What is the pH?

Question

Calculate the pH of 0.167 M NaC2H3O2 (sodium acetate)

C2H3O2-(aq) + H2O(l) ↔ HC2H3O2(aq) + OH-(aq) Ka for HC2H3O2= 1.75 x 10-5

Answer 1

.............C2H3O2-(aq) + H2O(l) ↔ HC2H3O2(aq) + OH-(aq)
I..............0.167 M................................0...........................0
C.................-x......................................x............................x
E............0.167-x...................................x............................x

Kb for C2H3O2^- = Kw/Ka for HC2H3O2 = (HC2H3O2)(OH^-)/(C2H3O2^-)
Kw/Ka = (x)(x)/(0.167-x). Plug in 1E-14 for Kw and 1.75E-5 for HC2H3O2 and solve for x = (OH^-). Then convert (OH^-) to pH.
Post your work if you get stuck.

Answer 2

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Answer 3

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Answer 4

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Answer 5

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Answer 6

I do!