4.7 Acid/base equilibria
Posted by: intechemistry on: September 23, 2010
Update 20 Sept 2011: James Mungall videos: 1. Buffer solutions – pH of a buffer, and 2. Buffer solutions – change in pH of a buffer followed by 3. Buffer solutions – change in pH on adding same amount of HCl to water. James has an Acid/Base tutorial section on his website.
pKa is usually determined by adding a strong base to a weak acid, and reading off the pH at half the equivalence. If instead a weak acid is added to a strong base, the pKa would be the pH at TWICE the equivalence volume. I’m sure I said to some of you that it would be 1.5 times the volume and this is incorrect. In the latter senario, the [HA] and [A-] cancel at twice the volume of equivalence.
The practice questions are found in the file depository.
40 Responses to "4.7 Acid/base equilibria"
December 13, 2010 at 1:23 am
Salam.
There’s a couple few ways to make buffer solutions. One way is to mix a weak acid and its salt together in water. e.g. mix ethanoic acid (a weak acid) with sodium ethanoate (the salt* of ethanoic acid). On mixing in water, you have your buffer solution.
When a (different) acid is added, e.g. HCl(aq), the salt (which ionised in water – remember all sodium compounds are soluble and ionise in water) being CH3COO-, can accept the proton from the HCl(aq). CH3COOH is formed and because what’s formed is a weak acid, it doesn’t# release its proton into the aq solution. The effect is we added HCl(aq) to the buffer but the H+(aq) from the HCl(aq) was ‘consumed’, so there is no decrease in pH value (i.e. the solution doesn’t move to a lower pH value). The solution has resisted a change in pH when an acid was added to it.
Instead of adding HCl(aq) to the buffer, you might want(?) to add NaOH(aq) instead. If you did so, the NaOH(aq) base reacts with the weak acid. The weak acid gives up it’s H+ to the OH- and forms H2O, so here we added OH- ions to the solution but they were ‘consumed’, so the [OH-] didn’t increase hence the pH didn’t increase (i.e. the solution didn’t become more basic). The solution has resisted a change in pH when a base was added to it.
It’s a bit like LeChateliers principle, the buffer solution resists any pH chance you try to make to it.
Eventually however, if you keep adding aq. HCl (or aq.NaOH) then all the salt (or acid respectively) will be consumed and the solution will show changes in pH.
OK. There is some small change in pH when the aq.HCl or aq.NaOH is added but the change is a lot smaller than would be the case if HCl(aq) or NaOH(aq) were just added to water. The buffer resists changes in pH on additions of small amounts of acid (or base). For good buffer performance you need to mix almost equal concentrations of the weak acid and its salt.
Notes:
* There are many salts of ethanoic acid. e.g. potassium ethanoate or ammonium ethanoate, but pick the sodium salt. It makes life simpler 
# actually, to say it ‘doesn’t# release its proton’ is just an approximation. The CH3COO- doesn’t completely consume the H+ when it forms the weak acid CH3COOH, because the CH3COOH will ionise slightly. The point is, the CH3COOH doesn’t relase as much H+ compared to the equivalent number of moles of the strong acid HCl, so the pH change isn’t as much as it would otherwise be.
Does that help?
January 14, 2011 at 11:10 am
Nice sir..
Now i’m more clear of what buffer is by reading this…
=D
I have a question…
Does the auto-ionization of H2O molecule determines the pH of a solution?
January 14, 2011 at 9:36 pm
The auto-ionisation is responsible for the pH in neutral (e.g. a solution of NaCl) or basic (e.g. a soln of NaOH) conditions because in such cases, the water is the only source of H+.
But if an acid (pKa <<7) is added to water, then, the auto-ionisation isn't the most significant source of H+ (and acid you added is!) so doesn't determine the pH
January 16, 2011 at 8:47 pm
Salam, sir:) May I know why when weak acid is diluted by a factor of 10, its pH only increases by half a unit? Is it because the ionisation of weak acid is reversible, so adding water produces more hydrogen ions, making the pH to increase less than expected?
January 16, 2011 at 11:48 pm
That was crertainly my understanding.
A strong acid in water, e.g. HCl, is all H+(aq) and Cl-(aq). Diluting by a factor of 10 would mean the [H+] falls by a factor of ten and hence pH would increase by one unit.
