What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M NH3 with 10.00 mL of 0.10 M NH4Cl? Assume that the volume of the solutions are additive and that Kb = 1.8 × 10-5 for NH3.

Answer 1

NH₃ + NH₄Cl makes a buffer because you have a weak base (NH₃) and the conjugate acid (NH₄⁺)

The pH can be determined using the Henderson Hasselbalch equation:

pH = pKa + log [conj.base] / [acid] or

pOH = pKb + log [conj.acid] / [base] which is the one we will use.

moles NH₃ = 50.00 ml x 1 L / 1000 ml x 0.10 mol / L = 0.005 moles NH₃

moles NH₄⁺ = 10.00 ml x 1 L / 1000 ml x 0.10 mol / L = 0.001 moles NH₄⁺

Final volume = 50.00 ml + 10.00 ml = 60.00 mls = 0.060 L

[NH₃] = 0.005 moles / 0.06 L = 0.0833 M

[NH₄⁺] = 0.001 moles / 0.06 L = 0.0167 M

pKb = -log Kb = -log 1.8x10-5

pKb = 4.74

pOH = 4.74 + log (0.0167 / 0.0833)

pOH = 4.74 - 0.70

pOH = 4.04

pH = 14 - pOH = 14 - 4.04

pH = 9.96

If you prefer to use pH = pKa + log [conj.base] / [acid], we must first find pKa.

pKa + pKb = 14

pKb = 4.74 (see above calculation)

pKa = 14 - 4.74 = 9.26

pH = 9.26 + log (0.0833 / 0.0167)

pH = 9.26 0.70

pH = 9.96