Calculate the pH of a 6.00×10-1 M aqueous solution of ammonium chloride (NH4Cl). (For ammonia, NH3, Kb = 1.80×10-5.)?

Ka for NH₄⁺ = Kw / (Kb for NH₃) = (1.00 × 10⁻¹⁴) / (1.80 × 10⁻⁵) M = 5.56 × 10⁻¹⁰ M
Initial concentration [NH₄⁺]ₒ = 6.00 × 10⁻¹ M = 0.600 M

Consider the dissociation of NH₄⁺ ions :
___________ NH₄⁺(aq) __ ⇌ __ NH₃(aq) __ + __ H⁺(aq) ___ Ka = 5.56 × 10⁻¹⁰ M
Initial: _____ 0.600 M ________ 0 M _________ 0 M
Change: _____ -y M _________ +y M ________ +y M
At eqm: __ (0.600 - y) M ______ y M _________ y M
=

As Ka is very small, then the dissociation of NH₄⁺ ions is to a very small extent.
Hence, we can assume that 0.600 ≫ y, i.e. [NH₄⁺] at eqm = (0.600 - y) M ≈ 0.600 M

Ka = [NH₃] [OH⁻] / [NH₄⁺]
5.56 × 10⁻¹⁰ M = y² / 0.600
y = √[(5.56 × 10⁻¹⁰) × .600]
y = 1.83 × 10⁻⁵ (The assumption that 0.600 ≫ y is correct.)

pH = -log(1.83 × 10⁻⁵) = 4.74

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