Question

Calculate the pH at each point listed for the titration of 100.0 mL of 0.100 M cocaine (Section 8 4, Kb = 2.6 × 10-6) with 0.200 M HNO3. The points to calculate are Va = 0.0, 10.0, 20.0, 25.0, 30.0, 40.0, 49.0, 49.9, 50.0, 50.1, 51.0, and 60.0 mL. Draw a graph of pH versus Va.

Answer 1

Moles of cocaine (C) taken = 0.100 M* 100 ml = 10 millimol

Kb = 2.6 x 10-6

 

C + H2O → CH+ + OH-

 

Let 'x' M of cocaine gets hydrolyzed at equilibrium,

Kb = [CH+][OH-]/[C]

Kb = x*x/(0.100 - x)

 

As Kb is low, small amount of cocaine would be dissociated at equilibrium

Kb = x2/0.100

   x = √(Kb*0.100)

      = √(2.6 x 10-6 * 0.100)

     = 5.1 x 10-4 M

 

[OH-] = 0.00051 M

 pOH = 3.29

    pH = 14 - 3.29 = 10.71

 

Further when base is added, pH is calculated by Henderson Hasselbalch equation,

 

pH = pKa + log(base/conjugate acid)

Here pKa is of conjugate acid.

pKa = 14 - pKb

        = 14 - [-log(2.6 x 10-6)]

       = 14 - 5.58

       = 8.42

 

Volume of acid added (mL)	Millimoles of acid added = Molarity *volume	millimoles of conjugate acid = millimoles of acid added	millimoles of cocaine remaining = 10 - millimoles of conjugate acid	ratio: log(conjugate acid/cocaine remaining)	pH = 8.42 + ratio
10	2	2	8	0.6021	9.02
20	4	4	6	0.1761	8.60
25	5	5	5	0	8.42
30	6	6	4	-0.1761	8.24
40	8	8	2	-0.6021	7.82
49	9.8	9.8	0.2	-1.690	6.73
49.9	9.98	9.98	0.02	-2.70	5.72

 

At equivalence point i.e. 50 ml of acid added,

millimoles of conjugate acid formed = 10 millimol

Total volume = 150 ml

[Conjugate acid] = 10/150 = 0.067

Hydrolysis of conjugate acid will be

CH+ + H2O → C + H3O+

Ka = 10-8.42 = 3.8 x 10-9

[H3O+] = √(Ka*0.067)

              = (3.8 x 10-9 * 0.067)1/2

              = 1.60 x 10-5

       pH = 4.80

 

After equivalence point, pH depends directly on volume of acid added.

 

Volume of acid added (ml)	Volume actually added after equivalence point (ml)	millimoles of acid added = 0.200 M* volume actually added	total volume of solution = volume of base + acid	[H+ ]= millimoles/total volume	pH = log[H+]
50.1	0.1	0.02	150.1	0.000132	3.88
51.0	1.0	0.2	151.0	0.00132	2.88
60.0	10.0	2.0	160.0	0.0125	1.90

Plotted curve:

Volume of acid(mL)	pH
0	10.71
10	9.02
20	8.6
25	8.42
30	8.24
40	7.82
49	6.73
49.9	5.72
50	4.8
50.1	3.88
51	2.88
60	1.9

 

Moles of cocaine (C) taken = 0.100 M* 100 ml = 10 millimol

Kb = 2.6 x 10-6

 

C + H2O → CH+ + OH-