what is the solubility product of lead sulphate?

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what is the solubility product of lead sulphate?

the electrode potentiales are:
pbSO4+2e -------> pb+ SO42- -0.359V

2007-03-08 21:14:31 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry

2 answers

I guess the temperature is 25 C= 298 K

ΔG= -nFE and also
ΔG= -RTlnK
here K=Ksp so we have

nFE=RTlnKsp =>
Ksp= e^(nFE/RT)=
= e^( 2*96485* (-0.359)/(298*8.314) ) =7.2*10^-13

2007-03-08 22:52:59 · answer #1 · answered by bellerophon 6 · 0⤊ 0⤋

1.6 x 10-8 mol2dm-6

2007-03-08 21:42:07 · answer #2 · answered by Gervald F 7 · 0⤊ 0⤋