Releasing a batch of normal saline by Mohr titration
A district hospital in Birgunj prepares its own intravenous normal saline (labelled 0.90% w/v NaCl) and the pharmacy asks you to release or reject batch 14. You standardise silver nitrate at exactly 0.1000 M and titrate 10.00 mL of the batch by Mohr's method, using potassium chromate indicator, with the sample adjusted to pH 7.0. Three runs require 15.35, 15.40 and 15.45 mL. You notice that the brick-red endpoint fades if you wait, and that a blank of chloride-free distilled water consumes 0.05 mL. The batch log shows that in an earlier batch the dilution water had been acidified with a little hydrochloric acid "to stop bacterial growth", and a junior technician suggests replacing potassium chromate with methyl orange so that "the colour change is sharper". The pharmacy's release limit is 0.85–0.95% w/v NaCl.
1. Calculate the NaCl content of the batch as % w/v and as mmol/L, and state whether it should be released.
Read the worked answer
Step 1 - mean titre and blank correction. Mean = (15.35 + 15.40 + 15.45)/3 = 15.40 mL. Corrected volume = 15.40 − 0.05 = 15.35 mL.
Step 2 - moles of chloride. 1 mol Ag+ reacts with 1 mol Cl−, so n(Cl−) = 0.1000 mol/L × 0.01535 L = 1.535 × 10⁻³ mol in 10.00 mL.
Step 3 - mass and percentage. M(NaCl) = 58.44 g/mol, so mass = 1.535 × 10⁻³ × 58.44 = 0.0897 g in 10.00 mL, i.e. 8.97 g/L = 0.897% w/v.
Step 4 - molarity and osmolarity. 8.97 g/L ÷ 58.44 = 0.1535 mol/L = 153.5 mmol/L of Na+ and of Cl−, total 307 mmol/L, close to plasma osmolality.
Decision. 0.897% lies inside 0.85–0.95% w/v, so the batch passes and may be released, provided sterility and particulate testing also pass. Note how close the answer is to the label claim; the blank correction is small here but becomes decisive near a release limit.
2. Explain why Mohr's method needs pH 6.5–8.3, and why methyl orange is not a valid substitute for potassium chromate.
Read the worked answer
How the indicator works. Chromate is present in small amount. Silver chloride is far less soluble (Ksp ≈ 1.8 × 10⁻¹⁰) than silver chromate (Ksp ≈ 1.1 × 10⁻¹²), so AgCl precipitates first and the red Ag₂CrO₄ only appears once chloride is essentially exhausted - that is the endpoint.
Why the pH window. In acid, chromate is protonated and dimerises: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O. The free chromate concentration falls, so the red endpoint appears late or never, giving a falsely high result. In alkali above about pH 8.3, silver hydroxide/oxide (Ag₂O) precipitates as a brown-black solid, giving a false endpoint in the other direction. That is why the acidified dilution water in the earlier batch was a real manufacturing fault, not a convenience.
Why methyl orange fails. Methyl orange is an acid–base indicator changing colour around pH 3.1–4.4. Its colour has nothing to do with silver concentration; it would simply be bleached or adsorbed, and no precipitation endpoint would be detected. The titration needs a precipitation indicator, not a pH indicator.
3. Why does the endpoint overshoot, and how does the blank titration correct for it?
Read the worked answer
The overshoot. A visible precipitate of Ag₂CrO₄ requires a finite excess of silver ions beyond the equivalence point - you must add roughly one extra drop before enough chromate converts for the red colour to be seen. The measured titre is therefore systematically too large, and the calculated chloride content too high.
The blank. Titrating the same volume of chloride-free water with the same indicator addition measures exactly the reagent volume needed to produce a visible endpoint with no chloride present. Subtracting it (here 0.05 mL) removes this systematic bias. Without it, the result would be 15.40 mL → 0.900% w/v instead of 0.897%; insignificant for this batch, but enough to fail a sample sitting at the upper limit.
Fading endpoint. Ag₂CrO₄ slowly redissolves as fresh chloride diffuses out of the precipitate, so the colour fades. The correct endpoint is the first persistent pink, read consistently across all three runs, and the flask should be swirled vigorously throughout. This is a determinate error you control by technique, unlike the indeterminate scatter seen in the 15.35–15.45 mL range.