Question

Calculate the pH for the titration of 40.00ml of 0.1000M solution of ethylamine (C2H5NH2) with 0.1000M HCl for the following volumes of added HCl: Kb for ethylamine = 6.4X10-4

Volumes; 0.00ml; 20.00ml; 40.00ml and 50.00ml

Answer 1

40.00ml of 0.1000M solution of ethylamine (C2H5NH2) is titrated with 0.1000M HCl.

(i). Before adding HCl.

CH3CH2NH2 + H2O CH3CH2NH3⁺   + OH-

Initially 0.100 0 0
At equilibrium 0.10 - x x x

Kb = x*x / (0.10 - x)

6.4*10-4 = x*x / 0.10 (Since x is very small)

x*x = 6.4*10^-5

x = 8*10^-3

[OH-] = 8*10^-3 M

pOH = - log [OH-]

pOH = - log (8*10^-3)

pOH = 2.09

pH = 14 - pOH

= 14 - 2.09

pH = 11.91

(ii) When 20.00 mL HCl is added -

Initial moles of CHeCH2NH2 = 0.100 * 0.04 = 0.004

Moles of HCl added = 0.100 * 0.02 = 0.002

Moles of CH3CH2NH3+ formed = 0.002

Moles of CH3CH2NH2 left = 0.002

Total volume = 60 mL = 0.06 L

[CH3CH2NH3+] = [CH3CH2NH2] = 0.002 / 0.06 = 0.033 M

CH3CH2NH2 + H2O CH3CH2NH3⁺   + OH-

Initially 0.033 0.033 0
At equilibrium 0.033 - x 0.033 + x x

Kb = x*(0.033 + x) / (0.033 - x)

6.4*10-4 = 0.033x + x² / 0.033

2.11*10^-5 = 0.033x + x²

x² + 0.033x - 2.11*10^-5 = 0

x = 6.27*10^-4 M

[OH-] = 6.27*10^-4 M

pOH = - log (6.27*10^-4)

pOH = 3.20

pH = 14 - 3.20

pH = 10.8

(3). When 40.00 mL HCl is added -

Moles of HCl added = 0.100 * 0.04 = 0.004

This is the equivalence point. At this point all the CH3CH2NH2 is converted to CH3CH2NH3+.

[CH3CH2NH3+] = 0.004 / 0.08 = 0.05 M

Ka = 10^-14 / 6.4*10-4

= 1.56*10^-11

CH3CH2NH3⁺ + H2O      CH3CH2NH2 + H3O+
Initially 0.05 0 0
At equilibrium 0.05 - x x x

Ka = x*x / 0.05 - x

1.56*10^-11 = x*x / 0.05

x*x = 7.8*10^-13

x = 8.83*10^-7

[H3O+] = 8.83*10^-7 M

pH = - log [H3O+]

= - log (8.83*10^-7)

pH = 6.05

(4). When 50.00 mL HCl is added -

Moles of HCl added = 0.100 * 0.05 = 0.005

Moles of H+ ions left unreacted = 0.005 - 0.004

= 0.001

Total volume = 90.0 mL = 0.09 L

[H+] = 0.001 / 0.09

= 0.011

pH = - log 0.011

pH = 1.95