But in the case of a weak acid,
HA + H2O H3O+(aq) + A-(aq),
adding H2O ‘begins’ to push the eqm to the right, and the H+ increases. There is more aq. volume present, so the H+ has been diluted… [H+] falls, so pH gets larger {like the strong acid example above}, but there are are greater amount of H+ ions in the solution than before, so the [H+] doesn’t go down as much as if you only considered the dilution alone. Hence the pH doesn’t increase as much. But adding more water doesn’t push the eqm very far to the RHS; HA is a weak acid after all. It’s ionisation is pretty lousy.
January 18, 2011 at 5:38 pm
salam sir
I have question about the half equivalence point. Is there any exact reason why do we assume that at half equivalence point the [HA]=[A-]?
thank you sir
January 24, 2011 at 12:01 pm
Salam sir…
From the half-equivalence point, we could also determine the range of pH of the buffer, right?
thank you..
January 30, 2011 at 12:47 am
#7, sya, January 18, 2011 at 5:38 pm
Salam.
I’m not too sure I understand why you asking this question, so let me just say this:
The equivalence point is where the number of moles of base added exactly neutralises all the H+ from the acid originally present. So at the neutralisation point, all the HA has been consumed and will change into the salt, A-.
If the volume of base needed to do this is say, x, then at a volume of x/2 half the amount of acid is neutralised and the other half changed into salt. i.e. amount of HA = amount of A-
It’s “assumed” because its what happens.
I’m not sure if I’ve managed to address your inquiry properly. What did you have in mind when you asked it?
January 30, 2011 at 1:35 am
#8, amirul arif, January 24, 2011 at 12:01 pm
Salam.
You said “From the half-equivalence point, we could also determine the range of pH of the buffer, right?”
Well, we don’t usually talk about a ‘range’ for a buffer. You are right in that the volume at half equivalence (when adding a base to a weak acid) is very important with regards to buffers, ‘cos it’s at this point {i.e. half-equivalence} at which the maximum (or best) buffering capacity can be achieved.
Like I said, we don’t really discuss ‘ranges’ for buffers, although I guess one could do so if one had to. Perhaps a reasonable/practical ‘range’ for the buffer would be, at most, +/- 0.5 from the pKa of the weak acid (or +/- 0.5 of the pKb for a weak base), but I think best not to mention ‘ranges’ for buffers as one might confuse it with indicators, to whom ‘range’ has some meaning – i.e. the pH range over which they completely change colour.
February 2, 2011 at 5:08 pm
Salam sir..
Are there any assumptions can be made for a basic buffer solution?
Are they the same as in assumptions for the acidic buffer solution?
Thank you
February 2, 2011 at 10:54 pm
Salam Arif
The usual sssumptions are made for basic buffer (wB & conjugate acid) is that the salt is fully ionised and the weak component, wB in this case, is only slightly ionised so that [wB] initial = [wB] at eqm and the autionisation of water is relative smaller so can be neglected in evaluating concentration of species.
It’s the same for acid buffer solutions but swap considerations from wB to wA instead.
March 5, 2011 at 1:24 am
According to George Facer pg 100, regarding [Acid]eq = [Acid]final, GF states ‘However, with weak acids such as HF and HNO2, which are stronger than propanoic acid, the approximation should not be made’
But in pg 118, Q13, it made the assumption above and the answer with the assumption is 4.56×10-4 M. Without the assumption, the answer should be 4.788×10-4 M (assuming of course i did not calculate wrongly).
So we do meet such questions in exam, when do we or when do we not make that assumption ? What if the answer scheme defers with our opinions whether the assumption is justified ?
March 5, 2011 at 10:54 am
You are not expected to be able to look at a collection of weak acids and then decide where the approximation can be used or not.
You will be expected to use the [H+]final = [Acid ompound]initial approximation.
You may however be asked about the assumptions used when doing that calculation. Actaully you are almost certainly going to be asked about the assumptions when weak speciaes are involved.
Why then is Facer doing what he does about GF etc? When applying scientific models it’s important to understand the limitations those models have. Facer, for the sake of accuracy, is showing where this model breaks down. In my opinion, Facer, and a couple of other authors of edexcel material, is/are good at looking one step beyond the ‘all encmpassing’ model.
