Buffer Q3
Chemistry discussion, revision, exam and homework help.
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Buffer Q3This is from a old thread I copied here:
Two things I didn't understand
1)Final answer has to be 3 s.f. ?
2) What is the significance of doing this
[A-]/[HA] = 1.245
So in your final solution there must be more salt than acid,
Now you have to do a bit of stoichiometry.
you need to prepare a buffer solution of pH 3.987 from 10mL of 0.254 M solution of a weak acid whose pka is 3.892. what volume of 0.238 M NaOH would you need to add? (answer in mL)
Quite a difficult question...
ka = [H+][A-]/[HA] = 0.000103 * [A-]/[HA] = 0.000128
[A-]/[HA] = 1.245
So in your final solution there must be more salt than acid,
Now you have to do a bit of stoichiometry.
HA + NaOH --> NaA + H2O
initial moles of acid = 0.01 * 0.254 = 0.00254 mol
initial moles of salt = 0
let moles of acid reacted = x
from the equation after reaction moles of acid = 0.00254 - x
moles of salt formed = x
and x/(0.00254 - x) = 1.245
x = 1.245(0.00254 - x)
x = 0.00316 - 1.245x
2.245x = 0.00316
x = 0.00141
Therefore moles of acid reacting = 0.00141 and so moles of sodium hydroxide added = 0.00141
molarity of NaOH = 0.238
volume = 0.00141/0.238 = 0.0059
Therefore 5.9ml of NaOH must be added.
Like I said ... not straightforward!
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Re: Buffer Q3We need
![\frac{[A^- (aq)]}{[HA]}](http://www.thestudentroom.co.uk/latexrender/pictures/c5/c5a8830163bfaaa95512e99d6004976a.png "\frac{[A^- (aq)]}{[HA]}")
because it will enable us to find the answer. Its best to find this ratio first.
If you start with the pH you want your buffer to be, and the Ka of the weak acid, then you can find the ratio from either of these equations:
![K_a = \frac{[H^+ (aq)][A^- (aq)]}{[HA]}](http://www.thestudentroom.co.uk/latexrender/pictures/f7/f776ac6e1704e5934d892041bbeadace.png "K_a = \frac{[H^+ (aq)][A^- (aq)]}{[HA]}")
![pH=pK_a + \log\frac{[A^- (aq)]}{[HA]}](http://www.thestudentroom.co.uk/latexrender/pictures/98/9867cef538715939f8fcdbf156f19741.png "pH=pK_a + \log\frac{[A^- (aq)]}{[HA]}")
(The second is formed from the first by taking logs to base ten.)
The ratio is dependent on the volume of NaOH added in two ways:
NaOH will react with the HA, removing some of it,
In the process it will also release A^-(aq) ions from the ionic salt NaA, formed during the neutralisation reaction.
NaOH + HA -> NaA + H2O (2)
NaA + aq -> Na^+(aq) + A^-(aq)
What follows is an explanation of the method of the poster.
We need a little bit of algebraic manipulation to proceed. Let one of the unknown quantities be x and write the other in terms of it.
So the poster has found the moles of acid they started with and looked at how many will remain after the NaOH has been neutralised. This will be the initial number - number that have reacted.
Calling the number of HA that have reacted x, they can deduce [A^-(aq)]=x from (2).
So![\frac{[A^- (aq)]}{[HA]}= \dfrac{[\frac{x}{V}]}{[\frac{\mathrm{initial\ no.\ available} - x}{V}]}=\dfrac{[x]}{[\mathrm{initial\ no.\ available} - x]}= \mathrm{value\ calculated\ earlier}](http://www.thestudentroom.co.uk/latexrender/pictures/8c/8cb64e55226a3037709648010a7ff73c.png "\frac{[A^- (aq)]}{[HA]}= \dfrac{[\frac{x}{V}]}{[\frac{\mathrm{initial\ no.\ available} - x}{V}]}=\dfrac{[x]}{[\mathrm{initial\ no.\ available} - x]}= \mathrm{value\ calculated\ earlier}")
Also, we need not worry about the volume of solution since there's the same number of substances in the numerator and denominator of the fraction; the volumes will cancel so we may use moles directly rather than concentration.
Notice that we have assumed all the A^-(aq) formed is due to the neutralisation reaction and not the normal dissociation of the acid. This is because the acid is weak.
Once x is found we may find NaoH by using (2):
The moles of NaOH needed will be the same as the moles of acid that reacted.
Using the concentration of NaOH we may find V.Last edited by Killjoy-; 06-05-2012 at 01:12. -
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Re: Buffer Q3Yh I get that - we need the ratio to do the calculation but was asking if ratio was 0.75(a value less than1) for instance would it make any difference in calculation?!(Original post by Killjoy-)
We need because it will enable us to find the answer. Its best to find this ratio first.
If you start with the pH you want your buffer to be, and the Ka of the weak acid, then you can find the ratio from either of these equations:
(The second is formed from the first by taking logs to base ten.)
The ratio is dependent on the volume of NaOH added in two ways:
NaOH will react with the HA, removing some of it,
In the process it will also release A^-(aq) ions from the ionic salt NaA, formed during the neutralisation reaction.
NaOH + HA -> NaA + H2O (2)
NaA + aq -> Na^+(aq) + A^-(aq)
What follows is an explanation of the method of the poster.
So we can find the volume with a little bit of algebraic manipulation once the ratio is found by letting one of the unknown quantities be x and writing the other in terms of it.
