pH Question

[will74]

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?

[stonecold]

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...

that is what I got also.

generally you assume [H+]=[X-]

edit: wrong...

[jasoN-]

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

[Potter]

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..

[will74]

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

[stonecold]

Aww shit.  This is what Mao was trying to explain to me last night.   i also completely over looked the additional CH3COO-

...

[stonecold]

So for these Q's do we just generally assume that Ka=H+  ?

[jasoN-]

Nothing assumed here, it is taking into account the acid/base reaction
But generally for weak acids
Ka = [H3O+]^2 / [weak acid]

[will74]

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!!!

[stonecold]

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.

[mroberts]

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

[stonecold]

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.

[will74]

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

[mroberts]

no worries, and yes this is a one of situation as the number of mole of NaOH is half the number of initial CH3COOH