F325 Hard chemistry Q help

Ok well...

Moles of Butanoic acid = 0.250 x 50 / 1000 = 0.0125
Moles of NaOH = 0.05 x 50 / 1000 = 2.5 x 10^-3

CH3CH2CH2COOH + NaOH --> CH3CH2CH2COONa + H2O

so 0.0125 moles of butanoic acid react with 2.5 x 10^-3 moles of NaOH to form 2.5 x 10^-3 moles of CH3CH2CH2COONa and water (but water is not needed here) because butanoic acid is in excess and its a 1 to 1 reaction. all the CH3CH2CH2COONa dissociate into CH3CH2CH2COO- to give you 2.5 x 10^-3 moles of CH3CH2CH2COO-

So 2.5 x 10^-3 moles of butanoic acid are used up to give: 0.0125 - 2.5x 10^-3 = 0.01 moles of Butanoic acid

So from the Ka equation:

[H+] = Ka [ CH3CH2CH2COOH] / [CH3CH2CH2COO-]
= (1.51 x 10^-5) x (0.01) / (2.5 x 10^-3)
=6.04 x 10^-5

pH = 4.22 (2.d.p)

Haha no prob

Basically u have 0.0125 moles of butanoic acid reacting with 2.5*10^-3 moles of NaOH, to give you your products. This is a 1 to 1 reaction, where you have 1 mole of everything in the chemical reaction. So if you use up 'X' number of moles of butanoic acid, u must also use 'X' moles of NaOH. Here you use all your NaOH because you have a lot more moles of butanoic acid (hence it is in excess) to react and the reaction goes to completion.

So if 2.5*10^-3 moles of NaOH are used up, that leaves you with 0 moles of it after the reaction; this means 2.5*10^-3 moles of butanoic acid are also used up to give you (0.0125 - 2.5*10^-3) moles of butanoic acid after the reaction, which is 0.01 moles.

You then use the rest in the Ka equation with 0.01 as ur [CH3CH2CH2COOH] value.

Haha no prob

Basically u have 0.0125 moles of butanoic acid reacting with 2.5*10^-3 moles of NaOH, to give you your products. This is a 1 to 1 reaction, where you have 1 mole of everything in the chemical reaction. So if you use up 'X' number of moles of butanoic acid, u must also use 'X' moles of NaOH. Here you use all your NaOH because you have a lot more moles of butanoic acid (hence it is in excess) to react and the reaction goes to completion.

So if 2.5*10^-3 moles of NaOH are used up, that leaves you with 0 moles of it after the reaction; this means 2.5*10^-3 moles of butanoic acid are also used up to give you (0.0125 - 2.5*10^-3) moles of butanoic acid after the reaction, which is 0.01 moles.

You then use the rest in the Ka equation with 0.01 as ur [CH3CH2CH2COOH] value.

I find the explanation a bit weird though. You're subtracting the moles of NaOH from the moles of Butanoic acid, but how are you using the MOLES of butanoic acid to find the [H+]? Don't you need to use the concentration of butanoic acid, not the moles? (since Ka is made from the concentration of products over reactants) Baffleddd

You do use the concentrations....

He subtracted the moles of NaOH from the moles of Butanoic acid and used that value...unless he converted the moles back to conc or I misread something. I haven't done this question myself but it was discussed in class

No offence to the guy above but what he said was wrong you dont use the moles....

You use the remaining moles of acid which he found correctly and the number of moles of the salt of the acid and convert these in to concentrations and bang them in to the equation.

It took me 40 minutes to do this question and only after those 40mins and help from a great chemist @charco on here I figured it out..

Yes this questions was hard here is a quote from the examiners report

Only the very best candidates calculated the concentrations
of butanoic acid and butanoate ions in the resulting buffer solution and obtained a correct pH
value of 4.22 for all marks.

So he DID use the mole value. I know you use the concentration in the Ka equation hence why I was really confused. Well, at least now we kinda know what to expect if something difficult like this comes up...