If he didn’t mention the HF thing, people would think that the approximation was always valid, which it’s not.
So he used the approximation in the calculation for Q13 – as is the expectation for exams, but as you demonstrate perfectly, the students should realise that it wasn’t valid here.
March 5, 2011 at 11:19 am
Awright.. thanks for the explanation… btw, this General Question area rocks =)
March 5, 2011 at 12:33 pm
Your Q was on a specific topic so I shifted it here. Glad you like the General Questions page though There is quite a bit of ‘clarification’ elements there.
March 5, 2011 at 10:53 pm
salam
sir, in the case of dilution of acids:
1) when strong acid is diluted by a factor of 10, the pH increases by one unit
2) when weak acid is diluted by a factor of 10, the pH increases by 0.5 unit
we can prove the statement thru calculation to find pH of strong acid and acid dissociation constant.
but, i confused when it comes to explanation:
- addition of water decreases the [H+]
- but at the same time causes the equilibrium to shift to the RHS, producing more H+ ions (i understand up to this point)
- pH tends to decrease
now, why the explanation contradicts the calculation?
March 6, 2011 at 3:59 pm
Salam
- addition of water decreases the [H+] => yes, so the pH increases
adding water to a strong acid cannot increase [H+] as all h+ was already ionised, so only a proportionate fall in [H+] and increase in pH can occur.
Adding water to the weak acid will also dilute the H+ hence lower the [H+] but because there is still HA molecules still present (remember HA only slightly ionised initially) then some more H+ will dissociate from HA. But because it is a weak acid the amount of extra H+ produced is small yet the dilution was big. so overall the [H+] falls and the pH increases.
If you were confused about this it’s likely because you didn’t know the relative degree (or significance) of the diltuion factor compared to the increase of H+.
Does it make sense now?
March 6, 2011 at 4:05 pm
so, from the explanation we could say that although the equilibrium shifts to the RHS and producing H+ ions,but overall [H+] falls and thus the pH increases.
? is that how we explain the increase in pH after dilution?
March 6, 2011 at 4:15 pm
Yes.
It shifts to the RHS, so more H+ is present, but it’s present in a more significant amoubt of water Hence the [H+] drop i.e. pH increase isn’t as much as we would expect.
If only dultion happened. a 10x dilution would cause pH dor increase by 1.
A 10x dilution followed by more ionisation would cause the pH to increase by less than one
April 20, 2011 at 11:40 pm
A question from M2 Lyn…
pH increases as the acid is diluted. This is cos conc of H+ decreases. But at the same time the eqm shifts to the right ‘cos of de presence of water. So does that mean conc of H+ is more influential?
I’m not too sure what you mean Lyn.
wA + H2O conj base of wA + H3O+
Adding H2O pushes the eqm to the RHS yes, but only marginally. The wA is still just a weak acid, it’s not now going to fully ionize (i.e. suddenly become a strong acid).
The shift to the RHS isn’t very much, plus in most cases there already was a lot of water present, adding a bit more to already large number wont make much change. So the the effect is on dilution of a weak acid, the pH increases, but slightly less than one might expect if considering dilution in isolation.
Errrm. Is that of any help?
April 29, 2011 at 10:19 pm
salam alayk,
sir… how small a value of concentration of acid have dissociated so that we can use the approximation of the concentration of the acid in equilibrium is the same as initially?
i referred to GF A2 book pg 100
thank you sir
April 30, 2011 at 3:53 pm
Salam Nadzriah.
I’m sure that at A-level, you will always use the approximation [acid initial] = [acid at eqm] and for all weak acids. I more than suspect that GF is just trying to tell you (as he does once in a while) that the simplifications we make are not always applicable in all cases.
If you ever had to abandon the approximation, the question almost certainly will guide you in such a way so that you should realise that you shouldn’t use it, but such a scenario is not going to be common.
The models at A-level are usually good enough to work the vast majority of times and with good accuracy. There isn’t much point is spending a disproportionate amount of time on ‘fringe’ or specialist chemistry.
At this level, it’s general principles that they want you to learn, so by far most of the time the general chemistry is what you need to know. On rare occasions there may be some ‘fringe’ chemistry to distinguish distinguish a grade A capable student from one of a lower standard, but that’s so unlikely because they can distinguish chemistry knowledge by asking ordinary conventional(standard) questions.
