Oh no ok let me re-draw everything out.
For Qn 12 (same paper), could you please take a look at my
working? Where did I go wrong?
http://imgur.com/XnR22rs
I put B as my answer, because it's slightly more than 10cm3.
Maleic acid, being a weak diprotic acid, can be represented by H2A.
Note that the x-axis specifies volume of weak acid added, not
strong base added.
At 20cm3 of H2A added, you have 1st equivalence point, hence you
only have HA-, which being amphiprotic, *can* function as a
buffer.
(Fyi, 2nd equivalence point to exactly fully deprotonate all of
H2A, is at 10cm3 of H2A added against 20cm3 of NaOH.)
However, at 40cm3 of H2A added (ie. 20cm3 excess H2A), you have
equal moles hence equal molarities of HA- and H2A, hence you have
*maximum* buffer capacity. Concordantly, D is the answer.
Options A and B are obviously wrong, because you have a
significant excess of the strong base NaOH, which means all of the
1st acidic protons have been removed, and most of the 2nd acidic
proton as well. In fact, for option A, all of H2A is completely
deprotonated to A2-, which means it is only basic, hence not a
buffer at all). While for option B (which at 12cm3 of H2A against
20cm3 of OH-, leaving behind only a little bit of HA- left) is only
effective at buffering against incoming strong acids, but almost
totally useless at buffering against incoming strong bases. Thus
options A and B are obviously wrong.
