Calculate the PH of 0.2 M solution potassium acetat. Write a Balanced equation for all reactions and equilibrium that describe this solution?

Refer to: http://www.chegg.com/homework-help/questions-and-answers/weak-acid-ka-weak-base-kb-ch3cooh-acetic-acid-18-x-10-5-nh3-ammonia-176-x-10-5-c6h5cooh-be-q8963943
Ka(CH₃COOH) = 1.8 × 10⁻⁵
Then, Kb(CH₃COO⁻) = Kw / Ka(CH₃COOH) = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)

__________ CH₃COO⁻(aq) + H₂O(l) ⇌ __ CH₃COOH(aq) __ + __ OH⁻(aq) ___ Kb = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)
Initial: ______ 0.2 M ___________________ 0 M ____________ 0 M
Change: _____ -y M ___________________ +y M ___________ +y M
At eqm: __ (0.2 - y) M __________________ y M ____________ y M

As Ka is very small, the dissociation of CH₃COO⁻ ion would be to a very small extent.
It is assumed that 0.2 M ≫ y M, i.e. [H₂A] at equilibrium = (0.2 - y) M ≈ 0.2 M

Ka = [CH₃COOH] [OH⁻] / [CH₃COO⁻]
(1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = y² / 0.2
y = √[0.2 × (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)] = 1.05 × 10⁻⁵

pOH = -log[OH⁻] = -log(1.05 × 10⁻⁵) = 5.0
pH = pKw - pOH = 14.0 - 5.0 = 9.0