Ammonia Buffer: Capacity Versus Chloride Limit
A student must prepare 500.0 mL of a pH 9.00 buffer for a colorimetric assay. The stockroom has solid NH4Cl (M = 53.49 g/mol), concentrated aqueous ammonia (25.0% NH3 by mass, density 0.910 g/mL), 1.00 M HCl and 1.00 M NaOH. For NH4+, Ka = 5.6 × 10^-10 (pKa = 9.25) at 25 °C. The protocol requires the assay pH to stay between 8.90 and 9.10 even after 0.010 mol of strong acid is released by the reaction, and it also requires the chloride concentration to stay below 0.20 M. A technician suggests dissolving 0.100 mol NH4Cl and 0.100 mol NH3 in 500 mL, arguing that 'equal amounts give pH = pKa, which is close enough'. Another student proposes starting from NH4Cl and adding NaOH instead of using the ammonia solution.
1. Calculate the mass of NH4Cl and the volume of concentrated ammonia needed for 500.0 mL of pH 9.00 buffer with total ammonia species 0.200 M. Check the chloride limit.
Read the worked answer
Henderson–Hasselbalch: pH = pKa + log([NH3]/[NH4+]). 9.00 = 9.25 + log ratio ⇒ log ratio = −0.25 ⇒ ratio = 10^−0.25 = 0.562.
With total ammonia species 0.200 M (0.100 mol in 0.500 L): [NH3] = 0.200 × 0.562/(1 + 0.562) = 0.0720 M, [NH4+] = 0.128 M.
Moles: NH3 = 0.0360 mol, NH4+ = 0.0640 mol. NH4Cl mass = 0.0640 × 53.49 = 3.42 g.
Ammonia stock: 1 L has mass 910 g and NH3 mass 0.250 × 910 = 227.5 g, i.e. 227.5/17.03 = 13.35 mol, so 13.35 M. Volume = 0.0360/13.35 = 2.70 mL.
Chloride = 0.0640 mol/0.500 L = 0.128 M < 0.20 M, so this recipe satisfies the chloride limit.
2. Show whether the technician's equal-amount buffer meets the 8.90–9.10 requirement before and after 0.010 mol of strong acid is added.
Read the worked answer
Equal amounts: [NH3] = [NH4+] = 0.200 M, so pH = pKa = 9.25, already outside 8.90–9.10.
After 0.010 mol HCl: NH3 = 0.100 − 0.010 = 0.090 mol, NH4+ = 0.110 mol, ratio = 0.818, pH = 9.25 + log 0.818 = 9.16, still above 9.10.
The technician's buffer fails both the initial and the after-acid requirement, so 'close enough' is not close enough for this assay.
3. Find the minimum total ammonia concentration that keeps pH ≥ 8.90 after the acid load, and check it against the chloride limit. What do you recommend?
Read the worked answer
Let n be total moles of ammonia species. At pH 9.00, NH3 = 0.3597n and NH4+ = 0.6403n. After adding 0.010 mol H+: ratio = (0.3597n − 0.010)/(0.6403n + 0.010). The lower limit pH 8.90 needs ratio ≥ 10^(8.90 − 9.25) = 0.4467.
Solve: 0.3597n − 0.010 ≥ 0.4467(0.6403n + 0.010) ⇒ 0.3597n − 0.010 ≥ 0.2860n + 0.00447 ⇒ 0.0737n ≥ 0.01447 ⇒ n ≥ 0.196 mol, i.e. total concentration ≥ 0.393 M.
Then NH4+ = 0.6403 × 0.196 = 0.126 mol, so chloride = 0.126/0.500 = 0.251 M, which exceeds the 0.20 M limit. The two constraints conflict for an NH4Cl/NH3 buffer. Recommendation: supply the ammonium as (NH4)2SO4 (no chloride), or reduce the acid load/dilute and use a buffer whose pKa is closer to 9.0, then recheck both limits.