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OCR F325 June 2012 Question 3 Help

I'm a bit stuck on question 3 c) ii) when I tried it I got one of the common mistakes and an answer of 4.12, and I understand now that the moles given are for the start and not at equilibrium, but how do I know why the moles at equilibrium are what they are?
(edited 6 years ago)
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Original post by abiril
I'm a bit stuck on question 3 c) ii) when I tried it I got one of the common mistakes and an answer of 4.12, and I understand now that the moles given are for the start and not at equilibrium, but how do I know why the moles at equilibrium are what they are?



This is making a buffer by partial neutralisation of a weak acid.
It goes like this
Moles of butanoic acid before reaction = 0.0125
Moles of NaOH added = 0.0025

When the NaOH and butanoic acid react they form the salt (conjugate base) which is butanoate. The reaction is all 1:1 so the moles of butanoate formed is the same as moles of NaOH added. = 0.0025
The number of moles of butanoic acid has now decreased as it reacted with the NaOH (as it is a 1:1 reaction the moles of butanoic acid that reacted = moles of NaOH = 0.0025) Leaving 0.0125 - 0.0025 = 0.0100 in the buffer.

So the buffer contains 0.0100 moles of butanoic acid and 0.0025 moles of butanoate in 100 cm^3.

So concentrations = butanooic acid = 0.100 moldm^3 and butanoate = .025moldm^3

So [H+] = 1.51x10^-5 x 0.100/.025
= 6.04x10^-5

pH = 4.22

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