calculate yhe pH of 0.1 m NH3 solution . calculate the pH after 25 mL of 0.1 m HCl is added to 50 mL of the solution .Kb for NH3 =1.77*10^5
Too find the pH of NH3 solution we need either the conc of H+ ion or OH- ions.NH3 on reaction with water generate OH- ions .We will calculate at first pOH by finding the concentration of OH- ion in solution.As shown below:
NH3 + H2O --> NH4+ + OH-
Make an ICE table to find out the concentration of OH- in solution
Kb= [NH4+][OH-]/ [NH3]= x2/ 0.1-x
Kb= 1.8 x 10-5 = x2/ 0.1-x
note : for easy calculation Kb is taken as 1.8 x 10-5
x can be omitted from denominator we get for easy
1.8 x 10-5 = x2/0.1
1.8 x 10-6= x2
x =0.0013
x= [OH-] = 0.0013
pOH= -log [OH-]= -log(.0013) = 2.87
pH = 14- pOH = 14- 2.87
pH= 11.13
2)
Moles NH3 = 0.1 x 0.05L =0.005
Moles HCl = 0.1 x 0.025L = 0.0025
total volume = 0.075 L
Ammonia will react with HCl according to following reaction
NH3 + H+ NH4+
moles NH3 after reacting HCl = 0.005 - 0.0025 = 0.0025
Concentation NH3 = 0.0025 / 0.075 =0.03 M
Moles NH4+ = 0.0025
Concentration NH4+ = 0.0025 / 0.075 =0.03 M
NH3 + H2O <---> NH4+ + OH-
pOH = pKb + log [NH4+]/[NH3] = 4.74 + log [0.03]/[0.03]
note pKb = -log Kb
pOH= pKb= 4.74
pH = 14-4.74 = 9.26
NH3 + H2O --> NH4+ + OH-
Make an ICE table to find out the concentration of OH- in solution
| NH3 | NH4+ | OH- | |
| INITIAL | 0.1 mole | 0 | 0 |
| CHANGE | -x | +x | +x |
| EQUILIBRIUM | 0.1-x | x | x |
Kb= [NH4+][OH-]/ [NH3]= x2/ 0.1-x
Kb= 1.8 x 10-5 = x2/ 0.1-x
note : for easy calculation Kb is taken as 1.8 x 10-5
x can be omitted from denominator we get for easy
1.8 x 10-5 = x2/0.1
1.8 x 10-6= x2
x =0.0013
x= [OH-] = 0.0013
pOH= -log [OH-]= -log(.0013) = 2.87
pH = 14- pOH = 14- 2.87
pH= 11.13
2)
Moles NH3 = 0.1 x 0.05L =0.005
Moles HCl = 0.1 x 0.025L = 0.0025
total volume = 0.075 L
Ammonia will react with HCl according to following reaction
NH3 + H+ NH4+
moles NH3 after reacting HCl = 0.005 - 0.0025 = 0.0025
Concentation NH3 = 0.0025 / 0.075 =0.03 M
Moles NH4+ = 0.0025
Concentration NH4+ = 0.0025 / 0.075 =0.03 M
NH3 + H2O <---> NH4+ + OH-
pOH = pKb + log [NH4+]/[NH3] = 4.74 + log [0.03]/[0.03]
note pKb = -log Kb
pOH= pKb= 4.74
pH = 14-4.74 = 9.26