Question

HNO₃ (aq) + CH₃NH₂ (aq) → CH₃NH₃⁺ (aq) + NO₃⁻(aq)                          K_b of CH₃NH₂ = 4.4 x 10^-4

CH₃NH₃⁺ (aq) + H₂O (l) ↔ H₃O⁺ (aq) + CH₃NH₂ (aq)                                K_a of CH₃NH₃⁺ = 2.3 x 10^-11

Exactly 100 mL of 0.10 M methylamine (CH₃NH₂) solution are titrated with a 0.10 M nitric acid (HNO₃) solution. Calculate the pH for:

A:The initial solution

B:The point at which 30 mL of the acid has been added

C:At the half-equivalence point

D:The equivalence point

E.The point at which 110 mL of the acid has been added

Answer 1

Kb = 4.4 x 10^-4

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

0.1 -x                                            x                 x

Kb = x^2 / 0.1 -x

4.4 x 10^-4 = x^2 / 0.1-x

x = 6.42 x 10^-3

[OH-] = 6.42 x 10^-3 M

pOH = -log [OH-] = 2.19

pH + pOH = 14

pH = 11.81

part B)

millimoles of base = 100 x 0.1 = 10

millimoles of acid = 30 x 3 =7

CH3NH2 + H+ ----------------> CH3NH3+

10                3

7            0                               3

pOH = pKb + log(salt /base)

pOH = 3.36 + log (3/7)

pOH = 2.99

pH = 11.01

par D ) at half equivalenc epoint pOH = pKb

pOH = 3.36

pH + POH = 14

pH = 10.64

d) at equivalence point :

[salt ] =C= 100 x 0.1 / 200 = 0.05 M

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

pH = 5.97

e)

miimoles of acid = 110 x 0.1 = 11

millimoles of base = 100 x 0.1 = 10

acid > base

acid millimoles remains = 11 - 10 = 1

acid concentration = millmoles / total volume

                              = 1 / (100 + 110)

                               = 4.76 x 10^-3

[H+] = 4.76 x 10^-3

pH = -log [H+]

pH = 2.32