Question

Calculate the pH of 100.0 mL of a buffer that is 0.0650 M NH4Cl and 0.185 M NH3 before and after the addition of 1.00 mL of 5.90 M HNO3.

Answer 1

i) Before addition of HNO3

Henderson- Hasselbalch equation is

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

[ Base ] = [ NH3 ] = 0.185M

[ conjucate acid ] = [ NH4+ ] = 0.0650M

pKb of NH3 = 4.75

Therefore,

pOH = 4.75 + log(0.0650M/0.185M)

= 4.75 - 0.45

= 4.30

pH = 14 -pOH

= 14 - 4.30

= 9.70

ii) After addition of HNO3

HNO3 react with the base NH3

NH3 + HNO3<-----> NH4+ + OH-

So, after addition of HNO3 , [ NH3 ] will decrease and [NH4+] will increase

No of mole of NH3 at initial =( 0.185mol/1000ml)×100ml = 0.0185mol

No of mole of NH4+ at initial = (0.0650mol/1000ml) ×100ml = 0.00650

No of mole of HNO3 added = (5.90mol/1000ml)× 1ml = 0.0059

0.0059 mole of HNO3 react with 0.0059mole of NH3 to form 0.0059mole of NH4+

So, After addition

mole of NH3 = 0.0185 - 0.0059 = 0.0126

mole of NH4+ = 0.00650+ 0.0059 = 0.0124

Total volume = 100ml + 1ml =101ml

[NH3] = (0.0126mol/101ml)×1000ml = 0.1248M

[ NH4+ ] = (0.0124mol/101ml)×1000ml =0.1228M

Now , applying the Henderson - Hasselbalch equation

pOH = 4.75 + log ( 0.1228M/0.1248M)

= 4.75 - 0.007

= 4.74

pH = 14 - 4.74

= 9.26