Question

A 50.0 mL sample of 0.150 M ammonia (NH3, Kb=1.8×10−5) is titrated with 0.150 M HNO3. Calculate the pH after the addition of each of the following volumes of acid.

A. 0.0 mL

B. 25.0 mL

C. 50.0 mL

Answer 1

millimoles of NH3 = 50 x 0.15 = 7.5

A) initially as NH3 is weak base

pOH = 1/2 [pKb - logC]

pKb = - log Kb = -log [1.8 x 10^-5] = 4.74

pOH = 1/2 [4.74 - log0.15]

pOH = 2.52

pH = 14 - 2.52

pH = 11.48

B) 25 mL HNO3 added

millimoles of HNO3 = 25 x 0.15 = 3.75

half way of equivalent point

pOH = pKb

pOH = 4.74

pH = 14 - 4.74

pH = 9.26

C) millimoles of HNO3 added = 50 x 0.15 = 7.5

at equivalent point

[salt] = 0.075 M

pOH = 1/2 [pKw + pKb + logC]

pOH = 1/2 [14 + 4.74 + log 0.075]

pOH = 8.81

pH = 14 - 8.81

pH = 5.19