What is the pH of a 0.139 M NH3 solution at 25 degrees Celsius?

To calculate the pH of a 0.139 M ammonia (NH₃) solution, we start by acknowledging that ammonia is a weak base. The base dissociation constant (K_b) for ammonia can be calculated from the provided pK_b of 4.74 using the formula:

K_b = 10^-pK_b = 10^-4.74 ≈ 1.82 x 10^-5

Next, we can set up an equilibrium expression for the dissociation of ammonia in water:

NH₃(aq) + H₂O(l) <=> NH₄⁺(aq) + OH^–(aq)

At equilibrium, if we let the change in concentration of NH₃ that dissociates be represented by ‘x’, the concentrations would be as follows:

• [NH₃] = 0.139 – x
• [NH₄⁺] = x
• [OH^–] = x

Substituting these into the K_b expression:

K_b = rac{[NH₄⁺][OH^–]}{[NH₃]} = rac{x imes x}{0.139 – x} ext{ and since } x ext{ will be small, we can approximate } (0.139 – x) ext{ as } 0.139.

This simplifies to:

1.82 x 10^-5 = rac{x²}{0.139}

Now, we can solve for x:

x² = (1.82 x 10^-5) × 0.139

x² ≈ 2.53 x 10^-6

x ≈ √(2.53 x 10^-6) ≈ 0.00159 M.

Since x corresponds to the [OH^–] concentration, we can now find the pOH:

pOH = -log[OH^–] ≈ -log(0.00159) ≈ 2.80.

Finally, to find the pH, we use the relationship:

pH + pOH = 14, thus:

pH = 14 – pOH = 14 – 2.80 = 11.20.

In summary, the pH of a 0.139 M NH₃ solution at 25 degrees Celsius is approximately 11.20.