chemistry of solutions?

Quantity of electricity passed = It = (11.3 A) × (73 × 60 s) = 49494 C

1 mol of e⁻ carries 96485 C of electricity.
No. of moles of e⁻ passed = (49494 C) / (96485 C/mol) = 0.513 mol

2H⁺ + 2e⁻ → H₂
Mole ratio H⁺ : e⁻ = 2 : 2 = 1 : 1
No. of moles of H⁺ consumed = 0.513 mol
[H⁺] remained = (0.552 M) - {(0.513 mol) / (1.25 L)} = 0.142 M

pH after 73 minutes = -log[H⁺] = -log(0.142) = 0.85

Half reaction: 2 H+ + 2 e- --> H2(g)

Coulombs = ampere X seconds = 11.3 A X 73 min X 60 s/min = 49494 C
moles electrons = 49494 / 96485 = 0.513 mol e-
Moles H+ consumed = 0.513 mol e- X (2 mol H+ / 2 mol e-) = 0.513 mol H+ consumed

Moles H+ initially present = 1.25 L X 0.552 mol/L = 0.690 mol H+
H+ remaining = 0.69 - 0.513 = 0.177
[H+] = 0.177 mol / 1.25 L = 0.142 M
pH = -log (0.142) = 0.85

Electrochemistry.....

Current is the rate of flow of charge and is given in amperes (a derived unit) which is the same as coulombs per second. As charge flows through the electrolytic cell the water in the solution is electrolyzed forming hydrogen gas and bromine. The loss of hydrogen during the formation of hydrogen gas will increase the pH (making it less acidic).

73 minutes is 4380 seconds
1 mol of electrons has a charge of 96,485 coulumbs (1 Faraday)

H+ + 2Br- + H2O --electricity--> H2(g) + Br2(aq) + OH-

4380 s x (11.3 C / s) x (1 mol e- / 96485C) x (1 mol H2 / 2 mol e-) x (1 mol H+ / 1 mol H2) = 0.2565 mol H+

Initial moles of H+ = 1.25L x (0.552 mol H+ / 1L) = 0.690 mol H+

0.690 mol H+ - 0.2565 mol H+ = 0.4335 mol H+ remaining after 73 minutes
The concentration of H+ = 0.4335 mol H+ / 1.25L = 0.3468 M H+
pH = -log[H+] = -log(0.3468) = 0.460