So the poster has found the moles of acid they started with and looked at how many will remain after the NaOH has been neutralised. This will be the initial number - number that have reacted.
Calling the number of HA that have reacted x, they can deduce [A^-(aq)]=x from (2).
So![\frac{[A^- (aq)]}{[HA]}= \frac{[x]}{[\mathrm{initial\ no.\ available} - x]}](http://www.thestudentroom.co.uk/latexrender/pictures/8d/8d6ced4348e357f183ce858ebeb8e284.png "\frac{[A^- (aq)]}{[HA]}= \frac{[x]}{[\mathrm{initial\ no.\ available} - x]}")
Also, we need not worry about the volume of solution since there's the same number of substances in the numerator and denominator of the fraction; the volumes will cancel so we may use moles directly rather than concentration.
Notice that we have assumed all the A^-(aq) formed is due to the neutralisation reaction and not the normal dissociation of the acid. This is because the acid is weak.
I got hung up about
this
To me it sounds like the poster somehow uses the fact that the ratio is bigger than 1 !So in your final solution there must be more salt than acid,
Now you have to do a bit of stoichiometry. -
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Re: Buffer Q3I'm sure the calculation will be the same regardless.
If the ratio is greater than one that says more than half the acid has reacted with the alkali (since all the A- is from the acid.)
The ratio determines the pH for a given buffer system, and the ratio may be varied by the moles of alkali used.
And this in turn is dependent on the volume of alkali used for a given concentration of it. -
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Re: Buffer Q3That's what I was looking for
shall +rep you tmr
Pls explain a bit more - also H+ is from the acidIf the ratio is greater than one that says more than half the acid has reacted with the alkali (since all the A- is from the acid.)
Salt is from A-
or am I talking nonsense! -
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Re: Buffer Q3I didn't say H+ wasn't from the acid.(Original post by arvin_infinity)
Pls explain a bit more - also H+ is from the acid
Salt is from A-
or am I talking nonsense!
A- is from the salt which is formed from the acid.
Remember earlier I explained that if you looked at the equations and molar ratios boiled down to ?
Well you can see from here that if
less than half of the acid would remain at equilibrium. (More than half of the acid has reacted when equilibrium has reached.)
The ratio would therefore be more than one. -
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Re: Buffer Q3I dunno if my calculations is wrong or what - cuz when I substitue some numbers into(Original post by Killjoy-)
I didn't say H+ wasn't from the acid.
A- is from the salt which is formed from the acid.
Remember earlier I explained that if you looked at the equations and molar ratios boiled down to ?
Well you can see from here that if less than half of the acid would remain at equilibrium. (More than half of the acid has reacted when equilibrium has reached.)
The ratio would therefore be more than one.
it just doesn't work!
also couldn't figure out where the 2 comes from! -
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Re: Buffer Q3The two is present because you want the amount of acid present at equilibrium to be less than half of its initial amount so that the ratio is greater than one.(Original post by arvin_infinity)
I dunno if my calculations is wrong or what - cuz when I substitue some numbers into
it just doesn't work!
also couldn't figure out where the 2 comes from!
I've put numbers in, it works.
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Re: Buffer Q3Now I can see how you got 2 - I just equate the whole thing to 1 and then x=number of moles/2(Original post by Killjoy-)
The two is present because you want the amount of acid present at equilibrium to be less than half of its initial amount so that the ratio is greater than one.
I've put numbers in, it works.
but still didn't get what you are trying to say !
Having said that we didn't need to do this to prove the ratio is greater than one! the ratio is greater than because we calculated it! -
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Re: Buffer Q3That's exactly it. (I assumed you wanted to know when the ratio will be greater than one.)(Original post by arvin_infinity)
Having said that we didn't need to do this to prove the ratio is greater than one! the ratio is greater than because we calculated it!
Whether or not the ratio is greater than one matters not when answering the question. -
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Re: Buffer Q3DO you reckon the answer should be in 3 s.f.!?(Original post by Killjoy-)
That's exactly it. (I assumed you wanted to know when the ratio will be greater than one.)
Whether or not the ratio is greater than one matters not when answering the question. -
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Re: Buffer Q3In an exam I'd give an answer to 3s.f., which is usually what is asked.(Original post by arvin_infinity)
DO you reckon the answer should be in 3 s.f.!?
Normally you go with the accuracy of your least precise value. -
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Re: Buffer Q3Exactly and the least degree of accuracy (according to other values) is 3 s.f.! right(Original post by Killjoy-)
In an exam I'd give an answer to 3s.f., which is usually what is asked.
Normally you go with the accuracy of your least precise value.
I only say that cuz the poster's final answer is 2 s.f. -
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Re: Buffer Q3I put the answer as 5.9 ml (2s.f.) rather than 5.92 ml because I know not of any apparatus in common use that will allow me to measure volumes of liquid to an accuracy of hundredths of a millilitre.(Original post by arvin_infinity)
Exactly and the least degree of accuracy (according to other values) is 3 s.f.! right
I only say that cuz the poster's final answer is 2 s.f.
It was a common-sense answer. -
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Re: Buffer Q3Oh yeh that makes sense now!(Original post by charco)
I put the answer as 5.9 ml (2s.f.) rather than 5.92 ml because I know not of any apparatus in common use that will allow me to measure volumes of liquid to an accuracy of hundredths of a millilitre.
It was a common-sense answer.
but then for exam purpose we dont need to think that far! or do we!? I guess in exam purposes they are only interested in the answer! and not so much on the possibility of actually measuring that vol in lab
shall +rep you tmr
