In this question, you’re making a phosphate buffer.
There is a LOT of prerequisite knowledge required, so please do reach out if any of these steps aren’t making sense to you.
First, it helps to identify the strong and weak acids and bases.
In this problem, HCl is the strong acid, and Na3PO4 (that is, PO4^3-) is the weak base.
More specifically H+ is the strong acid and PO4^3- is the weak base. Cl- and Na+ are spectator ions, and we’re going to ignore them. (This is because both the strong acid, HCl, and the strong electrolyte, Na3PO4 COMPLETELY dissociate into their respective ions when in solution, as they are in this problem.)
Orienting like this helps us write the correct acid/base reaction (again, ignoring the spectator ions).
H+ (aq) + PO4^3- (aq) -> HPO4^2- (aq)
How many moles of strong acid (H+) are reacting with how many moles of weak base (PO4^3-):
0.1 M HCl * 0.050 L * (1mol H+/ 1mol HCl) = 0.005 mol H+
0.1 M Na3PO4 * 0.020 L * (1 mol PO4^3- for every 1 mole of Na3PO4) = 0.002 mol PO4^3-
H+ (aq) + PO4^3- (aq) -> HPO4^2- (aq)
0.005 0.002 0
-0.002 -0.002 + 0.002
0.003 0.002
We still have strong acid left over! Reaction continues:
H+ (aq) + HPO4^2- (aq) -> H2PO4^1- (aq)
0.003 0.002 0
-0.002 -0.002 +0.002
0.001 0 0.002
We still have strong acid left over! Reaction continues:
H+ (aq) + H2PO4^2- (aq) -> H3PO4 (aq)
0.001 0.002 0
0.001 0.001 0.001
0 0.001 0.001
So this buffer will be governed by the equilibrium between [H3PO4] and [H2PO4^2-], which is pKa1 = 2.16
Use Henderson hasselbalch:
pH = pKa + log [base]/[acid]
We can see that in this case the [base] = [acid] = 0.001 moles***. This means that [base]/[acid] = 1 and log of 1 is 0.
Therefore the pH of this solution will equal pka.
pH = pKa1 = 2.16.
(***I did do a short cut: I did not bother converting moles to concentration. THis is because all species will be diluted into the same volume, since they are mixed into a single new solution. If they have the same # of moles, they’ll also have the same molarity. Remember you technically have to convert to MOLARITY prior to using the HEnderson-Hasselbalch equation).