chem help?

As Kₐ₁ ≫ Kₐ₂, [H⁺] only depends on the first dissociation.

[H₂CO₃]ₒ = 1.55 × 10⁻² M = 0.0155 M

Consider the first dissociation of H₂CO₃ :
__________ H₂CO₃(aq) __ ⇌ __ HCO₃ ⁻(aq) __ + __ H⁺(aq) ___ Kₐ₁ = 4.5 × 10⁻⁷
Initial: ____ 0.0155 M _______ 0 M __________ 0 M
Change: _____ -y M ________ +y M _________ +y M
At eqm: _ (0.0155 - y) M _____ y M __________ y M

As Kₐ₁ is very small, the dissociation of H₂CO₃ would be to a very small extent.
It is assumed that 0.0155 ≫ y, i.e. [H₂CO₃] at equilibrium = (0.0155 - y) M ≈ 0.0155 M

Kₐ₁ = [HS⁻] [H⁺] / [H₂S]
4.5 × 10⁻⁷ = y² / 0.0155
y = √[0.0155 × (4.5 × 10⁻⁷)]
y = 8.35 × 10⁻⁵

pH = -log[H⁺] = -log(8.35 × 10⁻⁵) = 4.08