Calculate the pH of a 7.2x10^-3 -M solution?

The first dissociation of H₂SO₄ is complete.
H₂SO₄(aq) + H₂O(l) → HSO₄⁻(aq) + H₃O⁺(aq) ____ Kₐ₁ is very large
Hence, 0.0072 M (or 7.2 × 10⁻³ M) of HSO₄⁻ and 0.0072 M of H₃O⁺ are formed in the first dissociation of H₂SO₄.

Consider the second dissociation of H₂SO₄:
____________ HSO₄⁻(aq) + H₂O(l) ⇌ SO₄²⁻(aq) + H₃O⁺(aq) ____ Kₐ₂ = 1.2 × 10⁻²
Initial (M): ___ 0.0072 _____________ 0 ______ 0.0072
Change (M): ___ -y _______________ +y ________ +y
Eqm (M): ___ 0.0072 - y ___________ y ______ 0.0072 + y

At eqm: Ka = [SO₄²⁻] [H₃O⁺] / [HSO₄⁻]
1.2 × 10⁻² = y (0.0072 + y) / (0.0072 - y)
0.012 (0.0072 - y) = y (0.0072 + y)
0.0000864 - 0.012y = 0.0072y + y²
y² + 0.0192y - 0.0000864 = 0
y = [-0.0192 ± √(0.0192² + 4*0.0000864)] / 2
y = 0.00376 or y = -0.0230 (rejected)

pH = -log[H₃O⁺] = -log(0.0072 + 0.00376) = 1.96 ≈ 2.0