pH calculation help :S

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

Help please :s-smilie:

Do the stoichiometry of the reaction to find the exact constitution of the products and unreacted reactant(s)

Reply 2

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

??

Reply 3

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+]

Reply 4

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??

Reply 5

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.

Reply 6

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?

Reply 7

Original post by kiiten

Oh so moles of salt = 0.01 and moles of acid remaining = 0.0025?

Reply 8

Original post by charco

Ah thank you! :biggrin:

I see where i was going wrong now - i was using 0.0125 as moles of acid

Reply 9

Screenshot 2017-02-06 20.19.12.png43.0KB

Original post by charco

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Hey i have another pH calc question.

How did they get the part ive circled in red?

Reply 10

Original post by kiiten

Attachment not found

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

Reply 11

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 :smile:

2HCl + CaCO3

Reply 12

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-

Reply 13

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.

Reply 14

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-

Reply 15

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

Reply 16

Mina_

Original post by kiiten

Attachment not found

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

Reply 17