# Acid-base equilibria help

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#1
Q. What volume of 1.00 mol dm-3 sodium hydroxide must be added to 100 cm3 of 1.00 mol dm-3 ethanoic acid solution to make a buffer solution of pH = 4.44? (Ka for ethanoic acid = 1.80 x 10-5 mol dm-3)

The way I did it, my answer comes up to be 50 cm3, while the correct answer is supposed to be 33.33 cm3. I've attached the answer, but I don't understand why it states that one-third of the acid must react.
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6 years ago
#2
Do you understand why the ratio of salt (conjugate base) and acid is 1:2?

You have 1 part of base and 2 parts of acid. Three parts together. That means one third of the initial amount of acid was neutralized.
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#3
(Original post by Borek)
Do you understand why the ratio of salt (conjugate base) and acid is 1:2?

You have 1 part of base and 2 parts of acid. Three parts together. That means one third of the initial amount of acid was neutralized.
But um. Isn't the acid and the base reacting in a 1:1 ratio? Where do you get the 1:2 ratio from?
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6 years ago
#4
You can calculate ratio of concentrations of the acid and its conjugate base from Ka and pH (that yields 1:2).

I guess half of your problems start with confusing base used for acid neutralization with the acid conjugate base. Can you name them both? Can you say how they are related to the salt present in the solution?
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#5
(Original post by Borek)
You can calculate ratio of concentrations of the acid and its conjugate base from Ka and pH (that yields 1:2).

I guess half of your problems start with confusing base used for acid neutralization with the acid conjugate base. Can you name them both? Can you say how they are related to the salt present in the solution?
To be honest, I didn't even realise I need to consider the difference between the base used for acid neutralization and the acid conjugate base while doing these calculations. So the reaction is: NaOH + C2H5OH -> Na+ + C2H5O-

So here, NaOH is the base, and its conjugate acid is Na+.
And C2H5OH is the acid and its conjugate base is C2H5O-. And since the salt is C2H5ONa, the conjugate acid + conjugate base gives us the salt.

Is that correct?
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6 years ago
#6
(Original post by You-know-who)
Q. What volume of 1.00 mol dm-3 sodium hydroxide must be added to 100 cm3 of 1.00 mol dm-3 ethanoic acid solution to make a buffer solution of pH = 4.44? (Ka for ethanoic acid = 1.80 x 10-5 mol dm-3)

The way I did it, my answer comes up to be 50 cm3, while the correct answer is supposed to be 33.33 cm3. I've attached the answer, but I don't understand why it states that one-third of the acid must react.
As the sodium hydroxide is added, it reacts with the ethanoic acid, effectively converting it into sodium ethanoate. You need to add enough sodium hydroxide to convert a third of the ethanoic acid to sodium ethanoate, leaving the other two thirds unreacted.

This will leave the sodium ethanoate and the ethanoic acid in a 1:2 ratio.
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6 years ago
#7
(Original post by You-know-who)
So here, NaOH is the base, and its conjugate acid is Na+.
And C2H5OH is the acid and its conjugate base is C2H5O-. And since the salt is C2H5ONa, the conjugate acid + conjugate base gives us the salt.
To be an acid (conjugate pair or otherwise) it has to be a proton donor. Na+ therefore cannot be an acid.

Can you work out what the conjugate acid is?
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6 years ago
#8
(Original post by You-know-who)
To be honest, I didn't even realise I need to consider the difference between the base used for acid neutralization and the acid conjugate base while doing these calculations. They usually come in the same amounts, but you need to know what you are doing to not make a stupid mistake.

C2H5OH is the acid
What is formula of the ethanoic acid?
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#9
As the sodium hydroxide is added, it reacts with the ethanoic acid, effectively converting it into sodium ethanoate. You need to add enough sodium hydroxide to convert a third of the ethanoic acid to sodium ethanoate, leaving the other two thirds unreacted.

