# pH calculations

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

Report

#2

(Original post by

Calculate the pH of the solutions resulting from the following reactions.

25cm³ of 0.1M HCL reacting with 5cm³ of 0.15M NaOH

For the i got and for i got final pH i got was 1.03 to 2dp

Where did it all go wrong?

**will'o'wisp**)Calculate the pH of the solutions resulting from the following reactions.

25cm³ of 0.1M HCL reacting with 5cm³ of 0.15M NaOH

For the i got and for i got final pH i got was 1.03 to 2dp

Where did it all go wrong?

mol acid = 0.025 x 0.1 = 0.0025 mol

mol base = 0.005 x 0.15 = 0.00075 mol

hence after reaction mol acid remaining = 0.0025 - 0.00075 = 0.00175 mol

total volume = 25 + 5 = 30 ml

new acid concentration = 0.00175/0.030 = 0.0583 M

pH = 1.23

0

reply

Report

#3

**will'o'wisp**)

Calculate the pH of the solutions resulting from the following reactions.

25cm³ of 0.1M HCL reacting with 5cm³ of 0.15M NaOH

For the i got and for i got final pH i got was 1.03 to 2dp

Where did it all go wrong?

Firstly, you work out the moles of acid and alkali.

ACID = 2.5x10^-3

ALKALI =7.5x10^-4

But the acid is in excess, so you need to find how many unreacted moles are in the resulting solution.

EXCESS ACID= ACID-ALKALI

=1.75x10^-3

Now, work out the concentrations using the final volume, 30cm^3.

ACID= 0.583

Then, use -logH+ to find the pH.

0

reply

(Original post by

Calculate the mol excess of the acid:

mol acid = 0.025 x 0.1 = 0.0025 mol

mol base = 0.005 x 0.15 = 0.00075 mol

hence after reaction mol acid remaining = 0.0025 - 0.00075 = 0.00175 mol

total volume = 25 + 5 = 30 ml

new acid concentration = 0.00175/0.030 = 0.0583 M

pH = 1.23

**charco**)Calculate the mol excess of the acid:

mol acid = 0.025 x 0.1 = 0.0025 mol

mol base = 0.005 x 0.15 = 0.00075 mol

hence after reaction mol acid remaining = 0.0025 - 0.00075 = 0.00175 mol

total volume = 25 + 5 = 30 ml

new acid concentration = 0.00175/0.030 = 0.0583 M

pH = 1.23

(Original post by

You've forgotten to work out the moles of excess acid.

Firstly, you work out the moles of acid and alkali.

ACID = 2.5x10^-3

ALKALI =7.5x10^-4

But the acid is in excess, so you need to find how many unreacted moles are in the resulting solution.

EXCESS ACID= ACID-ALKALI

=1.75x10^-3

Now, work out the concentrations using the final volume, 30cm^3.

ACID= 0.583

Then, use -logH+ to find the pH.

**typicalvirgo**)You've forgotten to work out the moles of excess acid.

Firstly, you work out the moles of acid and alkali.

ACID = 2.5x10^-3

ALKALI =7.5x10^-4

But the acid is in excess, so you need to find how many unreacted moles are in the resulting solution.

EXCESS ACID= ACID-ALKALI

=1.75x10^-3

Now, work out the concentrations using the final volume, 30cm^3.

ACID= 0.583

Then, use -logH+ to find the pH.

0

reply

Report

#5

(Original post by

So i never needed to work out hydrogen ion concentration??

How do i know acid is in excess and not the base?

**will'o'wisp**)So i never needed to work out hydrogen ion concentration??

How do i know acid is in excess and not the base?

0

reply

(Original post by

I

**charco**)I

**t's a 1:1 stoichiometric ratio in the balanced equation**and there are**more moles of acid in the experiment.**How do i know this? Do i guess? Is it because the moles of acid>mol of alkali?

0

reply

Report

#7

(Original post by

What is this supposed to tell me???

How do i know this? Do i guess? Is it because the moles of acid>mol of alkali?

**will'o'wisp**)What is this supposed to tell me???

How do i know this? Do i guess? Is it because the moles of acid>mol of alkali?

0

reply

Report

#8

**will'o'wisp**)

So i never needed to work out hydrogen ion concentration??

How do i know acid is in excess and not the base?

In this case, the excess acid was a strong one so the [HA] is taken as the [H+], but if it was H2SO4, or any other diprotic acid, the moles would need to be doubled.

If you had an excess of a weak acid, then Ka would come in and that's when you rearrange for H+.

0

reply

(Original post by

Do you understand the significance of a balanced equation?

**charco**)Do you understand the significance of a balanced equation?

(Original post by

Because the acid will have more moles. Sometimes the base is in excess and Kw=[OH-][H+] will be needed, where you rearrange for H+.

In this case, the excess acid was a strong one so the [HA] is taken as the [H+], but if it was H2SO4, or any other diprotic acid, the moles would need to be doubled.

If you had an excess of a weak acid, then Ka would come in and that's when you rearrange for H+.

**typicalvirgo**)Because the acid will have more moles. Sometimes the base is in excess and Kw=[OH-][H+] will be needed, where you rearrange for H+.

In this case, the excess acid was a strong one so the [HA] is taken as the [H+], but if it was H2SO4, or any other diprotic acid, the moles would need to be doubled.

If you had an excess of a weak acid, then Ka would come in and that's when you rearrange for H+.

