One litre buffer solution was prepared by adding 0.10 mol each of NH₃ and NH₄ Cl 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 : pK_b of NH₃=4.745 and log ₁₀ 3=0.477

Solution

Chemistry ma’am · 4 steps

The acid turns some NH3 into NH4+, so update both amounts and the pH change comes straight from log 3.

1. Before the acid. For a basic buffer, pOH = pK_b + log[NH₄⁺]/[NH₃]. With 0.10 mol of each the ratio is 1, so pOH = 4.745 and pH = 14 - 4.745 = 9.255.
2. Acid reacts. The added H⁺ reacts with the base: NH₃ + H⁺ → NH₄⁺. 0.05 mol of NH₃ is used up and 0.05 mol of NH₄⁺ forms. That leaves 0.10 - 0.05 = 0.05 mol NH₃ and 0.10 + 0.05 = 0.15 mol NH₄⁺.

   pOH = 4.745 + log 0.15/0.05 = 4.745 + log 3 = 4.745 + 0.477 = 5.222

3. New pH. pH = 14 - 5.222 = 8.778. The pH falls from 9.255 to 8.778, a change of 0.477.
4. So. The change is 0.477 = 47.7 × 10⁻², which rounds to 48.

The common slip: 18. Adding the 0.05 mol to NH₄⁺ but forgetting to remove it from NH₃ gives a ratio of 1.5 and a change of log 1.5 = 0.176, so 18 instead of 48.

The idea to keep. When acid or base goes into a buffer, first do the reaction in moles: one component goes down and the other goes up by the same amount. Only then put the new ratio into Henderson's equation.

Answer from the NTA final answer key. Solution by PrepWiser.