Question

What is the pH of a 0.300 M NH₃ solution that has Kb = 1.8 × 10⁻⁵ ? The equation for the dissociation of NH₃ is: NH₃ (aq) + H₂O (l) ⇄ NH₄⁺ (aq) + OH⁻ (aq)

Answer 1

Answer:

11.4

Explanation:

Step 1: Given data

• Concentration of the base (Cb): 0.300 M

• Basic dissociation constant (Kb): 1.8 × 10⁻⁵

Step 2: Write the dissociation equation

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

Step 3: Calculate the concentration of OH⁻

We will use the following expression.

$[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8 \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M$

Step 4: Calculate the pOH

We will use the following expression.

$pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6$

Step 5: Calculate the pH

We will use the following expression.

$pH+pOH=14\\pH = 14-pOH = 14-2.6 = 11.4$