Question

Calculate the pH of 100.0 mL of a buffer that is 0.0500 M NH4Cl and 0.125 M NH3 before and after the addition of 1.00 mL of 5.40 M HNO3.

pH before =

pH after =

Answer 1

i) Befor addition

Henderson - Hasselbalch equation is

pOH = pKb + log([Conjucate acid ]/ [ Base] )

= 4.75 + log ( [ NH4+ ]/[NH3])

= 4.75 + log(0.0500M/0.125M)

= 4.75 - 0.40

= 4.35

pH + pOH = 14

pH = 14 - pOH

= 14 - 4.35

= 9.65

b) After addition

Initial mole of NH3 =(0.125mol/1000ml)×100ml = 0.0125mol

Initial mole of NH4+= (0.050mol/1000ml)×100ml =0.0050mol

Mole of HNO3 added = (5.40mol/1000ml)×1ml = 0.0054mol

HNO3 react with base NH3 to form NH4+

NH3 + HNO3 ------> NH4+ + NO3-

0.0054mol of HNO3 react with 0.0054mole of NH3 to form 0.0054 mole of NH4+

After addition

No of mole of NH3 = 0.0125 - 0.0054 = 0.0071

No of mole of NH4+ = 0.0050 + 0.0054 = 0.0104

Total volume = 101ml

[ NH3] = (0.0071mol/101ml)×1000ml = 0.0703M

[ NH4+ ] = ( 0.0104mol/101ml)×1000ml = 0.1030M

Applying Henderson equation

pOH = 4.75 + log(0.1030M/0.0703M)

= 4.75 + 0.17

= 4.92

pH = 14 - 4.92

= 9.08