For 5) f) i)
A student sets out to make a buffer solution. The student measures out 27 cm3 of
0.050 mol dm–3 HA solution and reacts it with one-third of the volume of 0.10 mol dm–3 sodium
hydroxide needed for complete neutralisation.
HA + NaOH --------> NaA + H2O
So first of all the moles of Acid HA.
n = C X V / 1000
n = (0.05 X 27) / 1000
n = 1.35X10-3 moles
As the volume added was a 1/3 of the require volume to neutralise, the amount of moles in the volume actually added will be the moles required / 3.
So...
actual n = (1.35X10-3) / 3
actual n = 4.5X10-4
V added = (n X 1000) / C where C is the concentration of sodium hydroxide
V added = ( 4.5X10-4 X 1000) / 0.1
V added = 4.5 cm3
5))ii)
Here you use a lot of the values you calculated above ^^^
moles of HA added to the solution = 1.35X10-3 moles
moles of A- added to the solution = 4.5X10-4 moles
For part 2 I have 0.2 less than the mark scheme so going to have a quick check of the ol' solutions before I post this.
And for 1)c) if you draw out the structure you'll find the there is in fact an oxygen between the CH3 and the H bonded to the oxygen. This there is no splitting.