In: Chemistry
What is the pH of 0.1 M FeSO4 please show all steps
FeSO4 ------> Fe+2 +SO42-
SO42- + H2O -----> HSO4- + OH_
I 0.1 0 0
C -x +X +x
E 0.1-x +x +x
ka of HSO4- is 0.012
Kb = Kw/ka
= 1*10-14/0.012
= 83.3*10-14
Kb = [HSO4-][OH-]/[SO42-]
83.3*10-14 = x*x/0.1-x
x2 = 83.3*10-14
x = 9.12*10-7
[OH-] = x =9.12*10-7 M
POH = -log9.12*10-7
= -log9.12+7log10
= -0.9604 +7 = 6.0396
PH = 14-POH
= 14-6.0396
= 7.9604