# Calculating the pH of acid, base AND WATER

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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!

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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#2

When you add water, you change the volume of the solution, so the concentration changes.

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#3

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)

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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(Original post by

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)

**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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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)

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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#6

(Original post by

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)

**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)

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#7

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.

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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#10

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.

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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#11

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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#12

(Original post by

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.

**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.

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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Thank you

(Original post by

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.

**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

Try using the Henderson-Hasselbach equation.

**ounce**)Try using the Henderson-Hasselbach equation.

(Original post by

But you know

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.

**Borek**)But you know

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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#14

**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)

0

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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

Can anyone kindly quickly confirm?

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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#16

**Borek**)

But you know

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.

0

reply

Report

#17

(Original post by

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!

**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!

0

reply

Report

#18

(Original post by

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

Can anyone kindly quickly confirm?

**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?

0

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(Original post by

what's the ka of acetic acid?

**oversizedcarrot**)what's the ka of acetic acid?

The pka is 4.7

So ka = 1.995 x 10-5

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#20

(Original post by

Sorry, I should have mentioned it.

The pka is 4.7

So ka = 1.995 x 10-5

**gooner1886**)Sorry, I should have mentioned it.

The pka is 4.7

So ka = 1.995 x 10-5

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