Help with calculating the molarity of cation, anion, and pH?

(a) the molarity of the Na+ cation
The Na+ cation comes from dissociation of the NaOH
Final volume = 21.2mL + 48.4mL = 69.6mL
Molarity of Na+ in final solution = 0.193*21.2*69.6 = 0.0588M

(b) the molarity of the Cl anion
The Cl- anion comes from the dissociation of the HCl:
Molarity of Cl- in final solution = 0.131*48.4/69.6 = 0.0911M

(c) the pH of the final solution
Mol NaOH in 21.2mL of 0.193M solution = 21.2/1000*0.193 = 0.00409 mol NaOH
Mol HCl in 48.4mL of 0.131M solution = 48.4/1000*0.131 = 0.00634 mol HCl
HCl reacts with NaOH in 1:1 molar ratio
There will be 0.00634 - 0.00409 = 0.00225 mol HCl in excess in 0.0696L solution
Molarity of HCl in solution = 0.00225/0.0696 = 0.0323 M HCl unreacted
HCl disociates completely
[H+] = 0.0323M
pH = -log [H+]
pH = -log 0.0323
pH = 1.490

(d) the pOH of the final solution
pOH = 14.00 - pH
pOH = 14.00 - 1.490
pOH = 12.510

I have quoted all answers to 3 significant digits.

(a)
Final volume = (21.2 + 48.4) mL = 69.6 mL

Na⁺ ion is spectator ion.
Initial molarity of Na⁺ cation = Initial molarity of NaOH = 0.193 M
Final molarity of Na⁺ cation = (0.193 M) × (21.2/69.6) = 0.0588 M


(b)
Cl⁻ ion is also spectator ion.
Initial molarity of Cl⁻ anion = Initial molarity of HCl = 0.131 M
Final molarity of Cl⁻ anion = (0.131 M) × (48.4/69.6) = 0.0911 M


(c)
Initial number of moles of OH⁻ ion = (0.193 mol/L) × (21.2/1000 L) = 0.00409 mol
Initial number of moles of H⁺ ion = (0.131 mol/L) × (48.4/1000 L) = 0.00634 mol

H⁺ + OH⁻ → H₂O
1 mole of H⁺ ion reacts with 1 mole of OH⁻ ion.
As (Initial number of moles of H⁺ ion) > (Initial number of moles of OH⁻), H⁺ is in excess.
Number of moles of unreacted H⁺ = (0.00634 - 0.00409) mol = 0.00225 mol
Final [H⁺] = (0.00225 mol) / (69.6/1000 L) = 0.0323 M

pH = -log[H⁺] = -log(0.0323) = 1.5


(d)
pOH = pKw - pH = 14.0 - 1.5 = 12.5