Ammonia Buffer: Working Out How Much Acid It Can Really Absorb
A technician prepares 1.00 L of buffer solution containing 0.100 mol of NH₃ and 0.100 mol of NH₄Cl dissolved and made up to the mark with distilled water. For NH₄⁺, Ka = 5.6 × 10⁻¹⁰ (pKa = 9.25 at 25 °C). As a demonstration, the technician adds 10.0 mL of 1.00 mol/L HCl to the buffer with stirring and records the pH. For comparison, an identical 10.0 mL portion of the same HCl is added to 1.00 L of distilled water. Later, a reaction step must be kept above pH 9.00, so the technician needs to know how much of this acid the buffer can absorb before the pH drops below that threshold. Assume volumes are additive and the temperature remains 25 °C.
1. Calculate the pH of the buffer before any acid is added, and explain why this pH does not change appreciably on dilution with a small amount of water.
Read the worked answer
Henderson–Hasselbalch
pH = pKa + log([base]/[acid]) = 9.25 + log(0.100/0.100) = 9.25 + 0 = 9.25
Why dilution matters little
Adding water changes the total volume, so both [NH₃] and [NH₄⁺] fall by the same factor. The ratio [NH₃]/[NH₄⁺] - and therefore the log term - is unchanged, so the pH stays at 9.25. Diluting a buffer does not shift its pH much; what dilution ultimately destroys is the capacity (the absolute moles of acid or base that can be absorbed), because each addition consumes a larger fraction of the remaining NH₃.
(Strictly, activity coefficients shift slightly at low ionic strength, which is why the measured pH may drift a few hundredths of a unit.)
2. Calculate the pH after the 10.0 mL of 1.00 mol/L HCl is added to the buffer, and compare it with the pH of the water sample that received the same amount of acid.
Read the worked answer
Moles of acid added
n(HCl) = (0.0100 L)(1.00 mol/L) = 0.0100 mol
Stoichiometry (HCl is limiting and is consumed completely)
NH₃ + H⁺ → NH₄⁺
- n(NH₃) = 0.100 − 0.010 = 0.090 mol
- n(NH₄⁺) = 0.100 + 0.010 = 0.110 mol
pH of the buffer
pH = 9.25 + log(0.090/0.110) = 9.25 + log(0.818) = 9.25 − 0.087 = 9.16
The pH has barely moved (a 0.09-unit drop) even though a strong acid was added.
pH of the water sample
Total volume ≈ 1.01 L, n(H⁺) = 0.0100 mol → [H⁺] = 0.0100/1.01 = 9.9 × 10⁻³ mol/L
pH = −log(9.9 × 10⁻³) = 2.00
Comparison: the same amount of acid drives pure water from pH 7.00 to 2.00 (five orders of magnitude), while the buffer moves 0.09 units. The buffer works because the added H⁺ is consumed by NH₃ and converted into NH₄⁺ rather than remaining as free H⁺.
3. Calculate the volume of 1.00 mol/L HCl that can be added to the 1.00 L buffer before the pH falls to 9.00, and comment on the assumptions in the calculation.
Read the worked answer
Set up the equilibrium condition
At pH 9.00:
9.00 = 9.25 + log(n(NH₃)/n(NH₄⁺))
log(n(NH₃)/n(NH₄⁺)) = −0.25 → n(NH₃)/n(NH₄⁺) = 10⁻⁰·²⁵ = 0.562
Let x = moles of HCl added
n(NH₃) = 0.100 − x; n(NH₄⁺) = 0.100 + x
(0.100 − x)/(0.100 + x) = 0.562
0.100 − x = 0.0562 + 0.562x
0.0438 = 1.562x → x = 0.0280 mol
Volume of 1.00 mol/L HCl = 0.0280 mol / 1.00 mol/L = 0.0280 L = 28 mL
Comment on assumptions
- Volumes are additive and the volume change (28 mL in 1.00 L) is ignored; because the ratio of moles is used, not concentrations, this has no effect on the answer.
- The temperature stays at 25 °C, so Ka and pKa are constant.
- The HCl is treated as completely consumed by NH₃, which is valid until NH₃ is nearly exhausted - 0.028 mol is well below the 0.100 mol available.
- This is a small fraction (28%) of total capacity. The buffer would still have NH₃ left beyond pH 9.00, but the pH then falls rapidly, so pH 9.00 is a sensible practical limit for a process requiring that threshold.