I hope that helps and reassures.
May 4, 2011 at 7:33 am
salam sir
if a question asks about the end-point of titration, do we take the middle point on the vertical section of graph, or the last point marked on the graph?
October 25, 2011 at 6:05 pm
Hello Sir. Why does the dilution by x factor increases the pH of strong acid by one unit whilst for weak its ny half. and the questions under this spec requiremnt is ny entitled to cal of pH ( by certain factor dil) and reasonate why rite – edexcel?
I find the blog the quickest way to get good answers ,thanks Sir.
October 25, 2011 at 6:29 pm
and Sir, the reason why large reservoir of salt is needed (despite the fact that the weak acid already have the dual-function*) is to only to assist , in reacting with the added H + ions right?
*as A- ions also obtainable as weak acid dissociates .
October 26, 2011 at 2:40 pm
salam sir, if we sketch the graph of titration of weak acid against strong base, the vertical component of pH should increase by 4 unit like you said in the class right? so if we start at pH 6 it will stop at pH 10 right? next, when we plot the graph, should we finish it at pH above 12?
October 27, 2011 at 5:38 pm
#29, Magneto, October 25, 2011 at 6:05 pm
This question often comes up so I’ll talk about it in a bit more detail for the benefit of others.
The final pH depends on what x is. This is because pH is concentration dependent.
If x was 10, i.e. we diluted it by a factor 10 (which gives an easy to see example), we get this:
Say we had 0.1 mol dm-3 HCl, being a strong acid, essentially all the HCl(aq) has ioinsed into H+ and Cl-. The pH of this solution will therefore be 1. Diluting it by a factor of 10 gives the same amount of H+ in ten times the volume so the conc is now 0.01 mol dm-3 for which the pH of the solution will be 2. The only thing happening here is diluting the amount of H+.
For a weak acid (wA), e.g.ethanoic acid or the convenient symbol, HA, not only do we simply just consider the dilution, but we must also consider the effect upon the equilibrium of the weak acid.
In solution, most of the weak acid, HA, does not ionise.
We have to use the Ka equation to calc pH for weak acids.
pH = -0.5 lg (Ka * [HA]). Ka for ethanoic acid = 1.74 x 10^-5. Initial [HA] = 0.1
Assumption [H+] was = [A-] and [HA] at eqm is same as inition [HA] gives: pH = 2.88
If conc now goes to 0.01, repeating the calc gives pH=3.38.
The effect of diluting the HA by x10 gave a pH increase of 0.49, i.e. half the increase we got previously.
Mathematically: See the 0.5 part of the pH equation of the weak acid?
Before you multiply by 0.5, the Ka*Ha part has changed by a factor of 10 (when going from a 0.1 to a 0.01 molar solution). The log of a ten times difference is 1. Multiply 1 by a half, you will get a pH change of 0.5
That’s where the answer of 0.49 came from (allow the rounding error) The 0.5 in the equation came from the fact you assumed [H+} was = [A-] which generated a [H+]^2 term, which on rearrangement gave -lg(Ka * [HA])^2 and the ^2 came down to give -0.5 lg (Ka * [HA])
Chemically: when you dilute the wA, the [H+] falls. The eqm will shift to the RHS to produce more H+, (but as with all equilibrium shifts, it doesn’t shift “fully”). The eqm quotient on dilution no longer equals Ka, so the quotient adjusts to once again to = Ka. The [A-] Also changes in line with [H+} and there are two terms changing on the numerator yet only one on th denominator, so the small increase in H+ and the same small increase in A- along with the small decrease in HA will give the Ka value again.
A similar concept was discussed with the 3H2 + 2N2 = 2NH3 equilibria.
I hope that helps.
October 27, 2011 at 5:44 pm
#30, Magneto, October 25, 2011 at 6:29 pm
Yes. You need the large reservoir of salt because when the HA ionises, you only get a tiny bit of A-. If you relied on that amount of A- to react with any added H+ ions, very very quickly the A- would be fully consumed and the solution could no longer resist a decrease in pH when H+ ions are added. The buffer capacity would be really lousy if only the weak acid was used to provide the A-.
October 27, 2011 at 5:54 pm
#31, sakinah, October 26, 2011 at 2:40 pm
Salam.
