ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: will74 on November 10, 2010, 10:41:27 am
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Hey guys, bit unsure how to do this question
Calculate the pH of a buffer made by mixing 100 ml 0.1 M CH3COOH and 50ml 0.1 M NaOH
So this is what i did....
NaOH + CH3COOH > CH3COONa + H20
n(CH3COOH)intial=c*v=0.01 mol
n(NaOH)initial = 0.005 mol
Therefore, n(CH3COOH)reacted with NaOH =0.005 mol
Then c(CH3COOH)remaining = .005/.15=3.33*10^-2
Then to find the ph
Ka=(H+)(CH3COO-)/CH3COOH
from here my teacher said assume that there is no ionisation and that c(CH3COOH)=c(CH3COO-) due to the reaction with NaOH producing 0.005 mol of ethanoate ions.
Therefore Ka=c(H+)
Should we really just assume there was no ionisation, i thought it seemed a bit silly?
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i always thought reacting a weak acid with a strong base is 100%
This is what i would have done.
100mL 0.1M CH3COOH
[CH3COOH] = 0.1M
[H+] = 0.1M (ethanoic acid is monoprotic)
n(H+) = cV = 0.1 x 0.1 = 0.01 mol
50mL 0.1M NaOH
[OH-] = 0.1M
n(OH-) = cV = 0.1 x 0.05 = 0.005 mol
As we can see H+ is in excess according to the equation H+ + OH- <-> H2O
So n(H+ remaining) = 0.01-0.005
= 0.005 mol
Total volume = 100 + 50 mL = 0.15L
Therefore [H+] = n/v = 0.005 / 0.15
= 0.033333..
pH = -log10 (0.033333) = 1.47 = 1.5
incorrectly assumed 100% reaction because OH- runs out (it is not in excess)
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reacting a weak acid with a strong base is 100%
^this
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according to JACKS =D, that's how i learnt it
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Hahah maybe i shoulda kept going... i stopped during equilibrium cos he spent 5 or so lessons about LE CHATELIERS PRINCIPLE -_-
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You can assume there's no ionisation or you can make the concentration = to [CH3COOH -x]. In practice is doesn't make much difference to your results(unless you're very pedantic) and you don't get marked differently if you just assume the concentration doesn't change. It takes a lot longer to not assume since you get a quadratic and all so during the exam I wouldn't recommend it. Even in first year they assume that the concentration doesn't change so don't worry about it too much.
Edit - jasoN- it's alright to assume that it's 100% ionisation when the there's a mix of strong base and weak acid. However, I don't think this is the case when all the NaOH is neutralised. Think about why the ionisation is 100%(The NaOH is pushing the reaction forward) When there is no NaOH left the acid will not ionize fully, therefore you must use its constant.
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because they say 'buffer' they indicate it's a weak acid and weak base...yet they used a strong base so...bad question?
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The 'buffer' they're referring to in the question is the weak acid which slightly resists the change of pH.
Actually, after further reading.. The buffer that forms is actually sodium ethanoate.
http://www.chemguide.co.uk/physical/acidbaseeqia/buffers.html
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because they say 'buffer' they indicate it's a weak acid and weak base...yet they used a strong base so...bad question?
Yeh it is a lame question. But i think the buffer forms after the neutralisation reaction...therefore it can still form?
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You can assume there's no ionisation or you can make the concentration = to [CH3COOH -x]. In practice is doesn't make much difference to your results(unless you're very pedantic) and you don't get marked differently if you just assume the concentration doesn't change. It takes a lot longer to not assume since you get a quadratic and all so during the exam I wouldn't recommend it. Even in first year they assume that the concentration doesn't change so don't worry about it too much.
Edit - jasoN- it's alright to assume that it's 100% ionisation when the there's a mix of strong base and weak acid. However, I don't think this is the case when all the NaOH is neutralised. Think about why the ionisation is 100%(The NaOH is pushing the reaction forward) When there is no NaOH left the acid will not ionize fully, therefore you must use its constant.
Thankyou...That is really helpful...
Therefore does that mean
Ka=c(H+) and so pH=-log(Ka) of ethanoic acid?
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Not exactly, because what you'll have is...
