A litre of buffer solution contains 0.1 mole of each of NH₃ and NH₄Cl. On the addition of 0.02 mole of HCl by dissolving gaseous HCl, the pH of the solution is found to be _____________ × 10⁻³ (Nearest integer)
[Given : pK_b(NH₃)=4.745
log 2=0.301
log 3=0.477
T=298 K]

Solution

Chemistry ma’am · 4 steps

The HCl turns some NH₃ into NH₄⁺, so update both amounts first, then use Henderson.

1. HCl reacts first. The strong acid is used up by the base of the buffer: NH₃ + HCl → NH₄⁺ + Cl⁻. So 0.02 mol of NH₃ becomes 0.02 mol of NH₄⁺.
2. New amounts. NH₃: 0.1 - 0.02 = 0.08 mol. NH₄⁺: 0.1 + 0.02 = 0.12 mol. The volume is 1 L, so these are also the concentrations in mol L⁻¹.

   pOH = pK_b + log[NH₄⁺]/[NH₃] = 4.745 + log 0.12/0.08

3. Work out the log. 0.12/0.08 = 3/2, so log 1.5 = 0.477 - 0.301 = 0.176. That gives pOH = 4.745 + 0.176 = 4.921.
4. So. At 298 K, pH = 14 - 4.921 = 9.079 = 9079 × 10⁻³, so the answer is 9079.

The common slip: 9431. Turning the ratio upside down, log 0.08/0.12, makes pOH 4.569 and pH 9.431, giving 9431. For pOH the salt NH₄⁺ goes on top.

The idea to keep. When strong acid or base goes into a buffer, first let it react completely with the buffer component that opposes it. Then put the new mole ratio into Henderson.

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