This will leave the sodium ethanoate and the ethanoic acid in a 1:2 ratio.
Umm. WHY do we need to convert only a third of the ethanoic acid? (Original post by Pigster)
To be an acid (conjugate pair or otherwise) it has to be a proton donor. Na+ therefore cannot be an acid.

Can you work out what the conjugate acid is?
Right, so:

OH- + CH3COOH -> CH3COO- + H2O
So the conjugate acid is... H2O?

(Original post by Borek)
What is formula of the ethanoic acid?
Oops sorry, I meant CH3COOH.
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6 years ago
#10
(Original post by You-know-who)
Umm. WHY do we need to convert only a third of the ethanoic acid? Because [salt]/[acid] needs to be 1/2 (as shown in the attachment to your original post).

So the ratio [salt]:[acid] needs to be 1:2.

In other words, we need to end up with twice as much acid as salt in the solution.

At the beginning it is all acid, no salt. So we have to gradually convert some of the acid into salt (by adding NaOH solution).

By the time one hundredth of it is converted into salt, the other ninety-nine hundredths are still acid, so there is 99 times as much acid as salt. This is far too much acid, so we carry on converting more.

By the time one twentieth of it is converted into salt, the other nineteen twentieths are still acid, so there is 19 times as much acid as salt. This is still far too much acid, so we carry on converting more.

By the time one tenth of it is converted into salt, the other nine tenths are still acid, so there is 9 times as much acid as salt. This is still too much acid, so we carry on converting more.

By the time one fifth of it is converted into salt, the other four fifths are still acid, so there is 4 times as much acid as salt. This is still a bit too much acid, so we carry on converting more.

Eventually one third of it is converted into salt, the other two thirds are still acid, so there is twice as much acid as salt. This is what we want so we stop.
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#11
Because [salt]/[acid] needs to be 1/2 (as shown in the attachment to your original post).

So the ratio [salt]:[acid] needs to be 1:2.

In other words, we need to end up with twice as much acid as salt in the solution.

At the beginning it is all acid, no salt. So we have to gradually convert some of the acid into salt (by adding NaOH solution).

By the time one hundredth of it is converted into salt, the other ninety-nine hundredths are still acid, so there is 99 times as much acid as salt. This is far too much acid, so we carry on converting more.

By the time one twentieth of it is converted into salt, the other nineteen twentieths are still acid, so there is 19 times as much acid as salt. This is still far too much acid, so we carry on converting more.

By the time one tenth of it is converted into salt, the other nine tenths are still acid, so there is 9 times as much acid as salt. This is still too much acid, so we carry on converting more.

By the time one fifth of it is converted into salt, the other four fifths are still acid, so there is 4 times as much acid as salt. This is still a bit too much acid, so we carry on converting more.

Eventually one third of it is converted into salt, the other two thirds are still acid, so there is twice as much acid as salt. This is what we want so we stop.
The attachment is the answer. That part's obviously not given in the question. So is there any way I can know that I need to convert one third of the acid into salt JUST by looking at the question? Because without that information in the question, I worked it out the way I normally would, and according to the equation, the salt and acid react in a 1:1 ratio. And my answer comes out to be 50, while the correct answer is 33.33.
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6 years ago
#12
(Original post by You-know-who)
The attachment is the answer. That part's obviously not given in the question. So is there any way I can know that I need to convert one third of the acid into salt JUST by looking at the question? Because without that information in the question, I worked it out the way I normally would, and according to the equation, the salt and acid react in a 1:1 ratio. And my answer comes out to be 50, while the correct answer is 33.33.
But what you are doing is asking for an explanation of how to do something, for which you've already attached an explanation, and are asking how you would be able to do it without looking at an explanation!!!

The attachment is the answer. No, the attachment isn't just the answer – it is a step by step explanation of how to get the answer, which is what you are looking for!

That part's obviously not given in the question.
This is simply saying a step by step explanation of how to get the answer is not given in the question.

