You didn't substitute correctly for Keq. You should calculate
log Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+)
You made log Keq = 0.27. You should follow the instructions I gave.
(Pb) = 1
(Ni^2+) = 0.27
(Ni) = 1
(Pb^2+) = x so the equation is
0.020 = 0.12 - (0.0296)*log[(1)(0.27)/(1)(x)] and solve for x in moles/L.
I don't want to leave without acknowledging an error in what I wrote. It make no difference in the final answer becasue I corrected it later. Here is what I wrote for ONE of the steps.
Ecell = Eo cell - (2.303*RT/n) log Q and Q = Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+) but I omitted Faraday's constant. It should be
Ecell = Eo cell - (2.303*RT/nF) log Q and Q = Keq=(Pb)(Ni^2+)/(Ni)(Pb^2+)
Then if you substitute for R of 8.314, T = 298, F = 96,485 and 2 for n, then
(2.303*8.314*298/2*96,485) = 0.0296. Your post is now down below the line. You may want to repost at the top if you have additional quesitons.
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