Equilibria, Energetics and Elements (F325) - June 2011 Exam.

moles mno4 = 4.69 x 10-4
moles H2O2, you divide by 2 and times by 5 (from the previous part) = 1.1725x10-3

then you x by 10 because (250/25) = 10

so you get 1.1725x10-2 in 250cm3

x by 40 to get it into dm3

then x by the mr of H202 (34) to get the mass

= 1.1725x10-2 x 40 x 34 = 15.9gdm-3

Completely agree although it was a nightmare and it stayed that way for me .

OCR biol 2day was horrific, now i'm going to need to do really well in chem on wed

Does anyone mind explaining buffers to me?
/If it has already been covered here, could someone link me to the post?

Thank you very much.

hmm dilution always panics me :/

so you times by ten because the 250cm^3 has ten times the amount of substance, but this is also the amount of substance as the 25cm3 undiluted because theres not actually any more/less molecules?

could you also work this out by finding the conc of h2o2 and dividing by 10?

buffer solution needs high concentrations of weak acid and high concentrations of conjugate base
This can be obtained by an acid-alkali reaction with the acid in excess, this gives large concentration of acid and salt which can dissociate into conjugate base

weak acid dissociates CH3COOH---->CH3COO- + H+ Equilibrium
Salt dissociates CH3COO-Na+ ------>CH3COO- + Na+

If acid is added [H+] this reacts with the conjugate base and the equilibrium shifts left to try and remove the [H+]

If an alkali is added [OH-] this reacts with the [H+] from the weak acid H+ + OH- ----> H20
equilibrium shifts right to replace reacted [H+]

The point of a buffer is to minimise small changes in pH

Is the weak acid with its conjugate base a different buffer to the salt of the weak acid and its conjugate base? So a weak acid and its conjugate base could be used as a buffer as well as the salt of the weak acid and its conjugate base? (I don't know if that is clear, so sorry about that).

Thanks.

No a weak acid can not be a buffer on its own as a weak acid only partially dissociates meaning there will be not a high enough concentration of the conjugate base. That is why the salt is used to create a high concentration of the the conjugate base needed.

I think this is what you meant right?

chem5 will be similar to jan11

does anyone have any notes for this paper?

Ohhhh yeah, thanks. How does the salt of the weak acid produce high levels of H+?

it dosen't. it fully dissociates to produce the salt ion and the metal ion, which we can ignore as it doesn't effect the buffer.

Where does the high concentration of H+ come from? Or there isn't a high conc, only a weak one due to the weak acid?

Yea the buffer solution only contains high concentration of the weak acid and its conjugate base.
When an alkali is added [OH-] the small concentration of [H+] reacts with it. The equilibrium of the dissociation of the weak acid shifts to the right to try and restore the [H+] concentration

how do u know lol

somebody posted some a few pages back http://www.thestudentroom.co.uk/show...&postcount=892