pH Question

[will74]

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?

[jasoN-]

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)

[98.40_for_sure]

reacting a weak acid with a strong base is 100%

^this

[jasoN-]

according to JACKS =D, that's how i learnt it

[98.40_for_sure]

Hahah maybe i shoulda kept going... i stopped during equilibrium cos he spent 5 or so lessons about LE CHATELIERS PRINCIPLE -_-

[Potter]

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.

[masonnnn]

because they say 'buffer' they indicate it's a weak acid and weak base...yet they used a strong base so...bad question?

[Potter]

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

[will74]

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?

[will74]

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?

[Potter]

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.

[will74]

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...

[Potter]

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..

[will74]

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.

[jasoN-]

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