Question

What is the pH of 25 ml of 0.100 M ammonia sln. 1) before titration begins ; 2) after 12.50 ml of 0.100 M HCL is added; 3) after 14 ml of 0.100 M HCL is added; 4) after 25 ml of the HCL is added; and 5) after 35 ml of the HCl is added. pkb of ammonia = 4.74

Answer 1

1) NH3(aq) + H2O(l) <-------> NH4+ (aq) + OH-(aq)

Kb= [NH4+][OH-]/[NH3]= 1.8×10^-5

at equillbrium

x^2/(0.1 -x) = 1.8×10^-5

we can assume 0.1- x = 0.1

x^2 = 1.8×10^-6

x = 0.00134

[OH-] = 0.00134M

pOH = 2.87

pH = 14 - 2.87

= 11.13

2) NH3(aq) + HCl(aq) -----> NH4+ (aq) + Cl-(aq)

25ml of 0.1M NH3 require 25ml of 0.1M HCl

12.5ml is half equivalence point

at half equivalence point

pOH = pKb

pOH = 4.74

pH = 14 - 4.74

= 9.26

3) No of mole of HCl added = (0.100mol/1000ml)×14 = 0.0014

No of mole of NH3 = (0.100mol/1000ml)×25ml = 0.0025

No of mole after addition = 0.0025 - 0.0014 = 0.0011

No of mole of NH4+ formed = 0.0014

Total volume = 39ml

[NH3]= (0.0011mol/39ml)×1000ml = 0.0282M

[NH4+] = (0.0014/39ml)×1000ml = 0.0359M

Henderson - equation

pOH = pKb + log([NH4+]/[NH3])

= 4.74 + log (0.0359M/0.0282M)

= 4.74 + 0.10

= 4.84

pH = 14 - 4.84

= 9.16

4) 25ml is equivalence point

at equivalnce point

[ NH4+ ] = (0.0025mol/50ml)×1000ml = 0.05M

NH4+(aq) + H2O(l) <------> NH3(aq) + H3O+(aq)

Ka = [NH3][H3O+]/[NH4+] = 5.56×10^-10

x^2/0.05 = 5.56 ×10^-10

x^2 = 2.78×10^-11

x = 5.27×10^-6

[H3O+] = 5.27×10^-6 M

pH = 5.28

5) HCl -------> H+ + Cl-

After 35 ml of HCl addition

excess HCl addition = 10ml

No of mole of HCl = (0.100mol/1000ml)×10ml = 0.0010

total Volume = 60ml

[ HCl ] =( 0.0010mol/60ml)×1000ml = 0.0167M

[H+] = 0.0167M

pH = 1.78