Question

A 4.00*10^2 mL solution is 0.25 M in ethylamine (CH3CH2)NH2 and 0.40 M in ethyl ammonium chlorate (CH3CH)NH3ClO3

a) what is the pH of solution

b) what is the pH of the solution after adding 125mL of 0.64 M of HBr?

c) what is the pH of the solution after adding 0.050 mol of KOH to the original 400mL solution?

Answer 1

a) Henderson equation is

pOH = pKb + log([BH+]/[B])

= pKb + log ( [ CH3CH2NH3+]/[CH3CH2NH2] )

= 3.2 + log ( 0.40/0.25 )

= 3.2 + 0.20

= 3.4

pH = 14 - pOH

= 14 - 3.4

= 10.6

b) HBr react with CH3CH2NH2

CH3CH2NH2 + HBr -------> CH3CH2NH3+

This reaction is 1:1 reaction

no of mole of HBr added =( 0.64mol/1000ml)*125ml = 0.08mol

initial mole of CH3CH2NH2 = (0.25mol/1000ml)*400ml = 0.1mol

remaining mole of CH3CH2NH2 = 0.1-0.08 = 0.02

initial mole of CH3CH2NH3+ = (0.40/1000ml)*400ml =0.16

mole of CH3CH2NH3+ after addition = 0.16 + 0.08 = 0.24

final voleume = 525ml

[CH3CH2NH2] =( 0.02mol/525ml)*1000ml =0.0381M

[CH3CH2NH3+] = (0.24mol/525ml)*1000ml = 0.4571M

pOH = pKa + log( 0.4571/0.0381)

= 3.2 + 1.08

= 4.28

pH = 14- 4.28

= 9.72

c) KOH react with conjucate acid CH3CH2NH3+

CH3CH2NH3+ + OH- -------> CH3CH2NH2 + H2O

Mole of NaoH added = 0.05

after addition

mole of CH3CH2NH3+= 0.16 -0.05 = 0.11

mole of CH3CH2NH2 = 0.10 + 0.05 = 0.15

[ CH3CH2NH2] = (0.15mol/400ml) *1000 ml = 0.375M

[ CH3CH2NH3+] = (0.11mol/400ml)*1000ml = 0.275M

pOH = 3.2 + log(0.275/0.375)

= 3.2 - 0.13

= 3.07

pH = 14 - 3.07

= 10.93