So is there any way I can know that I need to convert one third of the acid into salt JUST by looking at the question? Yes there is a way – and it's explained in the attachment! The first six lines of the attachment explain how you can know the salt and acid have to be in a 1:2 ratio JUST by looking at the question. Edit: and my earlier posts in this thread carry on from there and explain why that means a third of the acid has to be converted into salt.

Because without that information in the question, That information is in the question but hidden. The first step is to find it, as explained in the first six lines of the attachment

I worked it out the way I normally would, and according to the equation, the salt and acid react in a 1:1 ratio. And my answer comes out to be 50, while the correct answer is 33.33

The way that you normally would is wrong. Bin it. Do it the way described in the attachment.

If you still don't understand, I think it's probably because you are not explaining your confusion clearly. I suggest there are two things you can try.

1. Actually clarify whether or not you understand the first six lines of the attachment which – as you may have noticed I'm trying to imply – answer the question you are asking here. If you don't understand those lines then, fair enough, say what the issues are and we can discuss them further.

2. Post how you've done the calculation yourself and we can discuss where you've gone wrong.
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#13

1. Actually clarify whether or not you understand the first six lines of the attachment which – as you may have noticed I'm trying to imply – answer the question you are asking here. If you don't understand those lines then, fair enough, say what the issues are and we can discuss them further.

2. Post how you've done the calculation yourself and we can discuss where you've gone wrong.
Okay, I think I get it now.

We know that the alkali and the acid react in a 1:1 ratio, but since we are FORMING a buffer solution where the weak acid must react with the salt once it forms, we need to find the ratio in which the acid and salt REACT once it forms.

Hence the 5th and 6th lines of the attachment showing the calculations. The calculations show that salt and acid react in a 1:2 ratio, i.e. 2 mol of the acid reacts with 1 mol of the salt. Now if considered together with the part of the acid reacting with the alkali, one-third of the acid reacts with the alkali, while two-third remains unreacted. Since the mol ratio AND the concentration of both the acid and the alkali are the same, we can calculate the volume of the alkali by finding one-third of the initial volume of acid added, i.e. one-third of 100 cm^3. Hence, we end up with 33.33 cm^3. Does that sound right?
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6 years ago
#14
(Original post by You-know-who)
Okay, I think I get it now.

We know that the alkali and the acid react in a 1:1 ratio, but since we are FORMING a buffer solution where the weak acid must react with the salt once it forms, we need to find the ratio in which the acid and salt REACT once it forms.

Hence the 5th and 6th lines of the attachment showing the calculations. The calculations show that salt and acid react in a 1:2 ratio, i.e. 2 mol of the acid reacts with 1 mol of the salt. Now if considered together with the part of the acid reacting with the alkali, one-third of the acid reacts with the alkali, while two-third remains unreacted. Since the mol ratio AND the concentration of both the acid and the alkali are the same, we can calculate the volume of the alkali by finding one-third of the initial volume of acid added, i.e. one-third of 100 cm^3. Hence, we end up with 33.33 cm^3. Does that sound right?
Excellent, that's almost there.

The remaining misconception is this: a buffer solution contains a weak acid and a salt (of the same weak acid), but these don't react with one another – they are merely present in the same solution. The salt can be produced by reacting (some of) the weak acid with NaOH. Let me rewrite your explanation but with corrections, to see if that makes it easier. I have added a few bits in brackets for clarity.

We know that the alkali and the acid react in a 1:1 ratio, but since we are FORMING a buffer solution where the weak acid must be present in the same solution as the salt once it forms, we need to find the ratio in which the acid and salt exist once it forms.

Hence the 5th and 6th lines of the attachment showing the calculations. The calculations show thatsalt and acid exist in a 1:2 ratio, i.e. 2 mol of the acid are present with (every) 1 mol of the salt. Now if considered together with the part of the acid reacting with the alkali, one-third of the acid reacts with the alkali, while two-third remains unreacted. Since the mol ratio (in which the acid reacts with the alkali is 1:1) AND the concentration of both the acid and the alkali are the same, we can calculate the volume of the alkali by finding one-third of the initial volume of acid added, i.e. one-third of 100 cm^3. Hence, we end up with 33.33 cm^3. Does that sound right?

