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How do you calculate the pH of a weak acid?

Lets say we have 0.89 mol dm-3 of CH3COOH, a weak acid, with a ka of 1.7X10^-5.

We want to find Ph, but to do this we first need to know [H+] because PH = -log10[H+].

Because CH3COOH is a weak acid, it only partially dissociates, 

               CH3COOH <--------> CH3COO- + H+

therefore it has an acid dissociation contast (ka)

ka= [CH3COO-] [H+] / [CH3COOH]

Because both of ur products are coming from the single reactant, and because of the molar ratios of the chemical equation, [CH3OO-] = [H+].

so we can simplify our expression to ka = [H+]^2/ [CH3COO]

With weak acids we are able to make an important assumption. Because dissociation is weak, we can assume [H+] is really small, so much so that 1- [H+] is roughly 1. S0 [H+} is roughly 0.

The amount of reactant we have at eqm in this reaction is equal to amount at start-amount of each product formed at eqm.

Therefore [CH3COOH] at the start of the reaction is roughly unchanged at equilibirum, its a bit like saying [CH3COOH]-0 is still [CH3COOH].

Therefore we can say ka= [H+]^2 / 0.89

Plug in the ka value: 1.7 x 10^-5 = [H+]^2 / 0.89

Remember we said we were looking to find [H+] so lets rearrange to make [H+] the subject:

[H+]^2= (1.7 x 10 ^-5 ) x 0.89

[H+] ^2 = 1.513x 10^ -5

[H+] = 3.89x 10^-3

ph = -log10(3.89 x 10^-3)

ph= 2.4

Amy H. GCSE Biology tutor, A Level Biology tutor, GCSE Chemistry tuto...

8 months ago

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