chemistry help please how do I do this?

Initial concentration of NH₃, [NH₃]ₒ = (0.050 mol) / (500/1000 L) = 0.10 M

_____________ NH₃(aq) ___ + ___ H₂O(l) ___ ⇌ NH₄⁺(aq) ___ + ___ OH⁻(aq) ______ Keq = 1.8 × 10⁻⁵
Initial: ________ 0.1 M _____________________ 0 M _____________ 0 M
Change: _______ -y M _____________________ +y M ____________ +y M
Equilibrium: __ (0.1 - y) M ___________________ y M _____________ y M

As Keq is very small, the dissociation of NH₃ is to a very small extent.
It can be assumed that 0.1 ≫ y and thus [NH₃] at equilibrium = (0.1 - y) M ≈ 0.1 M

At equilibrium:
Keq = [NH₄⁺] [OH⁻] / [NH₃]
1.8 × 10⁻⁵ = y² / 0.1
y = √(0.1 × 1.8 × 10⁻⁵) = 1.34 × 10⁻³

pOH = -log[OH⁻] = -log(1.34 × 10⁻³) = 2.9
pH = pKw - pOH = 14.0 - 2.9 = 11.1

show equation with concentrations
(0.050 mol / 0.500 mL) = 0.100 M)

NH3 + H2O → NH4+ + OH–
(0.100-X) , , , , , X , , , , , X
and
Ksp = [NH4+] [OH–] / [NH3]
1.8E-5 = X*X / (0.100-X)
1.8E-6 - 1.8E-5 X = X^2
X^2 + 1.8E-5 X - 1.8E-6 = 0
X = 1.33E-3 (the positive root)
and
pOH = - log[OH-] = -log(1.33E-3) = 2.88
then
pH = 14.00 - pOH = 14.00 - 2.88 = 11.12

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