Now I think it sounds about right. Note that the ratio of salt to acid is 1:2 only for a buffer of that particular pH. For a buffer of a different pH, the ratio will be different.

In fact you can see from the calculation that the ratio of salt to acid will always be

Ka :10-pH

which, in this case, is 1:2
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#15
Excellent, that's almost there.

The remaining misconception is this: a buffer solution contains a weak acid and a salt (of the same weak acid), but these don't react with one another – they are merely present in the same solution. The salt can be produced by reacting (some of) the weak acid with NaOH. Let me rewrite your explanation but with corrections, to see if that makes it easier. I have added a few bits in brackets for clarity.

We know that the alkali and the acid react in a 1:1 ratio, but since we are FORMING a buffer solution where the weak acid must be present in the same solution as the salt once it forms, we need to find the ratio in which the acid and salt exist once it forms.

Hence the 5th and 6th lines of the attachment showing the calculations. The calculations show thatsalt and acid exist in a 1:2 ratio, i.e. 2 mol of the acid are present with (every) 1 mol of the salt. Now if considered together with the part of the acid reacting with the alkali, one-third of the acid reacts with the alkali, while two-third remains unreacted. Since the mol ratio (in which the acid reacts with the alkali is 1:1) AND the concentration of both the acid and the alkali are the same, we can calculate the volume of the alkali by finding one-third of the initial volume of acid added, i.e. one-third of 100 cm^3. Hence, we end up with 33.33 cm^3. Does that sound right?

Now I think it sounds about right. Note that the ratio of salt to acid is 1:2 only for a buffer of that particular pH. For a buffer of a different pH, the ratio will be different.

In fact you can see from the calculation that the ratio of salt to acid will always be

Ka :10-pH

which, in this case, is 1:2
Thank you SO much for clearing that up! One last point, the concentrations of acid and alkali here are equal, but if they weren't, i.e. if the concentration of NaOH was 2 mol dm-3 while 100 cm3 of a 1 mol dm-3 ethanoic acid was used... would this be the volume of alkali then:

One-third of the acid would still react with the alkali, since the acid-alkali mol ratio is still 1:1, thus, one-third of 100 cm3 = 33.33 cm3.

Thus, the mol of acid that reacts with the alkali in 33.33 cm3 of acid is:
33.33 x 1 / 1000 = 0.03333 mol

The same mol of alkali reacts. Thus, in 2 mol dm-3 of alkali, the volume that reacts with the acid is:
0.03333 x 2 / 1000 = 16.665 cm3?
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6 years ago
#16
(Original post by You-know-who)
Thank you SO much for clearing that up! One last point, the concentrations of acid and alkali here are equal, but if they weren't, i.e. if the concentration of NaOH was 2 mol dm-3 while 100 cm3 of a 1 mol dm-3 ethanoic acid was used... would this be the volume of alkali then:

One-third of the acid would still react with the alkali, since the acid-alkali mol ratio is still 1:1, thus, one-third of 100 cm3 = 33.33 cm3.

Thus, the mol of acid that reacts with the alkali in 33.33 cm3 of acid is:
33.33 x 1 / 1000 = 0.03333 mol

The same mol of alkali reacts. Thus, in 2 mol dm-3 of alkali, the volume that reacts with the acid is:
0.03333 x 2 / 1000 = 16.665 cm3?
Good question. Yes, that is completely correct. The way you've done it ( i.e. the three steps you've written down) is a very nice and logical way of setting it out.
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#17
Good question. Yes, that is completely correct. The way you've done it ( i.e. the three steps you've written down) is a very nice and logical way of setting it out.
Thanks again for all your help! 0
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