Wait, is this asking "What additional information is needed to know/find the initial number of moles of H2CO3"? Or is it asking for what information you'd need, in addition to initial moles H2CO3 and 0.5M NaOH, in order to find the pH of the rainwater?
If it's to find the moles of acid, II + III aren't needed.
At the equivalence point, you've added a chemically equivalent amount of base (i.e., moles acid = moles base). So you start with X moles acid, and you start with 0 moles base.
If you know the volume of NaOH and the molarity of NaOH, then you can find the moles of acid in the sample:
Start with knowing:
Molarity = (moles/volume)
moles = (Molarity)(volume) (rearranged for moles)
So we apply that to this question:
(moles base) = (volume** base)(Molarity of base)
**volume of base added to get to equivalence point
And we know @ equivalence point:
(moles base) = (moles acid)
So,
(volume base)*(Molarity of base) = (moles base) = (moles acid)
which leads to:
(volume base)*(Molarity of base) = (moles acid)
So you only need to know the volume and molarity of the base to calculate the moles of acid. If we needed to find the molarity of the acid, then we'd need to know the volume of rain water. And for this problem, to find the moles of acid, then you don't need to use the Ka.
You don't need to use the Ka value for this reaction because you have a strong base, NaOH. A strong base always reacts completely. However many moles of strong base you add, it reacts with an equivalent number of moles weak acid.
So, if you start with 10 moles weak acid and you add 5 moles strong base, then 5 moles strong base have reacted and 5 moles of weak acid have reacted. This is the half-equivalence point (5 moles weak acid and 5 moles of its conjugate base present in mixture = 1:1 ratio = buffer). If you add 10 moles strong base, then 10 moles strong base have reacted and 10 moles weak acid have reacted. And, you're at the equivalence point. An equivalent number of moles of weak acid/strong base have been mixed in the solution.
And, since a strong base is used, it goes to completion and it's not a situation where you're dealing with an equilibrium between strong base/weak acid and their conjugate products. It reacts completely (however much strong base you add), giving the conjugate base for the weak acid and H2O.
The pH @ equivalence point for for a weak acid/strong base titration is >7 because the weak acid's conjugate base is basic (in H2O), while the strong base's conjugate acid isn't acidic (in H2O). The weak acid's conjugate base, since it's actually basic, goes on to react (hydrolysis) with the H2O formed from the initial weak acid/strong base reaction and produces hydroxide and the original weak acid. So that production of hydroxide makes the solution basic at the equivalence point; it's not because of the strong base. The moles weak acid neutralized by NaOH = moles NaOH added.
The Ka value is relevant to titration when you want to find the pH. At the 1/2 equivalence point in a titration of a weak acid/strong base, the ***pH = pKa.
This is because at the 1/2 eqivalence point you've formed a buffer in 1:1 ratio with the weak acid (HA) and its conjugate base (A-). buffers work best when the [HA]:[A-] is closer to 1:1. [HA] = [A-] at the 1/2 eq point, this is where they're in a 1:1: ratio (maximum buffering capacity <-> max ability to resist changes in pH <-> slope of graph closest to 0 <-> pH changing the least / with respect to volume added).
***pH = pKa is derived from the henderson-hasselbach equation:
pH = pKa + log([A-]/[HA])
and when [A-]:[HA] in 1:1 ratio, then log([A-]/[HA]) = log(1) = 0
So you just get:
pH = pKa + log(1)
pH = pKa + 0
pH = pKa
You generally use that when you're supposed to find that information from a graph. So, I guess the point of that is that's when Ka is relevant to titrations -- to find the pH from Ka (when you've got a buffer present in some way).
