Estimate the pH of a 0.1 M ethanoic acid solution using the equation below, where [CH3COOH]o is the nominal (i?

Question

Estimate the pH of a 0.1 M ethanoic acid solution using the equation below, where [CH3COOH]o is the nominal (initial) ethanoic acid conecntration.
Ka=[H3O+][CH3COO-]/[CH3COOH]o −[H3O2+] ≈ [H3O+]2/ [CH3COOH]o

pH = -log10[H3O+]

Hey guys, I'm supposed to answer this question, the problem is I have practically no idea what it means or how to do it, if somebody could show me the working that would be absolutely amazing! I'm really stuck and it needs to be done ASAP.
I did my best to lay the equation out in a way that makes sense so I hope it does, any help would be appreciated thanks in advance :)

HPV · Accepted Answer

The equation given is

Ka = [H3O+]^2 / [CH3COOH]o

Where does this come from? When you put the weak acid ethanoic (acetic) acid in water, it reacts to form small and equal amounts of H3O+ and CH3COO-. Set up an ICE chart.

Molarity . . . . . . .CH3COOH  +  H2O  <==>  H3O+  +  CH3COO-
Initial . . . . . . . . . . . . .0.1 . . . . . . . . . . . . . . .0 . . . . . . . .0
Change . . . . . . . . . . .-x . . . . . . . . . . . . . . . .x . . . . . . . .x
Equilibrium . . . . . . . .0.1-x . . . . . . . . . . . . . .x . . . . . . . .x

So Ka = [H3O+][CH3COO-] / [CH3COOH] = (x)(x) / (0.1-x)

Because Ka for CH3COOH is small (1.8 x 10^-5 --- I looked it up), then we can simplify the math by deleting the -x term after 0.1. We end up with

x^2 / 0.1 = 1.8 x 10^-5 . . .which is the equation that you were given.
x^2 = 1.8 x 10^-6
x = 1.3 x 10^-3 = [H3O+]

pH = -log [H3O+] = -log (1.3 x 10^-3) = 2.87

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Estimate the pH of a 0.1 M ethanoic acid solution using the equation below, where [CH3COOH]o is the nominal (i?

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