Question

What is the ph of a 0.20 M solution of NH4Cl? [Kb(NH3) = 1.8 x 10^-5]

Answer 1

"NH4+ <----> NH3 + H+  

The constant of this equilibrium is: K = Kw / Kb = 1 x 10^-14 / 1.8 x 10^-5 =5.56 x 10^-10  

5.56 x 10^-10 = x^2 / 0.20-x  

x = [H+] =1.1 x 10^-5 M  

pH = 5.0"

Answer 2

Answer:

5.0

Explanation:

NH₄Cl is a strong electrolyte that ionizes according to the following equation.

NH₄Cl(aq) → NH₄⁺(aq) + Cl⁻(aq)

NH₄⁺ is a weak acid according to the following equation.

NH₄⁺(aq) ⇄ NH₃(aq) + H⁺(aq)

Given Kb for NH₃, we can calculate the Ka for NH₄⁺.

Ka. Kb = Kw

Ka = Kw / Kb = 10⁻¹⁴/1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰

For a weak acid, we can find [H⁺] using the following expression.

[H⁺] = √(Ca × Ka) = √(0.20 × 5.6 × 10⁻¹⁰) = 1.1 × 10⁻⁵ M

where,

Ca: concentration of the acid

The pH is

pH = -log [H⁺] = -log 1.1 × 10⁻⁵ = 5.0