What is [H3O+] for a buffer solution that is 0.200 M in acid and 0.500 M in the corresponding salt, if the weak acid's Ka = 5.50 × 10−7?

Method 1 :

____________ acid(aq) __ + H₂O ⇌ __ salt(aq) __ + __ H₃O⁺(aq) ___ Ka = 5.50 × 10⁻⁷
Initial: ______ 0.200 M ____________ 0.500 M _______ 0 M
Change: ______ -y M ______________ +y M ________ +y M
At eqm: ___ (0.200 - y) M _______ (0.500 + y) M ______ y M

As Ka is very small, the dissociation of the acid would be to a very small extent.
It is assumed that 0.500 > 0.200 ≫ y, i.e.
[acid] at equilibrium = (0.200 - y) M ≈ 0.200 M
and [salt] at equilibrium = (0.500 + y) M ≈ 0.500 M

At equilibrium :
Ka = [salt] [H₃O⁺] / [acid]
5.50 × 10⁻⁷ = 0.500 [H₃O⁺] / 0.200
[H₃O⁺] = (5.50 × 10⁻⁷) × (0.200/0.500) M = 2.20 × 10⁻⁷ M


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Method 2 :

pH = pKa + log([salt]/[acid])
pH = -log(5.50 × 10⁻⁷) + log(0.500/0.200)
[H₃O⁺] = 10^{-[-log(5.50 × 10⁻⁷) + log(0.500/0.200)]} M = 2.20 × 10⁻⁷ M