buffer calculation questions Watch

Chez 01
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https://filestore.aqa.org.uk/sample-...1-QP-JUN17.PDF

on question 2.2
so ive calculated the concentration of sodium hydroxide to be 0.014
i dont understand the reason for adding that to the concentration of ethanoate and subtracting it from concentration of ethanoic acid

please explain
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Pigster
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(Original post by Chez 01)
https://filestore.aqa.org.uk/sample-...1-QP-JUN17.PDF

on question 2.2
so ive calculated the concentration of sodium hydroxide to be 0.014
i dont understand the reason for adding that to the concentration of ethanoate and subtracting it from concentration of ethanoic acid

please explain
You don't do anything with the conc of NaOH.

You should know the amount of EtOOH and EtOO-, then when you add a certain amount of OH-, it will react with some of the EtOOH (partially neutralising it) producing that many more mol of EtOO-.
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wendychan
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This type of calculation is called calculating new pH of buffer solution after addition of small amount of alkali.

1. Find moles of alkali being added - here they give that to you directly - 0.007 mol

This is something you should know:
Addition of alkali increases moles of buffer salt
Addition of alkali decreases moles of buffer acid

2. Find moles of buffer salt before addition of alkali
3. Add 0.007 mol to this to get new moles of buffer salt after addition of alkali

4. Find moles of buffer acid before addition of alkali
5. Subtract 0.007 mol to this to get new moles of buffer acid after addition of alkali

6. Use the Ka expression for acidic buffers. Rearrange for [H+]

[H+] = Ka x [buffer acid] / [buffer salt]

You can find conc. of buffer acid and buffer salt if you want to, but you can simplify the above expression to Ka/1 x [buffer acid] / [buffer salt] - try writing it on paper will be easier to see.

The volumes of [buffer acid / [buffer salt] cancel out - times numerator and denominator by total volume. So you can just insert the moles after addition of alkali of acid and salt - it's kind of like in Kc where sometimes you can just insert moles because volumes cancel out.

7. - log10 [H+] to get your new pH after addition of this alkali

Fyi - there are generally three main types of calculations in acid and bases that you should know about:

1. Finding pH during a neutralisation reaction at 4 distinct stages - at the very start, when the acid is in excess, at neutralisation, when base is in excess. This is assuming that the base is in the burette and acid is in the conical flask.
The only neutralisation needed for AQA is between Strong acid - strong base and weak acid - strong base. Be aware they can add complexities to these questions by making the acid polyprotic or base polybasic - so you have to be careful with moles of H+ and OH-

2. Preparation of acidic buffers and finding their pH after preparing them

3. Finding new pH of an already prepared acidic buffer after small addition of base, acid or water



(Original post by Chez 01)
https://filestore.aqa.org.uk/sample-...1-QP-JUN17.PDF

on question 2.2
so ive calculated the concentration of sodium hydroxide to be 0.014
i dont understand the reason for adding that to the concentration of ethanoate and subtracting it from concentration of ethanoic acid

please explain
Last edited by wendychan; 2 months ago
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Chez 01
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(Original post by wendychan)
This type of calculation is called calculating new pH of buffer solution after addition of small amount of alkali.

1. Find moles of alkali being added - here they give that to you directly - 0.007 mol

This is something you should know:
Addition of alkali increases moles of buffer salt
Addition of alkali decreases moles of buffer acid

2. Find moles of buffer salt before addition of alkali
3. Add 0.007 mol to this to get new moles of buffer salt after addition of alkali

4. Find moles of buffer acid before addition of alkali
5. Subtract 0.007 mol to this to get new moles of buffer acid after addition of alkali

6. Use the Ka expression for acidic buffers. Rearrange for [H+]

[H+] = Ka x [buffer acid] / [buffer salt]

You can find conc. of buffer acid and buffer salt if you want to, but you can simplify the above expression to Ka/1 x [buffer acid] / [buffer salt] - try writing it on paper will be easier to see.

The volumes of [buffer acid / [buffer salt] cancel out - times numerator and denominator by total volume. So you can just insert the moles after addition of alkali of acid and salt - it's kind of like in Kc where sometimes you can just insert moles because volumes cancel out.

7. - log10 [H+] to get your new pH after addition of this alkali

Fyi - there are generally three main types of calculations in acid and bases that you should know about:

1. Finding pH during a neutralisation reaction at 4 distinct stages - at the very start, when the acid is in excess, at neutralisation, when base is in excess. This is assuming that the base is in the burette and acid is in the conical flask.
The only neutralisation needed for AQA is between Strong acid - strong base and weak acid - strong base. Be aware they can add complexities to these questions by making the acid polyprotic or base polybasic - so you have to be careful with moles of H+ and OH-

2. Preparation of acidic buffers and finding their pH after preparing them

3. Finding new pH of an already prepared acidic buffer after small addition of base, acid or water
thanks for the explanation, this is really helpful!

you said that:
Addition of alkali increases moles of buffer salt
Addition of alkali decreases moles of buffer acid

this is for addition of alkali, what about in the case of when an acid is added?
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wendychan
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Addition of acid:

Increases moles of buffer acid
Decreases moles of buffer salt


(Original post by Chez 01)
thanks for the explanation, this is really helpful!

you said that:
Addition of alkali increases moles of buffer salt
Addition of alkali decreases moles of buffer acid

this is for addition of alkali, what about in the case of when an acid is added?
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Chez 01
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(Original post by wendychan)
Addition of acid:

Increases moles of buffer acid
Decreases moles of buffer salt
okay so its just the oppositte, thankyou!
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