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Confusing buffer questions?

Hey all! I'm generally fine with buffers, however I don't quite understand the subtraction in some of the questions. Here's some examples.

I thought, alright, everytime you add acid to the buffer solution you'll have to minus the initial amounts, easy.
But what about this one?

Why do you subtract only the acid? Does it have to do with it being in excess?
And finally the last one, which made me go "wait what am I even doing here"

All of the answers are from the mark scheme. Can anyone please explain why you have to subtract some of the values to get the right answer? I feel like it's a very dumb question. ^^"

Reply 1

shengoc

Original post by sodiumchloride

Hey all! I'm generally fine with buffers, however I don't quite understand the subtraction in some of the questions. Here's some examples.

I thought, alright, everytime you add acid to the buffer solution you'll have to minus the initial amounts, easy.
But what about this one?

Why do you subtract only the acid? Does it have to do with it being in excess?
And finally the last one, which made me go "wait what am I even doing here"

All of the answers are from the mark scheme. Can anyone please explain why you have to subtract some of the values to get the right answer? I feel like it's a very dumb question. ^^"

I like to start from the weak acid equation for Ka. for a buffer, two assumptions
- your [conj base] approximates to [salt] since for an acidic buffer, u add in the salt of the weak acid
- the [h+] remains as it was fromweak acid, mostly undissociated, so approximates to [H+] initial

the trickiest qst is actually if total volume of solution remains the same or not (as concentration changes with volume of water present). hence it is important to note if you are adding solution to solution or solid to solution

The usual first step is usually find mole of acid, then mole of base, which one in excess...HOW much in excess is mole difference ..then conc whatever in excess is that excess mole divided by TOTAL volume of solution

THe rest should be plug and play