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# pH calculation help :S watch

1. Find the pH of a buffer solution made by mixing an excess of weak acid with strong base.

25cm3, 0.5moldm-3 methanoic acid (ka = 1.78x10^-4) mixed with 10cm3, 1moldm-3 NaOH

2. (Original post by kiiten)
Find the pH of a buffer solution made by mixing an excess of weak acid with strong base.

25cm3, 0.5moldm-3 methanoic acid (ka = 1.78x10^-4) mixed with 10cm3, 1moldm-3 NaOH

Do the stoichiometry of the reaction to find the exact constitution of the products and unreacted reactant(s)
3. (Original post by charco)
Do the stoichiometry of the reaction to find the exact constitution of the products and unreacted reactant(s)
CH3COOH + NaOH --------> CH3COONa + H2O

1:1 ratio

??
4. (Original post by kiiten)
CH3COOH + NaOH --------> CH3COONa + H2O

1:1 ratio

??
So calculate moles of acid and moles of base to find out which is in excess and how many moles of salt are produced.

Then use the Ka expression to find [H+]
5. (Original post by charco)
So calculate moles of acid and moles of base to find out which is in excess and how many moles of salt are produced.

Then use the Ka expression to find [H+]
Im confused...

There are 0.0125 moles of acid and 0.01 moles of base

acid is in excess. Then? - moles of salt are 0.01??
6. (Original post by kiiten)
Im confused...

There are 0.0125 moles of acid and 0.01 moles of base

acid is in excess. Then? - moles of salt are 0.01??
You need moles of salt formed and moles of acid remaining and then plug the values into the ka expression.
7. (Original post by charco)
You need moles of salt formed and moles of acid remaining and then plug the values into the ka expression.
Oh so moles of salt = 0.01 and moles of acid remaining = 0.0025?
8. (Original post by kiiten)
Oh so moles of salt = 0.01 and moles of acid remaining = 0.0025?
9. (Original post by charco)
Ah thank you!

I see where i was going wrong now - i was using 0.0125 as moles of acid
10. (Original post by charco)
Attachment 617716617718

Hey i have another pH calc question.

How did they get the part ive circled in red?
Attached Images

11. (Original post by kiiten)
Attachment 617716617718

Hey i have another pH calc question.

How did they get the part ive circled in red?
mol total acid - mol excess acid = mol acid used.

mol acid used /2 = mol carbonate reacted
12. (Original post by charco)
mol total acid - mol excess acid = mol acid used.

mol acid used /2 = mol carbonate reacted
Ohh yeah thanks i didnt write out the equation thats why. Thanks

2HCl + CaCO3
13. (Original post by charco)
mol total acid - mol excess acid = mol acid used.

mol acid used /2 = mol carbonate reacted
Why are the following not bronsted-lowry acid-base reactions?

NH3 + HCI ------> NH4+ + CI-
KF + PF5 -------> K+ + PF6-
14. (Original post by kiiten)
Why are the following not bronsted-lowry acid-base reactions?

NH3 + HCI ------> NH4+ + CI-
KF + PF5 -------> K+ + PF6-
The first one is, but the second does not involve proton transfer.
15. (Original post by charco)
The first one is, but the second does not involve proton transfer.
Ah of course why didnt i see that.

For this one is it because the base doesnt accept a proton etc.
OH- + CH3CI ---------> CH3OH + CI-
16. (Original post by kiiten)
Ah of course why didnt i see that.

For this one is it because the base doesnt accept a proton etc.
OH- + CH3CI ---------> CH3OH + CI-
yes
17. (Original post by kiiten)
Attachment 617716617718

Hey i have another pH calc question.

How did they get the part ive circled in red?
Where do you get these model answers from? Thanks
18. (Original post by Mina_)
Where do you get these model answers from? Thanks
Sorry for not answering earlier , theyre from chemsheets

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