Question

117.8 mL sample of 0.120 M methylamine (CH3NH2;Kb=3.7×10−4) is titrated with 0.250 M HNO3. Calculate the pH after the addition of each of the following volumes of acid

0.00 mL

28.3 mL

56.5 mL

84.8 mL

Answer 1

millimoles of CH3NH2 = 117.8 x 0.120 = 14.14

Kb = 3.7 x 10^-4

pKb = 3.43

a) 0.00 mL HNO3 :

CH3NH2   +   H2O ----------------> CH3NH3+   +   OH-

0.120 - x                                             x                   x

Kb = x^2 / 0.120 - x

3.7 x 10^-4 = x^2 / 0.120 - x

x = 6.48 x 10^-3

pOH = - log [OH-] = -log (6.48 x 10^-3 )

pOH = 2.19

pH = 11.81

b)

millimoles of HNO3 = 28.3 x 0.250 = 7.075

this is half - equivalence point. so

here pOH = pKb

pOH = 3.43

pH = 10.57

c) 56.5 mL HNO3 :

millimoles of HNO3 = 56.5 x 0.250 = 14.125.

this is equivalence point. here salt reamins.

salt concentraiton = 14.125/ 117.8 + 56.5

                             = 0.081 M

pH = 7 - 1/2 (pKb + log C)

     = 7 - 1/2 (3.43 + log 0.081)

pH = 5.83

d)

pH = 1.47