Ethanol Alligation: Meeting a 12-Litre Order Without an Indent
The pharmacy of a medical college must supply 12 litres of 60% v/v ethanol for preparing antiseptic hand rub. Stores hold 4 litres of 90% v/v ethanol, 7 litres of 40% v/v ethanol and an unlimited quantity of 20% v/v ethanol; purified water is also available. The storekeeper claims he can meet the order without raising an indent, simply by adding the 20% stock to what is already on the shelf. The laboratory's standard operating procedure treats all percentages as v/v and volumes as additive, and requires any purchase to be rounded up to the nearest 0.1 litre. Later in the week a ward returns 1.5 litres of a hand-rub batch labelled 60% v/v; an assay shows it is actually 48% v/v, and the batch record blames a water line left open during mixing. The batch cannot be discarded because the ward has no reserve.
1. Using alligation, find the 90%:40% ratio needed for a 60% v/v mixture and the volumes for the full 12-litre order. Then test the storekeeper's claim that the order can be met from 4 L of 90%, 7 L of 40% and unlimited 20%.
Read the worked answer
Alligation. The two strengths bracket 60%: 90% is 30 parts above target, 40% is 20 parts below. The ratio of 90% to 40% is therefore the cross-difference:
(60 − 40) : (90 − 60) = 20 : 30 = 2 : 3.
Five parts make 12 L, so each part is 2.4 L:
- 90% ethanol = 2 × 2.4 = 4.8 L
- 40% ethanol = 3 × 2.4 = 7.2 L
Check: (4.8 × 0.90) + (7.2 × 0.40) = 4.32 + 2.88 = 7.20 L of absolute ethanol = 60% of 12 L. Correct.
Testing the storekeeper's claim. Let x, y and z be litres of 90%, 40% and 20% used. Then x + y + z = 12 and 0.9x + 0.4y + 0.2z = 7.2. Substituting z = 12 − x − y and simplifying: 0.7x + 0.2y = 4.8, i.e. 7x + 2y = 48.
With x ≤ 4 and y ≤ 7, the largest achievable left-hand side is 7(4) + 2(7) = 42, which is below 48. Unlimited 20% cannot help, because 20% is weaker than the 60% target and dilutes the batch. The claim is false: the order cannot be met from stock.
2. Find the minimum additional 90% v/v ethanol that must be procured, and give the final mixing volumes after rounding to the nearest 0.1 L.
Read the worked answer
Use the constraint 7x + 2y = 48 with the 40% stock capped at y = 7 (using the maximum of the cheaper, weaker stock minimises the 90% needed):
7x + 14 = 48 → x = 34/7 = 4.857 L.
Rounded up to the nearest 0.1 L as the SOP requires, x = 4.9 L, so the extra quantity to procure is 4.9 − 4.0 = 0.9 L of 90% ethanol.
Final volumes with x = 4.9 L:
- From 7(4.9) + 2y = 48: 34.3 + 2y = 48 → y = 6.85 L of 40% ethanol.
- z = 12 − 4.9 − 6.85 = 0.25 L of 20% ethanol.
Verification of absolute ethanol: (4.9 × 0.90) + (6.85 × 0.40) + (0.25 × 0.20) = 4.41 + 2.74 + 0.05 = 7.20 L, which is exactly 60% of 12 L.
Note the judgement point: the indent is 0.9 L, not 0.857 L, because the SOP rounds purchases upward. Rounding downward to 4.8 L would leave the batch at 59.8% and out of specification.
3. How much 90% v/v ethanol must be added to the returned 1.5 litres of 48% v/v batch to bring the whole quantity to 60% v/v?
Read the worked answer
Let v be the litres of 90% ethanol added.
Ethanol already present: 1.5 × 0.48 = 0.72 L. Final volume: 1.5 + v litres, and this must contain 60% ethanol:
(0.72 + 0.90v) / (1.5 + v) = 0.60 0.72 + 0.90v = 0.90 + 0.60v 0.30v = 0.18 v = 0.6 L
Check: final volume = 2.1 L; ethanol = 0.72 + (0.6 × 0.90) = 0.72 + 0.54 = 1.26 L; 1.26 / 2.1 = 0.60, i.e. 60% v/v. Correct.
Judgement note. Adding 60% stock would not repair the batch - it would leave the strength unchanged at 48% - and adding water would make it weaker still. Only a strength above the target can correct a dilute batch, which is why 90% is the right stock here. The corrected 2.1 L must be re-labelled with the new volume and an assay entry, and the deviation recorded, because the returned batch has already been outside the labelled specification.