[intelibp1600]

A 10.0 mL solution of 0.0500M AgNO3 was titrated with 0.0250M NaBr in the cell:
S.C.E. || titration solution | Hg(s)

Find the cell voltage for the following volumes of titrant: 0.1, 10.0, 20.0, and 30.0 mL.

I don't know where to start and how to solve this. Greatly appreciated if someone can help me out!!! Thanks!

[Borek]

titration solution | Hg(s)

Hg(s)? Or Ag(s)?

Write reaction equation.

[intelibp1600]

I'm not sure if I'm doing this correctly.

Na+ + NO3- -> NaNO3(s)

(10.0mL)(0.05M)=(v)(0.025M)
Equivalence point V=20 mL

At 0.1 mL, 0.5% of NaO3- has precipitated and 99.5% remain in solution:
[NaO3-]=(0.995)(0.050)(10.0mL/20.1mL) = 0.0248M


[Na+][NO3-]=Ksp -> [Na+] = Ksp/NO3- = ... I'm stuck on this part. I'm not sure as to how to find the Ksp of Na+...

[Albert]

NaNO3 does not precipitate! What precipitate is AgBr.

Please, write how the cell looks like again and we'll be able to help you.

To quote Borek: S.C.E. || titration solution | Ag(s). This seems more possible.

[intelibp1600]

The lab manual stated Hg....I assume that's a typo. Sorry, I don't quite understand what you mean by write down how the cell looks like. The question was just given like that. That is why I'm so confused.  

[Albert]

I asked you to tell us how the cell is composed. It seems impossible to determine the potential energy of a cell where you perform an argentometric titration (Ag+ + Br- -> AgBr) and you have Hg as solid component of the electrode (moreover, in absence of Hg ions).

[Borek]

Mercury electrodes with dissolved metal are used quite often in voltammetric methods (CV, thin layer, stripping V and so on). Potential of such electrode depends on the Ag⁺ activity in the solution and activity of Ag dissolved in mercury. However, IMHO it doesn't make sense to use mercury electrode for argentometric tirations - as it is not used to measure potential, but concentration (through current).

[intelibp1600]

There is an example in the book similar to the problem.

This is what I have so far.

 Ag+ + Br- ->AgBr(s)

The equivalence point is 20mL.

[Br-]=(fraction remaining)(orginial concentration of Br-)(dilution factor)
       = (.995)(0.025M)(20.0mL/20.1mL)= 0.0248M

[Ag+][Br-]=> Ksp -> [Ag+] = (Ksp)/([Br-]) = 5.0 X 10^-13/ 0.0248 = 2.02 X 10^-11M

Cell Voltage using Nernst Equation
E=0.558 + 0.05916log(2.02X10^-11)= -0.07V


I'm not sure if I'm doing this correctly because the numbers doesn't seem right.

Also, for 30mL: at 20mL, how can I write the fraction remaining?




Sorry if this is confusing as this is confusing to me.

[Albert]

Cell Voltage using Nernst Equation
E=0.558 + 0.05916log(2.02X10^-11)= -0.07V

This means you're using a silver electrode, like the one I described: Ag⁺_(aq) | Ag(s).

[intelibp1600]

This means you're using a silver electrode, like the one I described: Ag⁺_(aq) | Ag(s).

According to the problem yea... this wasn't experimented. It was just a problem for us to calculate which I didn't quite understand.