what is the ph of 0.1 M nh4cl that hydrolyses in 1 dm3 h2o?Kb =1.8 × 10^-5; log7.42=0.87?

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) …… Kb = 1.8 × 10⁻⁵

__________ NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) …… Ka = Kw / Kb = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)
Initial: ______0.1 M ___________ 0 M ____ 0 M
Change: _____ -y M __________ +y M ____ +y M
At eqm: ___ (0.1 - y) M _________ y M ____ y M

Ka is very small. Then, the dissociation of NH₄⁺ is to a very small extent.
Then, we can assume that 0.1 ≫ y
Hence, [NH₄⁺] at eqm = (0.1 - y) M

Ka = [NH₃] [H₃O⁺] / [NH₄⁺]
y² / 0.1 = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)
y = √[0.1 × (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)]
y = 7.42 × 10⁻⁶

pH = -log[H₃O⁺] = -log(7.42 × 10⁻⁶) = -log(7.42) - log(10⁻⁶) = -0.87 + 6 = 5.13

much more easier:

pH = 0.5(pKa - log(0.1)) = 5.13

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