Question

2. Calculate the pH’s of the following mixtures:

a. 0.150 moles of ammonia (Kb = 1.8 x 10-5 ) in 1575 ml of water.

b. 35.0 ml of 150 mM ammonia combined with 50.0 ml of 100 mM ammonium chloride.

c. 500.0 ml of 300 mM ammonia to which has been added 70.0 mmoles of HCl. Assume no volume change occurs.

d. A solution formed by combining 500.0 ml of 0.100 M ammonia with 0.015 moles of NaOH. Assume no volume change occurs.

Answer 1

a) [ NH3 ] =( 0.150mol/1575ml)×1000ml = 0.0952M

NH3 + H2O <--------> NH4+ + OH-

Kb = [ NH4+ ] [ OH- ]/[NH3]= 1.8 ×10^-5

1.8×10^-5 = X^2/(0.0952 - X)

X = 1.29×10^-3

[ OH- ] = 1.29×10^-3M

pOH = 2.89

pH = 14 - 2.89

= 11.11

b) mole of NH3 = (0.150M/1000ml)×35ml = 0.00525

mole of NH4+ =( 0.100M/1000ml)×50ml =0.00500

Total volume = 35 + 50 = 85ml

[ NH3 ] = (0.00525mol/85ml)×1000ml = 0.0618M

[ NH4+] = (0.00500mol/85ml)×1000ml = 0.0588M

Applying Henderson equation

pOH = 4.75+ log ( 0.0588/0.0618 )

= 4.75 - 0.022

= 4.73

pH = 14 - 4.73

= 9.27

c) No of mole of Ammonia=(0.3mole/1000ml)×500ml= 0.15M

No of mole of HCl = 0.070 mole

NH3 + HCl ------> NH4+ + Cl-

Remaining mole of NH3 = 0.15 - 0.070= 0.080mole

Volume = 500ml

[ NH3 ] = (0.080mol/500ml)×1000ml = 0.16M

NH3 + H2O <------->NH4+ + OH-

1.8 × 10^-5 = X^2/(0.16 - X )

   X = 1.68×10^-3

   [ OH- ] = 1.68×10^-3M

   pOH = 2.77

   pH = 14 - 2.77

   = 11.23

d) 1.8 × 10^-5 = X^2/0.1-X

   X = 1.34×10^-3

   [OH- ] = 1.34×10^-3M

No of mole of OH- = 0.00134 + 0.01500 = 0.01634 mole

Volume = 500ml

[ OH- ] = (0.01634mole/500ml)×1000ml = 0.03268M

pOH = 1.49

  pH = 14 - 1.49

   = 12.51