Question

Consider titration of 100.0 mL of 0.100 M H2NNH2 (Kb = 3.5x10^-6) by 0.200 M HNO3? Calculate the pH of the resulting solution after the following volumes of HNO3 have been added.

0.0 mL

20.0 mL

25.0 mL

40.0 mL

50.0 mL

100.0 ml

Can anyone answer this thouroughly please? Please be sure to show every step of the math, that us were I tend to get lost. Thank you!

Answer 1

a) 0.0 mL

pOH = 1/2 [pKb - logC]

pOH = 1/2 [5.46 - log 0.1]

pOH = 3.23

pH + pOH = 14

pH = 10.77

b) 20 mL

millimoles of B = 100 x 0.1 = 10

millimoles of acid = 20 x 0.2 = 4

millimoles of B = 10-4 = 6

pOH = pKb + log [4 / 6]

pOH = 5.46 +  log [4 / 6]

pOH = 5.28

pH = 8.72

c) 25 mL

it is half equivalence point ; here pOH = pKb

pOH = 5.46

pH =8.54

d ) 40 mL

millimoles of acid = 40 x 0.2 = 8

B remains = 10-8 = 2

POH = 5.46 + log (8 / 2)

pOH = 6.06

pH = 7.94