Question

Determine the pH during the titration of 34.1 mL of 0.391 M ammonia (NH3, Kb = 1.8×10-5) by 0.391 M HNO3 at the following points. (Assume the titration is done at 25 °C.) Note that state symbols are not shown for species in this problem.

(a) Before the addition of any HNO3 ?

(b) After the addition of 15.0 mL of HNO3 ?

(c) At the titration midpoint ?

(d) At the equivalence point ?

(e) After adding 54.2 mL of HNO3 ?

Answer 1

(a) Before the addition of any HNO3

pKb =-log Kb = -log (1.8 x 10^-4) = 4.74

pOH = 1/2 (pKb -log C)

pOH = 1/2 (4.74 - log 0.391)

pOH = 2.57

pH + pOH =14

pH = 11.43

(b) After the addition of 15.0 mL of HNO3

millimoles of NH3 = 34.1 x 0.391 = 13.33

millimoles of HNO3 = 15 x 0.391 = 5.86

NH3 + HNO3 --------------------->NH4+

13.33 5.86 0

7.46 0 5.86

pH = 14 - {pKb + log [NH4+ / NH3]}

pH = 14 - {4.74 + log (5.86 / 7.46)}

pH = 9.36

(c) At the titration midpoint ?

here pOH = pKb = 4.74

pH = 9.26

(d) At the equivalence point ?

volume of HNO3 needed here = 34.1 mL

salt only remains

salt concentration = 13.33 / 2 x 34.1 = 0.195 M = C

pH = 7 - 1/2 [pKa + logC]

pH = 4.98

(e) After adding 54.2 mL of HNO3

pH = 1.05