100 mL of a buffer that consists of 0.20 M NH3 and 0.20 M NH4Cl is titrated with 25 mL of 0.20 M HCl. Calculate the pH of the resulting solution given that the Kb for NH3 is 1.8 x 10-5.

pH = 9.03

Explanation:

The equilibrium of the NH₄Cl / NH₃ buffer in water is:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

Initial moles of both NH₃ and NH₄⁺ are:

0.100L ₓ (0.20 mol / L) = 0.0200 moles

The NH₃ reacts with HCl producing NH₄⁺, thus:

NH₃ + HCl → NH₄⁺ + Cl⁻

That means, moles of HCl added to the solution are the same moles are consumed of NH₃ and produced of NH₄⁺

Moles added of HCl were:

0.025L ₓ (0.20mol / L) = 0.0050 moles of HCl. Thus, final moles of NH₃ and NH₄⁺ are:

NH₃: 0.0200 moles - 0.0050 moles = 0.0150 moles

NH₄⁺: 0.0200 moles + 0.0050 moles = 0.0250 moles.

Using H-H equation for bases:

pOH = pKb + log [NH₄⁺] / [NH₃]

Where pKb is -log Kb = 4.745.

Replacing:

pOH = 4.745 + log 0.0250mol / 0.0150mol

pOH = 4.967

As pH = 14- pOH

pH = 9.03