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Chem a level buffers question

State how a buffer solution can be made from solutions of potassium hydroxide and ethanoic acid.
Give an equation for the reaction between potassium hydroxide and ethanoic acid. State how this buffer solution resists changes in pH when a small amount of acid is added.

How buffer solution resists pH change
[3 marks]


MS: CH3COO– (from salt) reacts with (added) acid/H+

I said this, but I also said that because CH3COO- concentration is very low, the OH- ions from the KOH could react with H+ as well, is this idea wrong?..if not why does the markscheme only accept the first statement?


and also another question: when we say all ion concentrations to 1moldm-3 what ions are we regarding that for. E.g. I know for H2SO4 you would normally use 0.5M but surely you could say that the SO4- ions are not 1M anymore...?
(edited 8 months ago)
Original post by cloverleaf39
State how a buffer solution can be made from solutions of potassium hydroxide and ethanoic acid.
Give an equation for the reaction between potassium hydroxide and ethanoic acid. State how this buffer solution resists changes in pH when a small amount of acid is added.

How buffer solution resists pH change
[3 marks]


MS: CH3COO– (from salt) reacts with (added) acid/H+

I said this, but I also said that because CH3COO- concentration is very low, the OH- ions from the KOH could react with H+ as well, is this idea wrong?..if not why does the markscheme only accept the first statement?


and also another question: when we say all ion concentrations to 1moldm-3 what ions are we regarding that for. E.g. I know for H2SO4 you would normally use 0.5M but surely you could say that the SO4- ions are not 1M anymore...?


In a buffer the concentration of salt CH3COO- isn’t very low. It would be in a weak acid but that isn’t a buffer. Ideally the concentration of acid and salt in a buffer are the same so [HA]/[A-] = 1.
as for all ions at 1 moldm-3 it doesn’t include spectator ions just what’s reacting.
(edited 8 months ago)
Reply 2
thank you, but I'm still confused because if CH3COOH dissociates only slightly and it is the KOH that is providing the extra OH- ions as [A-], would the CH3COO- not still be very low? Usually in a buffer, it consists of e.g. a weak acid and its salt, so the CH3COO- anion is the same as provided by the acid and the salt..? I hope this makes sense
Original post by FurryFreakBro
In a buffer the concentration of salt CH3COO- isn’t very low. It would be in a weak acid but that isn’t a buffer. Ideally the concentration of acid and salt in a buffer are the same so [HA]/[A-] = 1.
as for all ions at 1 moldm-3 it doesn’t include spectator ions just what’s reacting.
No. The KOH is not part of the buffer system and is not providing OH- ions in the buffer. All the OH- ions from the KOH are used up reacting with the CH3COOH to make the salt, CH3COO-.
A buffer is a mixture of a weak acid and its salt. In this case we are partially neutralising the weak acid with a strong base to make the salt rather than adding the salt directly to the weak acid.
Reply 4
Original post by FurryFreakBro
No. The KOH is not part of the buffer system and is not providing OH- ions in the buffer. All the OH- ions from the KOH are used up reacting with the CH3COOH to make the salt, CH3COO-.
A buffer is a mixture of a weak acid and its salt. In this case we are partially neutralising the weak acid with a strong base to make the salt rather than adding the salt directly to the weak acid.

I understand now! thank you

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