“so if we start at pH 6 it will stop at pH 10 right? next, when we plot the graph, should we finish it at pH above 12?” – For the purposes of A-level, for a weak acid and a strong base, you should start to draw vertical sections from pH7, not pH6. Where did you get the notion of drawing it from pH6?
This ‘old syllabus’ page (75% relevance to your syllabus) gives a quick example of such a graph: http://tinyurl.com/3mcn89z
Where to stop drawing? If you are given data, you must use the data to calculate the final pH of the solution (remember to factor in the TOTAL volume of the acid/base mixture). If it’s just a sketch then a post-vertical tail going from about 11 to 13 would probably suffice, but for sketches, I don’t think that aspect is focused upon in terms of mark allocation.
May 19, 2012 at 9:57 pm
Sir, normally if we’re given a acid/base titration graph and the question asked for buffer section, it is the horizontal section of the graph before neutralisation point right. my question is:
If the graph given or we draw on previous question is a reflected one, say, weak acid titrated with strong alkali. Then we would start at pH 14 then to pH3. So where is the buffer region?? I’d put the horizontal region before neutralisation too, meaning around pH 13 dropping to pH11 that region. yes?no?
May 20, 2012 at 11:59 am
35, sleep deprived bug, May 19, 2012 at 9:57 pm
Yes it’s the most horizontal region (Note: there is no purely horizontal region as the pH will ALWAYS change to some degree) before pH7 if you have a weak acid and it’s salt as Edexcel usually ask you about.
If you had a weak base and it’s salt the buffer would be the most horizontal region after pH7 (I don’t think I’ve ever seen edexcel ask about these systems…. but they may do!)
“weak acid titrated with strong alkali” means you have a sample of weak acid and you’re adding NaOH. So the weak acid is the conical flask, therefore starting pH will be something like 3 to 5. That’s actually the usual situation edexcel present you with in their questions.
The inverted one will be a base (strong or weak) titrated with an acid, in which case the base is in the conical flask and acid is gradually added to it. In this case you would probably be given the conc. o fthe base (or the no of grams of it dissolved in a certain amount of water, which you could then calc the [] of the base) then you calc the [OH] (they might give you something like Ba(OH)2 !! then calc pOH, then calc pH and then you will know EXACTLY where to start plotting your graph from.
The buffer region is where you have the weak base and salt OR where you have salt and weak acid. So NaOH (strong base) with ethanoic acid can only have weak acid and it’s salt. This MUST happen in the acidic region (you need weak acid remember – so it must happen AFTER neutralisation). The maximum buffering point here will be 3/2 the volume of neutralization. i.e. in the second ‘almost horizontal’ part of the curve.
May 20, 2012 at 2:48 pm
Since you have mentioned it, for Ba(OH)2, pOH would be -log(2x[OH])?? or will it be the case like in H2SO4 where the second H is partially ionised? (so the second OH also partially ionised?) In my opinion is 2[OH] because Ba(OH)2—–> Ba2+ +2(OH)- yes?
back to my question, why the buffer solution MUST happen in acidic region? I understand that acidic region in strong base and weak acid is 3/2 volume of neutralization. but WHY cant we have it in the basic region, we have also weak acid and the salt in the flask…..!
May 20, 2012 at 5:49 pm
For Ba(OH)2 the pOH = -log ( 2 x [Ba(OH)2] )
as [OH-] = 2 x [ Ba(OH)2 ]
For Ba(OH)2, you would assume both OH-’s break away from the Ba2+
I thought you would ask this… In the basic region, there isn’t any weak component. You only have OH- and salt, so no buffer – remember you need a weak acid(or weak base) and it’s salt. You only get a weak component in the acidic region once neutralization has occurred and the weak acid starts to be added in excess.
You have no weak acid in the basic solution
May 20, 2012 at 7:25 pm
Oh shoot. I could have understood it without further asking for the second time if I didnt think of ‘weak acid in the flask and titrating it with strong alkali’ is the reflected graph. I mixed two things up. I really need a good sleep oh my!
Thanks anyway.
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December 12, 2010 at 11:36 am
salam sir…….
can you explain more about buffer solution…?
i’d read this topic few times and still confuse with this small section….other parts were quite ok…..haha