[Rueus_]

I really don't get how to go about this question so if someone could explain it to me I'd really appreciate it.
volume=10,000   L   
strong acid HCl and a weaker one phosphoric acid, H3PO4; pKa1 = 2.148, pKa2 = 7.20, pKa3 = 12.15   
At maximum capacity, chlorine and phosphorus concentrations in the wastewater are 0.5 M and 1.0 M, respectively.      
Assume (i) that chlorine and phosphorus come only from HCl and   phosphoric acid,   respectively, and
(ii) that no H+ was consumed (or neutralized) in the manufacturing process.
(a) Knowing that the technician has to bring the pH of   the   waste water back to a minimum   of 7.2 before   discharging it in   the stream, what is the minimum amount (in kg) of NaOH(s) (40g/mol)   that he has to add to the pool before discharging the   waste water in the stream?
(b) What will be   the pH of the wastewater following the addition of 200   kg of   NaOH to the initial wastewater?
(c) What will then be the number of moles of each phosphorus species (H3PO4, (H2PO4)-, (HPO4)2-and (PO4)3-)   in the pool   after   the addition of NaOH?
(d) What will be   the pH of the wastewater if the technician adds 10.0 kg of NaOH in excess to what is needed to bring the pH to 7.2?

[Borek]

Would it be easier for you to solve if you would have 10,000 L of just HCl solutin? Just H₃PO₄ solution?

They react independently, so the amount of base is just sum of individual amounts.