Question

Determine the pH during the titration of 35.8 mL of 0.233 M methylamine (CH3NH2 , Kb = 4.2×10-4) by 0.233 M HBr at the following points.

(a) Before the addition of any HBr

(b) After the addition of 13.7 mL of HBr

(c) At the titration midpoint

(d) At the equivalence point

(e) After adding 54.1 mL of HBr

Answer 1

Determine the pH during the titration of 35.8 mL of 0.233 M methylamine (CH3NH2 , Kb = 4.2×10-4) by 0.233 M HBr at the following points.

(a) Before the addition of any HBr

(b) After the addition of 13.7 mL of HBr

(c) At the titration midpoint

(d) At the equivalence point

(e) After adding 54.1 mL of HBr

pH before addition of HBr

kb = [OH-][CH3NH+]/[0.170-x]

here Kb = 4.2*10^-4

4.2*10^-4= x^2/ 0.233-x

X^2= 9.786*10^-5

X=9.89*10^-3

[OH-] = [C5H5NH+] = x

x = [OH-] = 9.89*10^-3M

[H+] = 10^-14/9.89*10^-3

= 1.01*10^-12 M
pH = -log(1.01*10^-12)

= 11.995

(b) After the addition of 13.7 mL of HBr

pKb= -log Kb=-log 4.2*10^-4

=3.38

HBr + CH3NH2= CH3NH3+ + Br-

Moles of CH3NH2= 0.0358 *0.233

=0.00834 moles CH3NH2

Moles HBr = 0.0137*0.233

= 0.00319

Moles of CH3NH3+ = 0.00319

Total volume = 0.0358 +0.0137=0.0495 L

Molarity of CH3NH3+ = 0.00319 /0.0495

=0.0644     M

Moles of CH3NH2 = 0.00834 moles -0.00319=

Molarity of CH3NH2 = 0.00515 /0.0495

=0.104       M

pOH = pKb + log(0.0644/0.104)

= 3.38– 0.21

=3.17

pH = 14-pOH

= 14-3.17

=10.83

(c) At the titration midpoint; [CH3NH3+]=[CH3NH2]

pOH = pKb + log 1

pOH = pKb

pOH = 3.38

pH = 14-pOH

= 14-3.38

=10.62

(d) At the equivalence point

moles HBr required to reach the equivalence point

0.0358 *0.233=0.00834 moles

Volume HBr = 35.8 m L
total volume = 35.8 m L +35.8 m L = 71.6 ml = 0.0716 L
moles CH3NH3+ formed = 0.00834
[CH3NH3+] = 0.00834 / 0.0716L

=0.116 M

CH3NH3+ <=> CH3NH2 + H+
Ka = Kw/Kb = 1.0 x 10^-14 / 4.2 x 10^-4

=2.38 x 10^-11

2.38 x 10^-11 = x^2 / 0.116-x

Small value of Ka we can write 0.116-x = 0.116

X^2=2.38 x 10^-11 *0.116

X^2=2.76x 10^-12

X= 1.66*10^-6

x = [H+]= 1.66*10^-6M
pH = 5.77

(e) After adding 54.1 mL of HBr

Total volume = 54.1+35.8= 89.9 ml= 0.0899L

Moles of CH3NH2= 0.0358 *0.233

=0.00834 moles CH3NH2

Moles HBr = 0.0541*0.2333= 0.0126 moles

Remaing moles of HBr = 0.0126-0.00834= 0.00427 moles

Molarirty = 0.00427 /0.0899L

= 0.0474 M

pH = - log H+

= - log 0.0474

=1.32