What is the pH of a 0.337 M solution of NH4NO3? The ionization constant of the weak base NH3 is 1.8 × 10−5 .?

Ka for NH₄⁺ = Kw / (Kb for NH₃) = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)

Consider the dissociation of NH₄⁺ ions :
____________ NH₄⁺(aq) __ ⇌ __ NH₃ (aq) __ + __ H⁺(aq) ___ Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)
Initial: _____ 0.337 M __________ 0 M _________ 0 M
Change: ______ -y M __________ +y M _________ +y M
At eqm: ___ (0.337 - y) M _______ y M __________ y M

As Ka is very small, the dissociation NH₄⁺ is to a very small extent.
It is assumed that 0.337 ≫ y, and thus [NH₄⁺] at eqm = (0.337 - y) M ≈ 0.337 M

Ka = [NH₃] [H⁺] / [NH₄⁺]
(1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = y² / 0.337
y = 1.4 × 10⁻⁵

pH = -log[H⁺] = -log(1.4 × 10⁻⁵) = 4.9

pH calculation.....

The hydrolysis of ammonium nitrate. NH4NO3 is soluble in water and produces NH4^+(aq) and NO3^-(aq). Only the ammonium ion hydrolyzes. Nitrate ion is the anion of a strong acid and does not hydrolyze.

NH4^+ <==> H+ + NH3(aq) .... Ka = Kw/Kb = 1.00x10^-14 / 1.8x10^-5 = 5.56x10^-10
0.337M ........0 ......0 ................ Initial
-x .................+x ... +x .............. Change
0.337-x ........ x ..... x ............... Equilibrium


Ka =[H+][NH3] / [NH4^+]
5.56x10^-10 = x² / (0.337 - x) ......... 0.337-x = 0.337 since x is very small
5.56x10^-10 = x² / 0.337
x = 1.37x10^-5
[H+] = 1.37x10^-5M
pH = -log[H+] = -log(1.37x10^-5) = 4.86

The answer is expressed to two significant digits. When taking the log of a number, only the digits to the right of the decimal reflect the precision of the original number.