Question

In: Chemistry

What is the pH of 0.1 M FeSO4 please show all steps

Expert Solution

FeSO4 ------> Fe⁺² +SO4^2-

SO4^2- + 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-]/[SO4^2-]

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

queen_honey_blossom answered 2 years ago

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