Question

A titration was performed between 25 mL of 0.100 M NH3 and 0.100 M HCl. Calculate the pH of NH3 solution at the following points during the titration.

Prior to addition of any HCl
After addition of 12 mL of 0.100M HCl
At the equivalence point
After the addiation of 31 mL of 0.100M HCl

Answer 1

a) Prior to addition of any HCl

NH3 + H2O --------------------------> NH4+   + OH-

0.1                                                  0              0

0.1 -x                                            x              x

Kb = [NH4+][OH-]/[NH3]

1.8 x 10^-5 = x^2 / 0.1-x

x^2 + 1.8 x 10^-5 x -1.8 x 10^-6 = 0

x = 1.33 x 10^-3

[OH-] = x = 1.33 x 10^-3 M

pOH = -log[OH-] = -log(1.33 x 10^-3)=2.88

pH+pOH =14

pH = 11.12

b) 12 ml HCl added

NH3 millimoles= 0.1 x 25 = 2.5

HCl millimoles = 0.1 x 12 =1.2

NH3 + HCl -------------------------> NH4Cl

2.5      1.2                                          0 ---------------------------initial

1.3         0                                           1.2 ------------------------equilibrium

in this mixture contain only salt NH4Cl + NH3 base . so it is buffer

Henderson-Hasselbalch equation

For Basic buffer  

pH = 14 –{ pKb + log[salt/ base]}

       = 14 - { 4.74 + log (1.2 /1.3) }

       = 9.29

pH = 9.29

c) at equivalence point

HCl millimoles = 25 x 0.1= 2.5

NH3 + HCl -------------------------> NH4Cl

2.5        2.5                                          0 ---------------------------initial

0         0                                             2.5 ------------------------equilibrium

salt only remains .

NH4Cl concentation = 2.5 / total volume = 2.5 / (25+25) = 0.05M

NH4Cl is a salt of weak base and strong acid . from salt hydrolysis

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

pH = 7 - 1/2 [4.74 + log0.05]

pH = 5.28

d ) 31 ml HCl added

HCl millimoles = 31 x 0.1 = 3.1

NH3 + HCl -------------------------> NH4Cl

2.5        3.1                                          0 ---------------------------initial

0            0.6                                        2.5------------------------equilibrium

HCl remains .so HCl concentarion = 0.6 / (25 + 31)

                                                        = 0.0107

[H+] = 0.0107 M

pH = -log[0.0107]

pH = 1.97