One litre buffer solution was prepared by adding 0.10 mol each of NH3 and NH4Cl in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is 10-2. (Nearest integer) Given : pKb of NH3 = 4.745 and 10 3 = 0.477
Answer
48
Solution
For a basic buffer: pOH = pKb + [NH4+][NH3]
Initially: pOH = 4.745 + 0.100.10 = 4.745
On adding 0.05 mol HCl: NH3 + H+ NH4+
NH3 = 0.10 - 0.05 = 0.05 mol, NH4+ = 0.10 + 0.05 = 0.15 mol
pOH = 4.745 + 0.150.05 = 4.745 + 3 = 4.745 + 0.477 = 5.222
pOH increases by 0.477, so pH decreases by 0.477
Change in pH = 0.477 = 47.7 10-2 48 10-2
Answer: 48