Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: sundrops on November 26, 2005, 12:55:40 AM
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Determination of Ksp of PbI2.
make PbI2 by adding:
1mL of 0.050M Pb(NO3)2
3mL of 0.050 M KI
fill one well with the PbI2 and an adjacent well with 0.050 M Pb(NO3)2
Use polished Pb electrodes in both cells.
Use a string soaked in KNO3 solution as a salt bridge.
So experimentally I found that the cell potential was 0.066 volts.
that the:
Pb(NO3)2 was the cathode, so the Pb2+
and the: PbI2 was the anode.
I need to determine the:
- [Pb2+] in the equilibrium with PbI
- [I-] in solution
- the calculated Ksp value
- the theoretical Ksp value
well the theoretical Ksp is easy - you just look it up and it is 8.3E-9
would Ksp be = [Pb2+][I-]2 ?
now having that, i just need the concentrations! :-\
now I know that I can determine the Pb2+ concentration using the nerst equation:
0.66V = Eo - (0.0591/2) * log( [Pb2+] / [anode] )
problem is - i dont know the concentration of PbI2 nor do I know what Eo could be.
is there anyone out there who could give me a hand?
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PbI2 activity is 1, but it doesn't matter.
You have two identical halfcells, with Pb electrode and Pb2+ in solution. The only difference is in the [Pb2+ ]. IIRC it is called gradient cell.
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ok I tried to determine the concentration of the PbI cell. Lets hope you can follow my reasoning...
my concentration of Pb2+ is 0.050M (from the Pb(NO3)2 soln)
Ecell = Eo(cathode) - Eo(anode)
0.066V = -0.13V - X
0.196V = X
so the voltage of the PbI2 half-cell is 0.196V
then I use the nernst equation:
0.066V = 0.196V - ((8.3145)(294.65) / (2)(96485)) * ln([anode]/[cathode])
0.066V = 0.196V - 0.0127 * ln([PbI2]/0.050)
-0.13 / -0.0127 = ln [PbI2] - ln (o.o5o)
10.24 = ln [PbI2] - ln (0.050)
ln(10.24) = e^ (PbI) - e^0.050
2.33 = e^(PbI2) - 1.0512
3.38 = e^(PbI2)
ln (3.38) = [PbI2]
1.217 = [PbI2]
does that look ok?
I sure hope so... ;D
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does that look ok?
No. At first sight, due to several reasons.
First, you are not interested in the PbI2 concentration, but in Pb2+ concentration.
Second, what is 0.0127? It should be close to 0.059/n=0.0295.
And third - as I already mentioned in my previous post activity of PbI2 is 1 - so it can't be 1.217 at the same time.
Try to write down Nernst equation for concentration cell (not a gradient cell, sorry for that). Check it here (http://scienceworld.wolfram.com/chemistry/ConcentrationCell.html).
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how do uyou know that the [PbI2] = 1 ?
also how did you determine the number of electrons ais 0.0295? wouldn't it be 2e-? and therefore n=2? I'm still very confused.
I'll try the question again, and hopefully this time around it'll be a bit better...
Ecell = Eo - 0.0591/n * log([cathode]/[anode])
0.066 = -0.13 -0.0591/2 * log(0.05/[anode])
0.22555 = log(0.05)-log(anode)
0.22555 = -1.30 - log[anode]
1.527 = -log[anode]
[anode] = 10^(1.527)
[anode] = 33.65 = [Pb2+] but that is not possible.... :-\
i'm getting beyond frusterated
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how do uyou know that the [PbI2] = 1 ?
PbI2 is solid, not dissolved. Activity of solids is always considered to be 1.
also how did you determine the number of electrons ais 0.0295? wouldn't it be 2e-? and therefore n=2? I'm still very confused.
Coefficient in the Nernst equation is 0.059/n - where n is number of electrons transferred during reaction. In this case n=2, so the coefficent is 0.059/2 = 0.0285. You used in your calculations some strange value of 0.0127.
Ecell = Eo - 0.0591/n * log([cathode]/[anode])
These are not [cathode]/[anode] but Pb2+ concentrations in both cells.
0.066 = -0.13 -0.0591/2 * log(0.05/[anode])
0.22555 = log(0.05)-log(anode)
Where did you get 0.22555 from? Check your math.
Besides, are you sure it was 0.066 and not -0.066?
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ok i see,
\These are not [cathode]/[anode] but Pb2+ concentrations in both cells
yes but i thought that might be a good way to distinguish the Pb2+ from solution from the Pb2+ from the PbI2 mixture.
now using Ecell = Ecathode - Eanode
0.066 = -.013-X
X=-0.196 = Eanode
so would Eo in the nernst equation be -0.196 instead of the -.13?
in that case:
0.066 = -.196 - 0.0591/2 * log(0.05/Pb2+)
-.2926 = log(0.05) - log[Pb2+]
-0.2926 = -1.3010 - log[Pb2+]
-1.0084 = log[Pb2+]
[Pb2+] = 10^(-1.0084)
[Pb2+] = 0.09808
does that look better? *crosses fingers*
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does that look better? *crosses fingers*
Please stop guessing, try to understand what you are doing first.
Potential in concentration cell is given by
E = E0 + 0.059/n log(c1/c2)
so only concentration ratio is important. Now put the numbers into equation:
0.066 = -0.13 + 0.059/2 log(c1/c2)
after rearrangement:
log(c1/c2) = 2(0.066 + 0.13)/0.0591 = 6.6
so
c1/c2 = 2.5*10-7
Now, I never remember which concentration is which - but in this case it is not a problem, as the concentration in the half cell with PbI2 must be much lower than the concentration in the second cell (the one with 0.05M Pb(NO3)2). If so:
c1 = 2.5*10-7 * 0.05M
is the only way to go.
Use this c1 to calculate Ksp in the PbI2 cell - [I- ] can be asily calculated from the initial data, assuming that the precipitation was stoichiometric (it almost was).
The problem is - if you calculate Ksp this way you will be off by several orders of magnitude.
However, if you did a classic mistake measuring potential, and what you wrote as 0.066 was in reality -0.066, error in your Ksp determination will be much lower (although still too large for me).
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thanks Borek, you've been incredibly patient :)
the 0.066volts was obtained experimentally and our machine did not go below zero, I don;t think it really matters as long as the mean class average had around the same voltage.
thank you again. :)
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the 0.066volts was obtained experimentally and our machine did not go below zero
It doesn't matter. If you switch connectors potential will be reversed.