What is the pH of a 0.0075 M solution of H2SO4 solution? Ka1 = 1.0x103, Ka2 = 1.2x10-2?

Kₐ₁ is very large, and thus first dissociation of H₂SO₄ almost goes complete.
First of all, 0.0075 M H₂SO₄ undergoes almost complete first dissociation to give 0.0075 M HSO₄⁻ and 0.0075 M H₃O⁺.

Consider the second dissociation of H₂SO₄ :
HSO₄⁻(aq) + H₂O(l) ⇌ SO₄²⁻(aq) + H₃O⁺(aq) …. Kₐ₂ = 1.2 × 10⁻²

Initial concentrations :
[HSO₄⁻]ₒ = [H₃O⁺]ₒ = 0.0075 M
[SO₄²⁻]ₒ = 0 M

Let y M be the concentration of HSO₄⁻ dissociated in the second dissociation.
Equilibrium concentrations :
[HSO₄⁻] = (0.0075 - y) M
[H₃O⁺] = (0.0075 + y) M
[SO₄²⁻] = y M

Kₐ₂ = [H₃O⁺] [SO₄²⁻] / [HSO₄⁻]
1.2 × 10⁻² = (0.0075 + y) y / (0.0075 - y)
0.000090 - 0.012y = 0.0075y + y²
y² + 0.0195y - 0.000090 = 0
y = {-0.0195 ± √(0.0195² + 4×0.000090)} / 2
Rejecting the negative answer, y = 0.0039

At equilibrium, [H₃O⁺] = (0.0075 + 0.0039) M = 0.0114 M
pH = -log[H₃O⁺] = -log(0.0114) = 1.9

H2SO4 + H2O -----------> H3O + + HSO4 2-
** H2SO4 is a strong acid so it 'll be ionized completely => [ H3O ]+ = [HSO4]2- = 0.0075
HSO4 2- + H2O ----------> H3O+ + SO4 2-
At the beginning [HSO4]2- = [ H3O]+ = 0.0075, HSO4 2- ionized x moles to produce x moles H3O+, so at EQ, [ HSO4]2- = 0.0075 - x ; [ H3O]+ = 0.0075 + x ; [ SO4]2- = x ** ka2 = [x(0.0075 + x)] / ( 0.0075 - x ) = 1.2 x 10^-2. Solve this quadratic equation to find x then [ H3O]+ = 0.0075 + x => pH