Calculating the pH of acid, base AND WATER

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gooner1886
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Hi,

I'm ok calculating the pH's when its just acids and bases - but when you're adding water too, what do you do?

e.g.

1) Calculate the pH of a solution prepared by mixing 25ml of 0.001 mol/l of H2SO4 and 20ml of 0.001 mol/l NaOH and 55 ml of water.


How would you go about doing this?

I'd greatly appreciate any help!
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username938249
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When you add water, you change the volume of the solution, so the concentration changes.
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oversizedcarrot
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The water doesn't interfere with the acid or alkali, it only increases the total volume of the solution, thus diluting it making the solution less concentrated.

So as a general rule here, you calculate which is in excess, the acid or the alkali in moles. Then you calculate the concentration of [H+] from this, taking into the account the increase in total volume due to adding 55ml of water. Then we can find PH.

So to begin:
Moles of H+ in H2SO4=CxVx2= 0.001x(25x10^-3)x2=0.00005moles of H+
Moles of OH- in NaOH=CxV= 0.001x(20x10^-3)= 0.00002moles of OH-

Moles of H+ in excess = 0.00005-0.00002=0.00003

Where V=25+20+55=100ml
Therefore, [H+]=Mol/V= 0.00003/(100x10^-3) = 0.0003moldm^-3

So, PH=-log[H+]
PH=-log(0.0003)
PH= 3.52 (2dp)
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gooner1886
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(Original post by oversizedcarrot)
The water doesn't interfere with the acid or alkali, it only increases the total volume of the solution, thus diluting it making the solution less concentrated.

So as a general rule here, you calculate which is in excess, the acid or the alkali in moles. Then you calculate the concentration of [H+] from this, taking into the account the increase in total volume due to adding 55ml of water. Then we can find PH.

So to begin:
Moles of H+ in H2SO4=CxVx2= 0.001x(25x10^-3)x2=0.00005moles of H+
Moles of OH- in NaOH=CxV= 0.001x(20x10^-3)= 0.00002moles of OH-

Moles of H+ in excess = 0.00005-0.00002=0.00003

Where V=25+20+55=100ml
Therefore, [H+]=Mol/V= 0.00003/(100x10^-3) = 0.0003moldm^-3

So, PH=-log[H+]
PH=-log(0.0003)
PH= 3.52 (2dp)

Ah, got it.

So basically you just add the water volume when calculating the moles of the excess acid/base.

Thank you very much.
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gooner1886
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Any chance you could help me with this, please?

I'm having a hard time with these


Calculate how many ml of 0.2M KH2PO4 solution should be mixed with 0.1M K2HPO4 in order to obtain a buffer pH of 7.7 (pKa of H2PO4 = 7.41)
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username938249
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(Original post by gooner1886)
Any chance you could help me with this, please?

I'm having a hard time with these


Calculate how many ml of 0.2M KH2PO4 solution should be mixed with 0.1M K2HPO4 in order to obtain a buffer pH of 7.7 (pKa of H2PO4 = 7.41)
Try using the Henderson-Hasselbach equation.
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ChemistryBud
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Henderson–Hasselbalch equation:

pH = pKa + log([A-]/[AH])

In your example
7.7= 7.41 + log ([K2HPO4]/[KH2PO4])
rearrange
0.29= log ([K2HPO4]/[KH2PO4])
10^(0.29)= [K2HPO4]/[KH2PO4]
1.95= the ratio of [K2HPO4]/[KH2PO4]

Hope this helps.
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oversizedcarrot
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We don't learn that in AQA o_0
I guess that explains a lot.
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rs232
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(Original post by oversizedcarrot)
We don't learn that in AQA o_0
I guess that explains a lot.
Im pretty sure it is in AQA
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oversizedcarrot
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It's not, the only equation's we know that's similar to that is that in AQA is:

PH=pKa

But not the +(log[A-]/[AH] part onto it. That equation isn't in the syllabus and has never come up in AQA exams.
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BCMFM16
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I'm doing AQA A level Chemistry next academic year. Is Chemistry as bad as people make it out to be?
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Borek
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(Original post by oversizedcarrot)
It's not, the only equation's we know that's similar to that is that in AQA is:

PH=pKa

But not the +(log[A-]/[AH] part onto it. That equation isn't in the syllabus and has never come up in AQA exams.
But you know

K_a = \frac {[H^+][A^-]}{[HA]}

That's the same equation, just rearranged. And it can be used to solve this problem, using Henderson-Haselbalch equation is just an equivalent approach, often faster.

