Calculate concentrations of the major species present in a buffer solution prepared by adding 1.4 grams of NaOH to 140ml of 0.750 M NH4Cl?

Molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol
Initial concentration of NaOH before reaction = [(1.4 g) / (40.0 g/mol)] / (140/1000 L) = 0.25 M
Initial concentration of NH₄Cl before reaction = 0.75 M

Consider the reaction between NH₄⁺ and OH⁻ ions, where Cl⁻ and Na⁺ are spectator ions.
NH₄⁺(aq) + OH⁻(aq) → NH₃(aq) + H₂O(l)
In reaction, mole ratio NH₄⁺ : OH⁻ = 1 : 1 : 1
Initial concentration of NH₄⁺ ions before reaction (0.75 M) > Initial concentration of OH⁻ ions before reaction (0.25 M)
Hence, NH₄⁺ ions is in excess, and OH⁻ ions almost completely reacts to form NH₃.

[NH₄⁺] = (0.750 - 0.250) M = 0.50 M
[NH₃] = 0.25 M

Since Cl⁻ and Na⁺ are spectator ions, then :
[Na⁺] = 0.25 M
[Cl⁻] = 0.75 M

Consider the dissociation of NH₃ :
___________ NH₃(aq) _ + _H₂O(l) _ ⇌ _ NH₄⁺(aq) _ + _ OH⁻(aq) ___ Kb = 1.8 × 10⁻⁵
Initial: _____ 0.25 M ______________ 0.50 M _______ 0 M
Change: ____ -y M ________________ +y M ________ +y M
At eqm: __ (0.25 - y) M ___________ (0.50 + y) M ____ y M

As Kb is very small and due to the common ion effect in the presence of NH₄⁺, the dissociation of NH₃ would be to a very small extent.
It is assumed that 0.50 > 0.25 ≫ y, i.e.
[NH₃] at equilibrium = (0.25 - y) M ≈ 0.25 M
and [NH₄⁺] at equilibrium = (0.50 + y) M ≈ 0.50 M

Kb = [NH₄⁺] [OH⁻] / [NH₃]
1.8 × 10⁻⁵ = 0.50 × [OH⁻] / 0.25
[OH⁻] = (1.8 × 10⁻⁵) × (0.25/0.50) M = 9.0 × 10⁻⁶ M
[H⁺] = Kw / [OH⁻] = (1.0 ×10⁻¹⁴) / (9.0 × 10⁻⁶) = 1.1 × 10⁻⁹ M

Concentrations of the major species present in the buffer soluton :
[NH₄⁺] = 0.50 M
[NH₃] = 0.25 M
[Na⁺] = 0.25 M
[Cl⁻] = 0.75 M
[OH⁻] = 9.0 × 10⁻⁶ M
[H⁺] = 1.1 × 10⁻⁹ M