Answer to Question #225164 in Chemistry for Junior

Question

Calculate the pH of 0.12 M NH3 solution. Kb NH3 = 1.8 x10-5 

Answer 1

Solution:

Ammonia (NH₃) is a weak base.

The equilibrium equation for its reaction with water:

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^–(aq)

From this, we can construct an ICE (Initial, Change, Equilibrium) Table:

The K_b expression for NH₃ is:

K_b = [ x ] × [ x ] / [ 0.12 - x ] = 1.8×10^-5

C(NH₃) / K_b = (0.12) / (1.8×10^-5) = 6666.67 >> 500

Since the ratio is greater than 500, we can remove (–x) from the equation.

The equation simplifies to:

K_b = [ x ] × [ x ] / [ 0.12 ]

K_b = x² / 0.12

Solving for x:

x² = (1.8×10^-5) × 0.12 = 2.16×10^-6

x = (2.16×10^-6)^0.5 = 0.00147

This means [OH^–] = [NH₄⁺] = x = 0.00147 M

From [OH^–], we can calculate pOH:

pOH = -log[OH^-] = -log(0.00147) = 2.83

 

From pOH, we can calculate pH:

pH + pOH = 14;

pH = 14 - pOH = 14 - 2.83 = 11.17

pH = 11.17

 

Answer: pH = 11.17