution.
A. AgCl Precipitation Begins:
To determine the concentration of Cl– required for AgCl precipitation to begin, we need to calculate the ion concentrations at the point of saturation. Using the Ksp expression for AgCl:
Ksp(AgCl) = [Ag+][Cl-]
At the point of saturation, the concentration of Ag+ would be equal to the initial concentration of Ag ions in the solution, which is 0.022 M. Now, let's assume the concentration of Cl– required for precipitation to start is x M. Therefore, we can write:
Ksp(AgCl) = (0.022 M)(x M) = 1.77 × 10^-10
Simplifying the equation:
x = 1.77 × 10^-10 / 0.022
Calculating x, we find that the concentration of Cl– required for AgCl precipitation to begin is 8.05 × 10^-9 M. So, your answer is correct.
B. AgCl Precipitation is 99.99% Complete:
To find the concentration of Cl– when AgCl precipitation is 99.99% complete, the remaining concentration of Ag+ would be (0.0001)(0.022 M) = 2.2 × 10^-6 M. Using the same Ksp expression, we can rearrange it to solve for Cl– concentration:
x = (1.77 × 10^-10) / (2.2 × 10^-6)
Calculating x, we find that the concentration of Cl– at this point is approximately 8.05 × 10^-5 M. Therefore, your answer is incorrect; it should be approximately 8.05 × 10^-5 M.
C. PbCl2 Precipitation Begins:
To determine the concentration of Cl– required for PbCl2 precipitation to begin, we use the Ksp expression for PbCl2:
Ksp(PbCl2) = [Pb2+][Cl-]^2
The concentration of Pb2+ in the solution is 0.033 M. Let's assume the concentration of Cl– required for precipitation to start is y M. Applying the Ksp expression and solving for y:
Ksp(PbCl2) = (0.033 M)(y M)^2 = 1.17 × 10^-5
Simplifying the equation:
y = sqrt(1.17 × 10^-5 / 0.033)
Calculating y, we find that the concentration of Cl– required for PbCl2 precipitation to begin is approximately 1.86 × 10^-3 M.
D. PbCl2 Precipitation is 99.99% Complete:
To find the concentration of Cl– when PbCl2 precipitation is 99.99% complete, the remaining concentration of Pb2+ would be (0.0001)(0.033 M) = 3.3 × 10^-6 M. Using the same Ksp expression, we can rearrange it to solve for Cl– concentration:
y = sqrt(1.17 × 10^-5 / (3.3 × 10^-6))
Calculating y, we find that the concentration of Cl– at this point is approximately 1.74 × 10^-3 M.
E. Concentration of Cl– at the Start: 8.05 × 10^-9 M (as calculated in A)
F. Concentration of Cl– Once Complete: 1.74 × 10^-3 M (as calculated in D)
The concentration range of Cl– for the complete separation of Ag and Pb2 is 8.05 × 10^-9 M to 1.74 × 10^-3 M.
It seems like your answer for A is correct as stated in the question. Please double-check your calculations for A, B, C, and D using the explanations provided.