Chemistry help please?

(a)
CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) …. Ka

Initial concentrations :
[CH₃COOH]ₒ = 0.240 M
[CH₃COO⁻]ₒ = [H₃O⁺]ₒ = 0 M

At equilibrium :
Let x M be the molarity of CH₃COOH consumed.
[CH₃COOH] = (0.240 - x) M ≈ 0.240 M, assuming that 0.240 ≫ x
[CH₃COO⁻] = [H₃O⁺] = x M

Ka = [CH₃COO⁻] [H₃O⁺] / [CH₃COOH]
1.8 × 10⁻⁵ = x² / 0.240
x = 2.1 × 10⁻³
(Refer to the value of x. The assumption that 0.240 ≫ x holds.)

[H₃O⁺] = x M = 2.1 × 10⁻³ M
[OH⁻] = Kw / [H₃O⁺] = (1 × 10⁻¹⁴) / (2.1 × 10⁻³) M = 4.8 × 10⁻¹² M
pH = -log[H₃O⁺] = -log(2.1 × 10⁻³) = 2.7
pOH = 14 - pH = 14 - 2.7 = 11.3


(b)
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) …. Kb

Initial concentrations :
[NH₃]ₒ = 0.240 M
[NH₄⁺]ₒ = [OH⁻]ₒ = 0 M

At equilibrium :
Let y M be the molarity of NH₃ consumed.
[NH₃] = (0.240 - y) M ≈ 0.240 M, assuming that 0.240 ≫ y
[NH₄⁺] = [OH⁻] = y M

Kb = [NH₄⁺] [OH⁻] / [NH₃]
1.8 × 10⁻⁵ = y² / 0.240
y = 2.1 × 10⁻³
(Refer to the value of y. The assumption that 0.240 ≫ y holds.)

[OH⁻] = y M = 2.1 × 10⁻³ M
[H₃O⁺] = Kw / [OH⁻] = (1 × 10⁻¹⁴) / (2.1 × 10⁻³) M = 4.8 × 10⁻¹² M
pOH = -log[OH⁻] = -log(2.1 × 10⁻³) = 2.7
pH = 14 - pOH = 14 - 2.7 = 11.3