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# Acid + Base question watch

1. here's the question

Calculate the pH of the resultant solution, if 30ml of 0.1M NaOH is added to 50ml of 0.1M acetic acid (Ka=1.74x10-5)??

pH=pka+log(b/a)

I haven't a clue on how to do this??

2. I think it's about 3.38 but I'm not certain. If someone else verifies it I'd be more than happy to explain it to you if you wanted.
3. I got 10.12 :/

I wouln't have thought that it would be more acidic since Acetic acid is a weak acid and NaOH is a strong base.

I'm posting so I can reap the rewards of the help given by a much more knowledgable member :P

Edit: Im pretty sure my answer is wrong, sorry.

In my chemistry course we haven't learned about the significance of the volumes, but I can tell you:

[H+(aq)]= Square root of pKa*[HA] ---> 1.319x10^-3---> -log to get the pH

For the Base, you use:

[H+(aq)]= Kw/[OH-] ----> 1x10-14/0.1 = 1x10^-13 and then -log that to get the pH (13)

Im not really sure where to go after this I'm afraid.
4. hmmm i dunno. i havent learnt how to do this yet. (am in year 12) so i'm trying to work out how it works. could you check through my working please?

NaOH CH3COOH CH3COO-Na+ H2O

I 0.0375M 0.0625M 0 0

C -x -x x x

E 0.0375-x 0.0625-x x x

x^2/(0.0375-x)(0.0625-x) = 1.74x10-5
so x = 0.0002010772961397973
so [H+] = 0.0002010772961397973
so -log(0.0002010772961397973) = 3.6966

where have I gone wrong?
5. (Original post by CM23925)
I think it's about 3.38 but I'm not certain. If someone else verifies it I'd be more than happy to explain it to you if you wanted.
how did you get 3.38???

Ive realised that by adding the acid and base, you get a buffer.

I have tried something and seemed to get 4.93.

This is what I did, correct me if im wrong

acetic acid ka=1.74x10-5 , therefore -log to get pka, pka=4.76 (2dp)

by the way ml=cm3

The acetic acid is in excess, so when NaOH reacts with it produces sodium acetate and water.

moles of NaOH used are 0.1m/1000ml x 30ml = 0.003 moles.

initial acetic acid moles = 0.1m/1000ml x 50ml = 0.005 moles.

Then once the reaction takes place, the solution will contain acetic acid, sodium acetate and water.

So acetic acid concentraion after reaction = 0.005 - 0.003 = 0.002moles

Therefore the sodium acetate will equal = 0.003 moles.

Now our acid concentration is 0.002 moles in 80ml. Therefore using m=cxv/1000, c=0.025

conjugate base concentration ( sodium acetate ) = 0.003 moles in 80ml . c=0.0375

Now using the H-H equation

pH=pka+log(conjugate base concentration/acid concentration)

pH= 4.76 + log ( 0.0375/0.025)

pH = 4.94

Any ideas??????
6. (Original post by ansarali)
how did you get 3.38???

Ive realised that by adding the acid and base, you get a buffer.

I have tried something and seemed to get 4.93.

This is what I did, correct me if im wrong
Here we go then...

(Original post by ansarali)
acetic acid ka=1.74x10-5 , therefore -log to get pka, pka=4.76 (2dp)
I would work directly in ka

(Original post by ansarali)

The acetic acid is in excess, so when NaOH reacts with it produces sodium acetate and water.

moles of NaOH used are 0.1m/1000ml x 30ml = 0.003 moles.

initial acetic acid moles = 0.1m/1000ml x 50ml = 0.005 moles.

Then once the reaction takes place, the solution will contain acetic acid, sodium acetate and water.

So acetic acid concentraion after reaction = 0.005 - 0.003 = 0.002moles

Therefore the sodium acetate will equal = 0.003 moles.
No need to change to concentration as all components are in the same solution and volumes cancel out...

(Original post by ansarali)

Now our acid concentration is 0.002 moles in 80ml. Therefore using m=cxv/1000, c=0.025

conjugate base concentration ( sodium acetate ) = 0.003 moles in 80ml . c=0.0375
Using the acid equilibrium:

ka = [H+][A-]/[HA]

1.74 x 10-5 = [H+][A-]/[HA]

[H+] = [HA]/[A-] x 1.74 x 10-5 = 0.002/0.003 x 1.74 x 10-5

[H+] = 1.16 x 10-5

Therefore pH = 4.94
7. (Original post by charco)
Here we go then...

I would work directly in ka

No need to change to concentration as all components are in the same solution and volumes cancel out...

Using the acid equilibrium:

ka = [H+][A-]/[HA]

1.74 x 10-5 = [H+][A-]/[HA]

[H+] = [HA]/[A-] x 1.74 x 10-5 = 0.002/0.003 x 1.74 x 10-5

[H+] = 1.16 x 10-5

Therefore pH = 4.94

Thankyou for that, its seems I did get the right answer in the end. You seemed to use the exact same method, only difference was you used the unlogged version of the henderson hasselbach equation.

You are totally correct about not needing to divide by volumes, they do cancel out as you said.

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