Ka = [CH3COO-][H+]/[CH3COOH]
Lets say the concentration of H+ is equal to x and the concentration of H+ equals the concentration of CH3COO-(1:1 ratio)
Ka = x^2/[CH3COOH]
So your pH should = -log(Ka[CH3COOH])^1/2
If you work it out it comes out around 3.12
Which makes some sense as a 1M CH3COOH solution will have a pH of 2.38 or so.
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Not exactly, because what you'll have is...
Ka = [CH3COO-][H+]/[CH3COOH]
Lets say the concentration of H+ is equal to x and the concentration of H+ equals the concentration of CH3COO-(1:1 ratio)
Ka = x^2/[CH3COOH]
So your pH should = -log(Ka[CH3COOH])^1/2
If you work it out it comes out around 3.12
Which makes some sense as a 1M CH3COOH solution will have a pH of 2.38 or so.
But didn't we say that c(CH3COOH)=c(CH3COO-)
Therefore wouldn't Ka = [CH3COO-][H+]/[CH3COOH]= 1*c(H+)=c(H+)
How can you assume that c(H+)=c(CH3COOH)
because in the first neutralisation reaction there was 0.005 mol CH3COO-...Then there was 0.005 mol CH3COOH left over. Since ethanoic acid is a weak acid, it won't produce 0.005 mol H+, therefore, c(H+)=/=c(CH3COOH) even though there is a one to one ratio??
Sorry to keep asking...
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Nonono, I'm not assuming that [H+] equals [CH3COOH].
You have the ionization reaction which is..
CH3COOH = H+ + CH3COO-
What we're saying is that the amount of H+ and CH3COO- is the same. We want to find out what [H+] is so we say that the concentration of H+ is equal to x. Since there is as much CH3COO- as H+, the concentration of CH3COO- equals x too. The assumption that the [CH3COOH] doesn't change comes in here. Since the value of x is so small we can say that [CH3COOH-x] = [CH3COOH].
Then the pH is equal to -log(x).
Hope that helps..
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Nonono, I'm not assuming that [H+] equals [CH3COOH].
You have the ionization reaction which is..
CH3COOH = H+ + CH3COO-
What we're saying is that the amount of H+ and CH3COO- is the same. We want to find out what [H+] is so we say that the concentration of H+ is equal to x. Since there is as much CH3COO- as H+, the concentration of CH3COO- equals x too. The assumption that the [CH3COOH] doesn't change comes in here. Since the value of x is so small we can say that [CH3COOH-x] = [CH3COOH].
Then the pH is equal to -log(x).
Hope that helps..
How can you assume that c(H+)=c(CH3COO-)
because in the first neutralisation reaction there was 0.005 mol CH3COO-...Then there was 0.005 mol CH3COOH left over. Since ethanoic acid is a weak acid, it won't produce 0.005 mol H+, therefore, c(H+)=/=c(CH3COO-) even though there is a one to one ratio??........
I meant....How can you assume that c(H+)=c(CH3COO-)....typo...
I totally agree with you if it were just a buffer solution
But...The first reaction...
NaOH + CH3COOH > CH3COONa + H20
will produce 0.005 mol of CH3COO-...as it fully reacts...0.005 mol NaOH with 0.005 mol CH3COOH
Then the second reaction...
CH3COOH + H20 > CH3COO- +H3O+
we have 0.005 mol of CH3COOH left over from the first reaction. Since it will not fully ionise...It cannot produce 0.005 mol of H+ ions.
So the number of ethanoate ions we have = 0.005 (from first reaction) + any produced in ionisation
let x be the n(CH3COO-) produced in the ionisation of ethanoic acid.
Therefore n(CH3COO-)overall=x+0.005
n(H+)from ionisation=n(CH3COO-)from ionisation=x
Therefore n(H+)=x
n(CH3COO-)overall=x+0.005
n(H+)overall=x
so n(CH3COO-)>n(H+) not equal to?
Sorry again, i just don't get how they can be equal....
It just seems that any H+ produced will have a CH3COO- with it...But from the first reaction, we have 0.005 mol of CH3COO- still hanging around......
I really appreciate this.
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Hey good point there with the second equation
You have 0.005 mol left
The volume will be accumulative, ie. 0.15L
[CH3COOH unreacted] = n/V = 0.033333..
