What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M NH₃ with 25.00 mL of 0.10 M NH₄Cl?

Assume that the volume of the solutions are additive and that Kb = 1.8 * 10⁻⁵ for NH₃. Enter your answer in exponential (E) format (sample 1.23 E-4) with two decimal places and without units.

Answer: 9.56

Explanation:

First we need to know the moles of each species involved:

NH3: 50.00 * 10^-3 * 0.10 M = 5.0 * 10^-3

NH4Cl: 25.00 * 10^-3 * 0.10 M = 2.5 * 10^-3

Then we calculate the molarities:

5.0 * 10^-3 moles/75 * 10^-3 L = 0.67

2.5 * 10^-3 moles/75 * 10^-3 L = 0.33

Ka and Kb are related by:

pKa = 14 - pKb

If Kb is 1.8 * 10^-5

then pKb is - log (1.8 * 10^-5) = 4.74

Therefore

pKa = 14 - 4.74 = 9.26

Using the Henderson-Hasselbalch (H-H) equation,

pH = pKa + log (0.67) / (0.33) = 9.26 + log 2 = 9.26 + 0.30 = 9.56