Alevel chemistry buffers

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Thread starter 3 years ago
#1
Im really confused. Sometimes you have to work out which of the acid or salt is in excess and sometimes you dont.

An acidic buffer solution is obtained when sodium ethanoate is dissolved in aqueous
ethanoic acid.
(i) Calculate the pH of the buffer solution formed at 298 K when 0.125 mol of sodium
ethanoate is dissolved in 250 cm3 of a 1.00 mol dm–3 solution of ethanoic acid.
The acid dissociation constant, Ka, for ethanoic acid is 1.70 × 10–5 mol dm–3 at
298 K.

For this question you dont look at the excess why?

A buffer solution is formed when 2.00 g of sodium hydroxide are added to 1.00 dm3 of a
0.220 mol dm–3 solution of ethanoic acid.
Calculate the pH at 298 K of this buffer solution

for this question you do look at which is in excess why?

whats the difference and how do i know when to look at which is in excess
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Thread starter 3 years ago
#2
i can see for one it says the salt is added to the acid
the other the salt is dissolved in the acid
is this the decider?
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3 years ago
#3
(Original post by sohaail23)
i can see for one it says the salt is added to the acid
the other the salt is dissolved in the acid
is this the decider?
yes
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Thread starter 3 years ago
#4
(Original post by ChemChad)
yes
could you explain why please?
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3 years ago
#5
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3 years ago
#6
(Original post by ChemChad)
charco
There are two ways to make a buffer:

1. dissolve salt in weak acid
2. React weak acid with base making sure that the weak acid is in excess

Both of these result in a mixture of a weak acid and its salt.
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3 years ago
#7
base + acid => salt + water.
With buffers you need to find the concentration of the acid and the salt.

In the first question your given the details about the salt, but in the second question you need to find the concentration of the salt from equation. I hope this helped
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3 years ago
#8
(Original post by sohaail23)
could you explain why please?
The first example is not a reaction, you are just dissolving salt into acid to make your buffer so number of moles you have is number of moles you end up with.

In the second example you are making a buffer from a reaction between acid and base so a reaction is taking place and mole are being used up.
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3 years ago
#9
well good luck i hate buffers
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3 years ago
#10
(Original post by sohaail23)
Im really confused. Sometimes you have to work out which of the acid or salt is in excess and sometimes you dont.

An acidic buffer solution is obtained when sodium ethanoate is dissolved in aqueous
ethanoic acid.
(i) Calculate the pH of the buffer solution formed at 298 K when 0.125 mol of sodium
ethanoate is dissolved in 250 cm3 of a 1.00 mol dm–3 solution of ethanoic acid.
The acid dissociation constant, Ka, for ethanoic acid is 1.70 × 10–5 mol dm–3 at
298 K.

For this question you dont look at the excess why?

A buffer solution is formed when 2.00 g of sodium hydroxide are added to 1.00 dm3 of a
0.220 mol dm–3 solution of ethanoic acid.
Calculate the pH at 298 K of this buffer solution

for this question you do look at which is in excess why?

whats the difference and how do i know when to look at which is in excess
Btw if you find it difficult to distinguish between those 2 types of qs regardless of trhe explanations, I would recommend refering to the number of marks. Buffers qs are usually 6 marks whereas the qs where you need to find excess are usually 4...
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