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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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RowanF RowanF
wrote...
Posts: 9
Rep: 1 0
8 years ago
Hi there,

I'm not sure if I've done this problem right, I would really appreciate any help.

Quote
A sheet of lead is plunged in a 0.1 M solution of sulfuric acid. The following reaction takes place:

Pb  +  2 H+    Pb2+  +  H2

The potential of the solution is immediately measured. The result is E = -0.326 V. Knowing that E° of redox pair Pb2+/Pb is equal to - 0.126 V,

a) Calculate the concentration in Pb2+ of the solution.
b) Deduce the value of the solubility product Ks of PbSO4.

For the first part I plugged everything into the Nernst equation:

E = E° + (0.0592/n)log[ox]^a

-0.326 = -0.126 + 0.0592/2 log(x)

I got x = 1.74 x 10^-7


For the second part, I just squared that part, since the solubility product of lead sulfate should be:

[Pb2+][SO4 2-]

Thanks so much for any help !
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