Question

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 23.0 mL of HNO3. Express your answer numerically.

Answer 1

Here
Initial moles of NH3 = 75/1000 x 0.200 = 0.015 mol

Moles of HNO3 added = 23/1000 x 0.500 = 0.0115 mol
The equation is as follow-

NH3 + HNO3 => NH4+ + NO3-
So

Moles of NH3 left = 0.015 - 0.0115 = 0.0035 mol

Moles of NH4+ = 0.0115 mol

Ka(NH4+) = Kw/Kb(NH3)

= 10-14/1.8 x 10-5 = 5.556 x 10-10

Using Henderson-Hasselbalch equation:

pH = pKa + log([NH3]/[NH4+])

= -log Ka + log(moles of NH3/moles of NH4+) because the volume is same for both

= -log(5.556 x 10-10) + log(0.0035/0.0115)

= 8.74