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OCR F325 past paper question on buffer solutions

Someone please help me with this question! I'm stuck on question 3(c)(ii) in the june 2012 F325 paper (on page 7) (Available from this page: http://www.ocr.org.uk/Images/131289-question-paper-unit-f325-equilibria-energetics-and-elements.pdf)

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.
Original post by bben955
Someone please help me with this question! I'm stuck on question 3(c)(ii) in the june 2012 F325 paper (on page 7) (Available from this page: http://www.ocr.org.uk/Images/131289-question-paper-unit-f325-equilibria-energetics-and-elements.pdf)

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)

[video="youtube_share;29QttHjLiSE"]http://youtu.be/29QttHjLiSE[/video]
Reply 2
Avatar for bben955
bben955
OP
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.
Reply 3
Avatar for freetown
freetown
6
Original post by charco
Here's a work through of the question (no audio)

[video="youtube_share;29QttHjLiSE"]http://youtu.be/29QttHjLiSE[/video]


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

[video="youtube_share;29QttHjLiSE"]http://youtu.be/29QttHjLiSE[/video]


Why isn't the concentration of the acid just 0.0025? So the same as the NaOH due to the 1:1 ratio?
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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