Now we have the equation CH3COOH + H2O <--> CH3COO- + H3O+, as NaOH has been used up (limiting reactant)
We can assume [CH3COOH] = [CH3COO-] as ethanoic acid is a weak acid.
Ka = [H3O+]^2 / [CH3COOH]
Ka = 1.7 x 10^-5, from the data booklet
[CH3COOH] = 0.033333
therefore: 1.7 x 10^-5 = [H3O+]^2 / 0.0333333
[H3O+]^2 = 5.666666 x 10^-7
[H3O+] = 7.528 x 10^-4
therefore: pH = -log10 [H3O+] = -log10 (7.528 x 10^-4) = 3.12
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Hey good point there with the second equation
You have 0.005 mol left
The volume will be accumulative, ie. 0.15L
[CH3COOH unreacted] = n/V = 0.033333..
Now we have the equation CH3COOH + H2O <--> CH3COO- + H3O+, as NaOH has been used up (limiting reactant)
We can assume [CH3COOH] = [CH3COO-] as ethanoic acid is a weak acid.
Ka = [H3O+]^2 / [CH3COOH]
Ka = 1.7 x 10^-5, from the data booklet
[CH3COOH] = 0.033333
therefore: 1.7 x 10^-5 = [H3O+]^2 / 0.0333333
[H3O+]^2 = 5.666666 x 10^-7
[H3O+] = 7.528 x 10^-4
therefore: pH = -log10 [H3O+] = -log10 (7.528 x 10^-4) = 3.12
The bolded step essentially assumes [H+]=[CH3COO-] doesn't it?
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Not exactly, because what you'll have is...
Ka = [CH3COO-][H+]/[CH3COOH]
Lets say the concentration of H+ is equal to x and the concentration of H+ equals the concentration of CH3COO-(1:1 ratio)
Ka = x^2/[CH3COOH]
So your pH should = -log(Ka[CH3COOH])^1/2
If you work it out it comes out around 3.12
Which makes some sense as a 1M CH3COOH solution will have a pH of 2.38 or so.
sexual... :P
that is what I got also.
generally you assume [H+]=[X-]
edit: wrong...
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Okay this just got complicated.
The first reaction is 100% completed, therefore [CH3COONa] = 0.03333333..
So [CH3COO-] = [H3O+] + 0.03333333
Ka = 1.7 x 10^-5, from the data booklet
[CH3COOH] = 0.033333
therefore: 1.7 x 10^-5 = [H3O+]([H3O+] + 0.033333) / 0.0333333
[H3O+]([H3O+] + 0.033333) = 5.666666 x 10^-7
lolwut this is a quadratic equation.
Idk man I'm completely confused
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Ahhhh, I follow you now. That's a very interesting point that you make. I overlooked that completely.
I did it two ways...
First way i assumed that the [CH3COOH] was constant and it spat out a pH of around 7(intuitively incorrect)
The second time I made the concentration equal to [CH3COOH - x]
So I guess in this situation you must not assume anything.
so basically what we have is.. x(x+.005)/(.005-x) = Ka
Put it in the cas.. wait 5 minutes(my cas hasn't worked so hard for a while) and you'll get two solutions. Take the positive solution and log that. You get a pH of 4.77.
Which infact sounds a lot better for a buffer(compared to 3.12). Sorry for the misconception there. Also, there's no way vcaa will ask a question like this.. Forgot they're sick minded pricks..
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Ahhhh, I follow you now. That's a very interesting point that you make. I overlooked that completely.
I did it two ways...
First way i assumed that the [CH3COOH] was constant and it spat out a pH of around 7(intuitively incorrect)
The second time I made the concentration equal to [CH3COOH - x]
so basically what we have is.. x(x+.005)/(.005-x) = Ka
Put it in the cas.. wait 5 minutes(my cas hasn't worked so hard for a while) and you'll get two solutions. Take the positive solution and log that. You get a pH of 4.77.
Which infact sounds a lot better for a buffer(compared to 3.12). Sorry for the misconception there. Also, there's no way vcaa will ask a question like this.. Forgot they're sick minded pricks..
hahahaha...yeah they're sicko's, especially with the 2010 exams they have produced.
