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# Stuck on a pH calculation watch

1. calculate the pH of a mixture of 25 cm3 0.200 mol dm-3 ethanoic acid (pKa = 4.76) and 25 cm3 0.100 mol dm-3 NaOH

Hi everyone, for this question I worked out the excess mole of H+ ions and got 2.5*10^-3 and pH as 1.3. The answer is 4.7 however, any suggestions?

Thanks
2. (Original post by coconut64)
calculate the pH of a mixture of 25 cm3 0.200 mol dm-3 ethanoic acid (pKa = 4.76) and 25 cm3 0.100 mol dm-3 NaOH

Hi everyone, for this question I worked out the excess mole of H+ ions and got 2.5*10^-3 and pH as 1.3. The answer is 4.7 however, any suggestions?

Thanks
This is a buffers question. Post your calculation showing how you used the pKa
3. (Original post by TeachChemistry)
This is a buffers question. Post your calculation showing how you used the pKa
Hi, thanks for the help. I haven't learnt buffers yet. Can you just briefly explain what the pka value is used for in this question?

Thanks
4. (Original post by coconut64)
Hi, thanks for the help. I haven't learnt buffers yet. Can you just briefly explain what the pka value is used for in this question?

Thanks
If you haven't been taught about buffers yet why are you doing this question?
5. (Original post by TeachChemistry)
If you haven't been taught about buffers yet why are you doing this question?
No idea, it is part of my homework..
6. (Original post by coconut64)
No idea, it is part of my homework..
Have you been taught the relevance of half equivalence? In other words, when exactly half of the acid has been neutralised by the alkali?
7. (Original post by TeachChemistry)
Have you been taught the relevance of half equivalence? In other words, when exactly half of the acid has been neutralised by the alkali?
Curiously, I don't remember being taught it and I've never seen it in a textbook, but a handful of years ago I noticed it for myself and thought that it'd be good to teach it as it is a nice short cut.

It turns out that I wasn't the first person to spot it.
8. Thanks everyone for the help, I am going to ask about this question in lesson because it has not been taught yet.
9. This is a really easy buffer calc. You're on the right lines with working out the excess of [H+] but you have the wrong method. Just for a head start, i'll let you know how to do it :-)
There are 2 ways to make a buffer. The first is to mix the weak acid/base and its salt together and the other is to react chemicals that will leave you wil a weak acid/base and its salt. The method used depends on the situation, and in this case a reaction occurs so the ratio of moles will change. AQA only want acidic buffers, so that makes it easy.

Firstly, work out the moles of acid and alkali. Remember, the moles of alkali also correspond to the moles of A- in the expression for Ka.
n(acid)=0.025 x 0.2 = 5x10^-3
n(alkali)= 0.025 x 0.1 = 2.5x10^-3

You'll notice that the acid is in excess, but that value corresponds to [HA] not [H+]. So, work out how much acid you have left:
5x10^-3 - 2.5x10^-3 = 2.5x10^-3

Now, you have equimolar amounts of acid and alkali. This is important when you use the Ka expression. To find out Ka, rearrange pKa= -log[Ka] into Ka=10^-pKa to get 1.74x10^-5.

Now, plug the numbers into your expression and rearrange for [H+] as this is what determines pH:
Ka= [H+][A-]/[HA]
(1.74x10^-5)x(2.5x10^-3)/2.5x10^-3. [HA] and [A-] cancel out so [H+]=Ka, so the concentration wouldn't matter because it would cancel out anyway.

Use -log[H+] to find the pH= -log[1.74x10^-5]
=4.7594...
pH=4.76 (remember, pH is ALWAYS to 2dp)

Hope this helped!
10. (Original post by typicalvirgo)
This is a really easy buffer calc. You're on the right lines with working out the excess of [H+] but you have the wrong method. Just for a head start, i'll let you know how to do it :-)
There are 2 ways to make a buffer. The first is to mix the weak acid/base and its salt together and the other is to react chemicals that will leave you wil a weak acid/base and its salt. The method used depends on the situation, and in this case a reaction occurs so the ratio of moles will change. AQA only want acidic buffers, so that makes it easy.

Firstly, work out the moles of acid and alkali. Remember, the moles of alkali also correspond to the moles of A- in the expression for Ka.
n(acid)=0.025 x 0.2 = 5x10^-3
n(alkali)= 0.025 x 0.1 = 2.5x10^-3

You'll notice that the acid is in excess, but that value corresponds to [HA] not [H+]. So, work out how much acid you have left:
5x10^-3 - 2.5x10^-3 = 2.5x10^-3

Now, you have equimolar amounts of acid and alkali. This is important when you use the Ka expression. To find out Ka, rearrange pKa= -log[Ka] into Ka=10^-pKa to get 1.74x10^-5.

Now, plug the numbers into your expression and rearrange for [H+] as this is what determines pH:
Ka= [H+][A-]/[HA]
(1.74x10^-5)x(2.5x10^-3)/2.5x10^-3. [HA] and [A-] cancel out so [H+]=Ka, so the concentration wouldn't matter because it would cancel out anyway.

Use -log[H+] to find the pH= -log[1.74x10^-5]
=4.7594...
pH=4.76 (remember, pH is ALWAYS to 2dp)

Hope this helped!
Appreciate the help, I will have a look and try understand it. Thanks !

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