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buffer solution- really confused :S

The student adds 50 cm3 of 0.25 mol dm-3 butanoic acid to 50 cm3 of 0.05 mol dm-3 sodium hydroxide. A buffer solution forms. Calculate the hydrogen concentration of the buffer solution. The ka of butanoic acid is 1.51 x 10-5 mol dm-3.

I don't understand the working out:

MOLES
amount of CH3(CH2)2COOH = 0.0100 mol (<-- how did they get this I tried n= cV but I couldn't get this value)
amount of CH3(CH2)2coo- = 0.0025 mol (<-- how do we get this value when we aren't given any information for this compound)

CONCENTRATION
[CH3(CH2)2COOH] = 0.1 mol dm-3 (<-- where did they get this from??)
[CH3(CH2)2COO-] = 0.025 mol dm-3 (<-- same)

[H+] = 1.51 x 10-5 x (0.100/0.025) = 6.04 x 10-5 mol dm-3

(edited 11 years ago)

Reply 1

Oh Lola!

Ive worked it out, just need to type it up :smile:

Reply 2

username1003420

Original post by Oh Lola!

Ive worked it out, just need to type it up :smile:

Thank you! :biggrin:

Reply 3

Oh Lola!

You first have to find the excess moles which is: moles of butanoic acid- moles sodium hydroxide
(50/1000) x 0.25 - (50/1000) x 0.05 = 0.01
(xs/moles NaOH) x Ka = 6.04 x 10-5

Reply 4

username1003420

Original post by Oh Lola!

You first have to find the excess moles which is: moles of butanoic acid- moles sodium hydroxide
(50/1000) x 0.25 - (50/1000) x 0.05 = 0.01
(xs/moles NaOH) x Ka = 6.04 x 10-5

What about the mol for CH3(CH2)2COO-? Why isn't that in the working out, and also what about the concentrations? Why is the last bit divided by the mol of NaOH I thought it was supposed to be CH3(CH2)2COO-??

I'm sorry for all the questions it's just I really want to maximise my chances of getting the full 5 marks for these questions in the real exam and we're given 1 mark per line that's in the mark scheme.

(edited 11 years ago)

Reply 5

Oh Lola!

I have extremely bad memory and cant remember most of this, though I only did the exam in Jan. Ill try figure it out as I may need to resit this exam anyway

Reply 6

username1003420

Original post by Oh Lola!

I have extremely bad memory and cant remember most of this, though I only did the exam in Jan. Ill try figure it out as I may need to resit this exam anyway

Thank you, but don't worry about it if you can't work it out, I'll just ask my chemistry teacher. :smile:

Reply 7

Oh Lola!

Original post by tammie123

The student adds 50 cm3 of 0.25 mol dm-3 butanoic acid to 50 cm3 of 0.05 mol dm-3 sodium hydroxide. A buffer solution forms. Calculate the hydrogen concentration of the buffer solution. The ka of butanoic acid is 1.51 x 10-5 mol dm-3.

I don't understand the working out:

MOLES
amount of CH3(CH2)2COOH = 0.0100 mol (<-- how did they get this I tried n= cV but I couldn't get this value)
amount of CH3(CH2)2coo- = 0.0025 mol (<-- how do we get this value when we aren't given any information for this compound)

CONCENTRATION
[CH3(CH2)2COOH] = 0.1 mol dm-3 (<-- where did they get this from??)
[CH3(CH2)2COOH] = 0.025 mol dm-3 (<-- same)

[H+] = 1.51 x 10-5 x (0.100/0.025) = 6.04 x 10-5 mol dm-3

Right, I was also confused why you said NaOH instead of CH3(CH2)2coo-
I think it was meant to say CH3(CH2)2coo- instead of NaOH

So moles CH3(CH2)2coo- = 2.5 x 10-3
Excess moles (as I showed before) = 0.0100

by rearranging the formula: Ka = [H+][A-]/[HA]
H+ = 1.51 x 10 -5 X 0.0100/ 0.0025 =6.04 x 10-5

I think there are two methods for working these out and this one doesn't use the concentrations. hope this makes more sense..

(edited 11 years ago)

Reply 8

username1003420

Original post by Oh Lola!

Thank you, it does make more sense. But just one more thing why do we use the information given for NaOH (i.e. concentration and volume and stuff) to work out the mol for CH3COO-? How are the two related?

anyone?

Does anyone know the purpose of adding water to a buffer solution ? And why the buffer cannot resist change in pH upon large addition of acid or base?