# OCR F325 past paper question on buffer solutions

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#1

It asks you to work out the pH of a buffer solution and gives you the concentration and volume of butanoic acid and sodium hydroxide. But how do I get the amounts of [HA] and [A-] from this? (so that I can use them in the Ka equation to find out [H+])

I looked at the mark scheme and it shows values for the amount in moles of CH3(CH2)2 COOH and CH3(CH2)2 COO- but I have no idea how they got those values. When I did it, I just worked out the moles of butanoic acid and sodium hydroxide because they were the only ones that the concentration and volume were given for.
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8 years ago
#2
(Original post by bben955)

It asks you to work out the pH of a buffer solution and gives you the concentration and volume of butanoic acid and sodium hydroxide. But how do I get the amounts of [HA] and [A-] from this? (so that I can use them in the Ka equation to find out [H+])

I looked at the mark scheme and it shows values for the amount in moles of CH3(CH2)2 COOH and CH3(CH2)2 COO- but I have no idea how they got those values. When I did it, I just worked out the moles of butanoic acid and sodium hydroxide because they were the only ones that the concentration and volume were given for.
Here's a work through of the question (no audio)

5
#3
Wow did you just make that video for me?
Thank you so much!
It explains exactly the bit I didn't understand so its perfect, thanks again.
0
8 years ago
#4
(Original post by charco)
Here's a work through of the question (no audio)

Thank you, the video was incredibly helpful.
I would give rep but i have already given you some rep recently!!!
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7 years ago
#5
(Original post by charco)
Here's a work through of the question (no audio)

Why isn't the concentration of the acid just 0.0025? So the same as the NaOH due to the 1:1 ratio?
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7 years ago
#6
(Original post by Mutleybm1996)
Why isn't the concentration of the acid just 0.0025? So the same as the NaOH due to the 1:1 ratio?
The acid reacts with the base in a 1:1 proportion.

Initial moles of acid = 0.05 ml x 0.25 mol dm-3 = 0.0125 mol

Initial moles of NaOH = 0.05 ml x 0.05 mol dm-3 = 0.0025 mol

All of the base reacts to make the same moles of salt = 0.0025 mol

Moles of acid remaining = initial moles - moles of acid reacted (same as moles of base) = 0.0125 - 0.0025 = 0.001 mol

Not the same ...
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