0

reply

Report

#10

(Original post by

Yes

hmmmm, i don't understand >.> part of this is the teachers fault for not teaching me this >.>

**will'o'wisp**)Yes

hmmmm, i don't understand >.> part of this is the teachers fault for not teaching me this >.>

0

reply

(Original post by

It all depends on the situtation that then determines the calculation to use. pH calcs are so repetitive, so practice themand you'll see a pattern emerge :-) my teacher was poo too.

**typicalvirgo**)It all depends on the situtation that then determines the calculation to use. pH calcs are so repetitive, so practice themand you'll see a pattern emerge :-) my teacher was poo too.

0

reply

Report

#12

(Original post by

I'm gonna go with more moles of it calculated if that's not it i'll have to ask another chem teacher, shame really because i got the worst out of the bunch

**will'o'wisp**)I'm gonna go with more moles of it calculated if that's not it i'll have to ask another chem teacher, shame really because i got the worst out of the bunch

1. Calculate the moles of both the acid and the base.

2. The one with the larger number is in excess.

3. Subtract the smaller number from the larger number of moles and you have the excess moles.

4. If the acid is in excess, divide the moles excess acid by the volume to get the molarity. Use this to calculate pH

5. If the base is in excess, divide the moles excess base by the total volume to get the molarity of the base. Calculate pOH (same method as pH except using [OH-] instead of [H+], The subtract result from 14 to get pH.

NOTE: This is only this simple when the balanced equation is 1:1

0

reply

(Original post by

This is not rocket science.

1. Calculate the moles of both the acid and the base.

2. The one with the larger number is in excess.

3. Subtract the smaller number from the larger number of moles and you have the excess moles.

4. If the acid is in excess, divide the moles excess acid by the volume to get the molarity. Use this to calculate pH

5. If the base is in excess, divide the moles excess base by the total volume to get the molarity of the base. Calculate pOH (same method as pH except using [OH-] instead of [H+], The subtract result from 14 to get pH.

NOTE: This is only this simple when the balanced equation is 1:1

**charco**)This is not rocket science.

1. Calculate the moles of both the acid and the base.

2. The one with the larger number is in excess.

3. Subtract the smaller number from the larger number of moles and you have the excess moles.

4. If the acid is in excess, divide the moles excess acid by the volume to get the molarity. Use this to calculate pH

5. If the base is in excess, divide the moles excess base by the total volume to get the molarity of the base. Calculate pOH (same method as pH except using [OH-] instead of [H+], The subtract result from 14 to get pH.

NOTE: This is only this simple when the balanced equation is 1:1

0

reply

Report

#14

(Original post by

So if it's a 2:1 in favour of H+ ions then i multiply the value i get from step 4 by 2?

**will'o'wisp**)So if it's a 2:1 in favour of H+ ions then i multiply the value i get from step 4 by 2?

0

reply

(Original post by

No, you have to take this into account when working out which is in excess.

**charco**)No, you have to take this into account when working out which is in excess.

0

reply

Report

#16

**charco**)

Calculate the mol excess of the acid:

mol acid = 0.025 x 0.1 = 0.0025 mol

mol base = 0.005 x 0.15 = 0.00075 mol

hence after reaction mol acid remaining = 0.0025 - 0.00075 = 0.00175 mol

total volume = 25 + 5 = 30 ml

new acid concentration = 0.00175/0.030 = 0.0583 M

pH = 1.23

0

reply

Report

#17

(Original post by

i have a question.. dont we take into account the number of moles of conjugated base formed from the salt that would increase the PH???????

**pondsteps**)i have a question.. dont we take into account the number of moles of conjugated base formed from the salt that would increase the PH???????

This is only the case when dealing with weak acids or bases...

0

reply

Report

#19

(Original post by

Oh? I though you just multiply conc by 2 if there's a diprotic acid before you calculate the pH using pH=-log[H+]

**will'o'wisp**)Oh? I though you just multiply conc by 2 if there's a diprotic acid before you calculate the pH using pH=-log[H+]

For example if you react 25ml of 0.1M sulfuric acid with 10ml of 0.2M NaOH,

Mol H2SO4 = 0.0025

Mol NaOH = 0.002

BUT

2 mol of NaOH react with each mol of H

_{2}SO

_{4}

.

SO

0.002 mol of NaOH reacts with 0.001 mol of H

_{2}SO

_{4}

therefore the H

_{2}SO

_{4}is in excess by 0.0025 - 0.001 = 0.0015 mol

NOW

To find mol of H+ ions you double the sulfuric acid concentration value.

1

reply

(Original post by

No, you must factor in the 2:1 ratio when calculating the excess.

For example if you react 25ml of 0.1M sulfuric acid with 10ml of 0.2M NaOH,

Mol H2SO4 = 0.0025

Mol NaOH = 0.002

BUT

2 mol of NaOH react with each mol of H

.

SO

0.002 mol of NaOH reacts with 0.001 mol of H

therefore the H

NOW

To find mol of H+ ions you double the sulfuric acid concentration value.

**charco**)No, you must factor in the 2:1 ratio when calculating the excess.

For example if you react 25ml of 0.1M sulfuric acid with 10ml of 0.2M NaOH,

Mol H2SO4 = 0.0025

Mol NaOH = 0.002

BUT

2 mol of NaOH react with each mol of H

_{2}SO_{4}.

SO

0.002 mol of NaOH reacts with 0.001 mol of H

_{2}SO_{4}therefore the H

_{2}SO_{4}is in excess by 0.0025 -**0.001**= 0.0015 molNOW

To find mol of H+ ions you double the sulfuric acid concentration value.

Oh i see, so i have to adjust the mol when calculating excess too? Ok thanks! Didn't know that, i'll be sure to tag you when i get stuck on some chem ^-^

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top