Thanks for taking the time to read over it again...this question freaked me out.
so basically what we have is.. x(x+.005)/(.005-x) = Ka
makes so much sense to me now :)
Cheers
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Aww shit. This is what Mao was trying to explain to me last night. i also completely over looked the additional CH3COO-
...
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So for these Q's do we just generally assume that Ka=H+ ?
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Nothing assumed here, it is taking into account the acid/base reaction
But generally for weak acids
Ka = [H3O+]^2 / [weak acid]
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So for these Q's do we just generally assume that Ka=H+ ?
I don't think you can generally assume anything.....this is a wack question i doubt we well get asked anything like this....but maybe!
I think yeah, the only way you could do it without a CAS calculator or a foullllll quadratic formula application would be to assume that the two are equal...But i guess that is only the case because it was 0.005 initial and 0.005 left over, otherwise...with different amounts, the ratio would be different.
ah, CANT WAIT TO FINISH TOMORROW WOOOOOOOOOOOH!!!
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http://vcenotes.com/forum/index.php/topic,22981.msg347259.html#msg347259
Here is a similar question which Mao did. Explained it nicely.
It makes sense, and your teacher was right.
Have a go at it.
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actually in this question we cannot assume the [H30+]=[CH3COO-].
if you consider the addition of NaOH in terms of equilibrium what occurs is the NaOH reacts with the CH3COOH and so will reduce n(CH3COOH) to 0.005 mol and increase n(CH3COO-) to 0.005 mol.
from here one might expect that the .005 mol of CH3COOH would continue to ionize as it would prior to the addition of NaOH. however this is not the case.
as a result of the higher concentration of ethanoate ions which must be held constant at 0.033 M, the equilibrium of the remianing .005 mol of CH3COOH is forced backwards such that the concentration of ethanoate ions does not increase (hence why it is constant at 0.033 M) and hydronium ion concentration will not increase any further.
however while we can consider [CH3COO-] to be constant we cannot disregard it in the equilibrium equation to find pH.
This is why we cannot carry out the calculation using [CH3COO-] = [H30+]
so [CH3COOH]=[CH3COO-] and therefore [H3O+] = Ka = 1.7 10^-5, and pH=4.77
so yes your teacher is correct
if this seems a little difficult to follow then consider this.
prior to the addition of the NaOH the pH of 100ml 0.1M ethanoic acid is -log(1.7x10^-5 x 0.01) and this is equal to 3.37
so with addition of 50ml 0.1M NaOH the pH can only possibly increase, and so the resulting pH of the buffer solution could never be below 3.37
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Yeah, it ends up being [H+]=Ka, and then pH of 4.77 or something.
But I don't think that is a rule. Just worked out that way by luck that [H+] was equal to Ka.
Seeing as we know how to do this now, hopefully it comes up tomorrow.
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actually in this question we cannot assume the [H30+]=[CH3COO-].
if you consider the addition of NaOH in terms of equilibrium what occurs is the NaOH reacts with the CH3COOH and so will reduce n(CH3COOH) to 0.005 mol and increase n(CH3COO-) to 0.005 mol.
from here one might expect that the .005 mol of CH3COOH would continue to ionize as it would prior to the addition of NaOH. however this is not the case.
as a result of the higher concentration of ethanoate ions which must be held constant at 0.033 M, the equilibrium of the remianing .005 mol of CH3COOH is forced backwards such that the concentration of ethanoate ions does not increase (hence why it is constant at 0.033 M) and hydronium ion concentration will not increase any further.
however while we can consider [CH3COO-] to be constant we cannot disregard it in the equilibrium equation to find pH.
This is why we cannot carry out the calculation using [CH3COO-] = [H30+]
so [CH3COOH]=[CH3COO-] and therefore [H3O+] = Ka = 1.7 10^-5, and pH=4.77
so yes your teacher is correct
if this seems a little difficult to follow then consider this.
prior to the addition of the NaOH the pH of 100ml 0.1M ethanoic acid is -log(1.7x10^-5 x 0.01) and this is equal to 3.37
so with addition of 50ml 0.1M NaOH the pH can only possibly increase, and so the resulting pH of the buffer solution could never be below 3.37
thanks robbo :)
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no worries, and yes this is a one of situation as the number of mole of NaOH is half the number of initial CH3COOH