This "this equation is not in syllabus" approach will make you fail at some point. Not my problem though.
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gooner1886
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Thank you


(Original post by ChemistryBud)
Henderson–Hasselbalch equation:

pH = pKa + log([A-]/[AH])

In your example
7.7= 7.41 + log ([K2HPO4]/[KH2PO4])
rearrange
0.29= log ([K2HPO4]/[KH2PO4])
10^(0.29)= [K2HPO4]/[KH2PO4]
1.95= the ratio of [K2HPO4]/[KH2PO4]

Hope this helps.
(Original post by ounce)
Try using the Henderson-Hasselbach equation.
(Original post by Borek)
But you know

K_a = \frac {[H^+][A^-]}{[HA]}

That's the same equation, just rearranged. And it can be used to solve this problem, using Henderson-Haselbalch equation is just an equivalent approach, often faster.

This "this equation is not in syllabus" approach will make you fail at some point. Not my problem though.
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bakergirl
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(Original post by gooner1886)
Any chance you could help me with this, please?

I'm having a hard time with these


Calculate how many ml of 0.2M KH2PO4 solution should be mixed with 0.1M K2HPO4 in order to obtain a buffer pH of 7.7 (pKa of H2PO4 = 7.41)
We don't use the log(base/acid), so what I would do it convert the pKa back to Ka. Then work out the [H+]. Then you can use the buffer equation
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gooner1886
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Just attempting another question:

A buffer was prepared by mixing:

100 ml of 0.1 mol/l Acetic acid
70ml of 0.1 mol/l NaOH

Calculate the pH.

I'm getting for my answer 5.07.

Can anyone kindly quickly confirm?
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oversizedcarrot
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(Original post by Borek)
But you know

K_a = \frac {[H^+][A^-]}{[HA]}

That's the same equation, just rearranged. And it can be used to solve this problem, using Henderson-Haselbalch equation is just an equivalent approach, often faster.

This "this equation is not in syllabus" approach will make you fail at some point. Not my problem though.
Yeah I always just use this instead which is the same thing. Just a different approach.
K_a = \frac {[H^+][A^-]}{[HA]}
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Mike_123
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(Original post by gooner1886)
Hi,

I'm ok calculating the pH's when its just acids and bases - but when you're adding water too, what do you do?

e.g.

1) Calculate the pH of a solution prepared by mixing 25ml of 0.001 mol/l of H2SO4 and 20ml of 0.001 mol/l NaOH and 55 ml of water.


How would you go about doing this?

I'd greatly appreciate any help!
you work out the excess no. mol of acid in this case. so to work out conc of acid, you have to divide by the collective volumes of all 3 rather than just the acid and alkali
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oversizedcarrot
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(Original post by gooner1886)
Just attempting another question:

A buffer was prepared by mixing:

100 ml of 0.1 mol/l Acetic acid
70ml of 0.1 mol/l NaOH

Calculate the pH.

I'm getting for my answer 5.07.

Can anyone kindly quickly confirm?
what's the ka of acetic acid?
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gooner1886
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(Original post by oversizedcarrot)
what's the ka of acetic acid?
Sorry, I should have mentioned it.

The pka is 4.7

So ka = 1.995 x 10-5
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oversizedcarrot
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(Original post by gooner1886)
Sorry, I should have mentioned it.

The pka is 4.7

So ka = 1.995 x 10-5
Yeah I got PH as 5.07 also, well done mate!
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