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# pH calculations watch

1. 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?
2. (Original post by 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?
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
3. (Original post by 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?
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.
4. (Original post by 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
So i never needed to work out hydrogen ion concentration??
(Original post by 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.
How do i know acid is in excess and not the base?
5. (Original post by 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?
It's a 1:1 stoichiometric ratio in the balanced equation and there are more moles of acid in the experiment.
6. (Original post by charco)
It's a 1:1 stoichiometric ratio in the balanced equation and there are more moles of acid in the experiment.
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?
7. (Original post by 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?
Do you understand the significance of a balanced equation?
8. (Original post by 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?
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+.
9. (Original post by charco)
Do you understand the significance of a balanced equation?
Yes
(Original post by 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+.
hmmmm, i don't understand >.> part of this is the teachers fault for not teaching me this >.>
10. (Original post by will'o'wisp)
Yes

hmmmm, i don't understand >.> part of this is the teachers fault for not teaching me this >.>
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.
11. (Original post by 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.
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
12. (Original post by 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
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
13. (Original post by 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
So if it's a 2:1 in favour of H+ ions then i multiply the value i get from step 4 by 2?
14. (Original post by 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?
No, you have to take this into account when working out which is in excess.
15. (Original post by charco)
No, you have to take this into account when working out which is in excess.
Oh? I though you just multiply conc by 2 if there's a diprotic acid before you calculate the pH using pH=-log[H+]
16. (Original post by 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
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???????
17. (Original post by 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???????
No.

This is only the case when dealing with weak acids or bases...
18. (Original post by charco)
No.

This is only the case when dealing with weak acids or bases...
Oh right! thanks
19. (Original post by 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+]
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 H2SO4
.

SO

0.002 mol of NaOH reacts with 0.001 mol of H2SO4

therefore the H2SO4 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.
20. (Original post by 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 H2SO4
.

SO

0.002 mol of NaOH reacts with 0.001 mol of H2SO4

therefore the H2SO4 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.
Do i have to half the NaOH mol? or can i double the sulfuric acid mol? then divide that answer by 2?

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 ^-^

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