Question

A 100.0 mL sample of 0.100 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.

Part A: 0.0 mL

Part B: 20.0 mL

Part C: 40.0 mL

Part D: 60.0 mL

Answer 1

Part A

The dissociation of CH3NH2 is

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

Kb = [CH3NH3+] [OH-] /[CH3NH2] = 3.7×10^-4

at equillibrium,

[CH3NH3+] = x

[OH-] = x

[CH3NH2] = 0.100 - x

Therefore,

x²/1-x = 3.7×10^-4

solving for x

x = 0.0059

So,

[OH-] = 0.0059M

pOH = - log[OH-] = - log(0.0059)= 2.23

pH = 14 - pOH

= 14 - 2.23

= 11.77

Part B

the reaction between HNO3 and CH3NH2 is

HNO3 + CH3NH2 - - - - - > CH3NH3+ + NO3-

this reaction is 1:1 molar reaction

therefore,

V1×M1 = V2 × M2

V2 = V1×M1/M2

= 100ml × 0.1M/0.250M

= 40ml

so,

equivalent point = 40ml

20ml is half equivalence point

at half equivalance point

pOH = pKb

Kb = 3.7×10^-4

pKb = - logKb = - log(3.7×10^-4) =3.43

pOH = 3.43

pH = 14 - 3.43

= 10.57

Part C

40ml is equivalence point

at equivalence point all the CH3NH2 is converted into CH3NH3+

Total volume = 100ml + 40ml = 140ml

diltion factor for [CH3NH3+] = 140/100= 1.4

[CH3NH3+] = 0.100M/1.4 = 0.0714

CH3NH3+ is partly hydrolysed by water

CH3NH3+ + H2O < - - - - - > CH3NH2 + H3O+

Ka = [CH3NH2] [H3O+] /[CH3NH3+]

Ka = Kw/Kb = 1.00×10^-14/3.7×10^-4 = 2.70×10^-11

at equillibrium

[CH3NH2] = x

[H3O+] = x

[CH3NH2] = 0.0714 - x

x²/0.0714-x =2.70×10^-11

we can assume 0.0714 - x = 0.0714

x²/0.0714 = 2.70×10^-11

x² = 1.93×10^-12

     x = 1.39×10^-6

Therefore,

[H3O+] = 1.39×10^-6

pH = - log[H3O+]

= - log(1.39×10^-6)

= 5.86

Part D

Excess HNO3 added after equivalence point = 20ml

No of moles of HNO3 excessly added = (0.250mol/1000ml)×20ml = 0.005

Total Volume = 160ml

[HNO3] =( 0.005mol/160ml) ×1000ml = 0.03125M

HNO3 is strong acid

therefore,

[H3O+] = [H+]

[H+] = 0.03125M

pH = -log[H+]

= - log(0.03125)

= 1.51