Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: intelibp1600 on March 02, 2006, 10:30:33 PM
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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!
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titration solution | Hg(s)
Hg(s)? Or Ag(s)?
Write reaction equation.
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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+...
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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.
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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. :(
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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).
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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).
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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.
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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).
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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.