[martini]

I have the following problem to solve by Tuesday:

A solution prepared by mixing 47.1 mL of 0.4 M AgNO3 and 47.1 mL of 0.4 M TlNO3 was titrated with 0.8 M NaBr in a cell containing a silver indicator electrode and a reference electrode of constant potential 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TlBr is Ksp = 3.6 × 10–6 and the solubility constant of AgBr is Ksp = 5.0 × 10–13.


(a) Which of the following expressions shows how cell voltage depends on [Ag ]?

b) What is the cell voltage when the following volumes of 0.8 M NaBr have been added? (b) 1.0 mL (c) 13.5 mL (d) 22.6 mL (e) 23.5 mL (f) 23.9 mL (g) 37.0 mL (h) 47.1 mL (i) 50.5 mL

I found part a) to be:

E = .175-(.799-.05916log(1/[Ag+])

I tried doing part b) tons of times now and I can't figure it out.

I'm confused with the whole fact that the solution that is titrated is a mixture of 2 solutions. I'm not sure what the equivalence volume should be, or what total volume to use to calculate the concentration of Ag+.

I tried doing it a bunch of different ways, and for the (b) 1.00mL titrant added, I keep getting -.599 or somewhere close to it, but the program tells me this is wrong. If someone could help me that would be wonderful because I've been stuck on this problem for hours!

[Borek]

I'm not sure what the equivalence volume should be

There will be two equivalence points - first, when Ag⁺ was titrated, second, when Tl⁺ was titrated. Without calculating it is hard to predict whether you will see one, or two inflection points on the curve.

or what total volume to use to calculate the concentration of Ag+.

I am not sure what your problem is. Total volume after adding V mL of titrant is 47.1+47.1+V.

[martini]

I'm sorry i spent so much time on this problem that my brain stopped working, you're totally right about the total volume.

The equivalence volume is my biggest problem. To get the equivalence volume I figured, as usual, I have to use the MV = M2V2 equation. The problem I have is that I don't know what to plug in.

First, I plugged in just the volume and concentration of AgNO3 and concentration of NaBr:

47.1(.4) = (.8)Ve ---->Ve=equivalence volume
Ve = 23.55

then, solving the 1st titrant volume, 1mL I did the following:

[Ag+] = (Ve-volume titrant added)/Ve * [initial Ag+] * (initial volume/total volume)
[Ag+] = (23.55-1)/23.55 * (.4) * (94.2/95.2)
[Ag+] = .379

then I plugged that into the equation from part (a) E = .175-(.799-.05916log(1/[Ag+]))

E = .175 - (.799 - .05916 log(1/.379))
E = -.599

I plugged that into the program and it says the answer is wrong.

Then I thought that maybe since the two solutions in one mixture would precipitate simultaneously the equivalence volume would double, since the volume of titrant would be separated between precipitating the two solutions and you would need more titrant to reach both equivalence points. So I used 47.1 as my equivalence volume:

[Ag+] = (Ve-volume titrant added)/Ve * [initial Ag+] * (initial volume/total volume)
[Ag+] = (47.1-1)/47.1 * (.4) * (94.2/95.2)
[Ag+] = .387

then E = .175-(.799-.05916log(1/[Ag+]))

E = .175 - (.799 - .05916 log(1/.387))
E = -.600

I plugged that into the program, wrong again.

I have no idea what I'm doing wrong.

[Borek]

I don't get why you need the equivalence volume. Question - as listed -asks to calculate potential at

(b) 1.0 mL (c) 13.5 mL (d) 22.6 mL (e) 23.5 mL (f) 23.9 mL (g) 37.0 mL (h) 47.1 mL (i) 50.5 mL

I don't see how it is related to the equivalence point?

[martini]

When I entered an answer and it was wrong, a hint that the program gave me was:

"
For part (b), addition of 1.0 mL of 0.800 M NaBr will precipitate some Ag . Determine the concentration of unprecipitated Ag . The concentration of unprecipitated Ag is calculated via the equation
[Ag+] = (fraction Ag+ remaining) * (initial [Ag+]) * (dilution factor)
Use the concentration of unprecipitated Ag and the answer to part (a) to calculate the cell voltage.
"

Isn't (fraction Ag+ remaining) calculated: (equivalence vol. - volume titrant added)/(equivalence vol.)??

I know initial [Ag+] is .4M and that dilution factor is (47.1/(47.1+47.1+vol titrant added) but I guess I don't know how to find the fraction of Ag+ remaining.
 

[Borek]

Isn't (fraction Ag+ remaining) calculated: (equivalence vol. - volume titrant added)/(equivalence vol.)??

Can be calculated with such an approach, but there is no need for that.

I know initial [Ag+] is .4M and that dilution factor is (47.1/(47.1+47.1+vol titrant added) but I guess I don't know how to find the fraction of Ag+ remaining.

This is a simple stoichiometry. Forget it is a titration, forget about Tl presence. What was the initial amount of Ag⁺? How much reacted with Br^-? How much is left?

[martini]

Using your method and then further hints from the program I got the answer for all EXCEPT the last one. The hint that I got for last one from the program is:

"
For part (i), addition of 50.5 mL of 0.800 M NaBr will result in an excess of Br- since precipitation of both AgBr and TlBr are complete. Determine the concentration of excess [Br–]. The concentration of excess Br– is calculated via the equation
[Br-] = (dilution factor)*(initial [Br-])
Use [Br–] and the solubility product of AgBr to calculate [Ag ]. Then use [Ag ] and your answer to part (a) to calculate the cell voltage.
"

I'm don't know how to calculate the dilution factor for this. I tried doing:
50.5mL total Br- - 47.1 reacted Br- = 3.4mL excess Br-
then:
dilution factor= 3.4/(47.1+47.1+50.5) = .023
.023*.8M inital [Br-] = .0184M Br-
then: Ksp = 5e-13
[Ag+] = 5e-13/.0184 = 2.72e-11
Plug into equation E = .175-(.799-.05916log(1/[Ag+]))
E = .175-(.799-.05916log(1/2.72e-11)) = .001
I tried this answer and it said it was wrong.
I'm just not sure how to proceed with this one. I don't know if the dilution is correct,I have a feeling it's not, which would be my problem.

[Borek]

dilution factor= 3.4/(47.1+47.1+50.5) = .023
.023*.8M inital [Br-] = .0184M Br-

Don't round down intermediate results. It is not 0.23, but 0.234968.... - that's over 2% error.

Do the calculations using full precision, round down only the reported values.

I am not saying that's everything, but that's wrong for sure.

[martini]

Got all of them! Thank you so much for your *delete me*