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chemistry a2 acids and bases Question

A 0.210 mol dm–3 solution of potassium hydroxide was added from a burette to 25.0 cm3 of a 0.160 mol dm–3 solution of ethanoic acid in a conical flask.
Given that the value of the acid dissociation constant, Ka, for ethanoic acid is 1.74 × 10–5 mol dm–3, calculate the pH at 25 °C of the solution in the conical flask at the following three points:
before any potassium hydroxide had been added;
after 8.0 cm3 of potassium hydroxide solution had been added;
after 40.0 cm3 of potassium hydroxide solution had been added.


Can I get a detailed answer if any of you have the time as im really confused with this, thanks
I'm afraid we can't give full solutions. Have you attempted anything? :h:
Reply 2
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souhm
OP
10
Original post by Kvothe the arcane
I'm afraid we can't give full solutions. Have you attempted anything? :h:


Why not? I got the answers but I find it difficult to get to them
Original post by souhm
Why not? I got the answers but I find it difficult to get to them


It is considered to be not very helpful. What parts do you have trouble with?

You can write an equation for the neutralisation reaction.
Work out the moles of ethanoic acid and use stoichiometry to work out the moles of alkali. One will be in excess and this will be what is left in the buffer solution.
The volume will be the combined initial volumes so you can work out the concentration and then work out the PH of the unreacted acid or alkali.
Reply 4
Avatar for souhm
souhm
OP
10
For part b i did:

moles of the KOH added which is equal to the moles of CH3COO-
Then I found the moles of CH3COOH and rearranged the Ka formula to work out H+ conc then found the ph..
Whats wrong with this? (btw i found the concentrations of each before putting it into the formula)
Original post by Kvothe the arcane
It is considered to be not very helpful. What parts do you have trouble with?

You can write an equation for the neutralisation reaction.
Work out the moles of ethanoic acid and use stoichiometry to work out the moles of alkali. One will be in excess and this will be what is left in the buffer solution.
The volume will be the combined initial volumes so you can work out the concentration and then work out the PH of the unreacted acid or alkali.

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