Fertiliser Batch Release Decision
A fertiliser plant must decide whether to release a 12,000 kg batch labelled 15.0% nitrogen by mass. A 17.65 g sample is distilled with excess sodium hydroxide; the ammonia produced is absorbed in 200.0 mL of 1.00 M hydrochloric acid. The unreacted acid requires 22.4 mL of 0.500 M sodium hydroxide for neutralisation. The specification is 14.5-15.5% N. A second sample taken from the top of the tank gives 13.9% N, while a sample from the bottom gives 15.1%. The tank has no agitator, and the customer's lorry is waiting. The plant chemist suspects stratification and wants to release the batch after taking one more sample. The quality manager argues that the first result is within specification and that further sampling will delay delivery. The plant manager asks for a recommendation.
1. Calculate the nitrogen percentage from the first sample and state whether it meets the specification. Show the stoichiometry.
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Reactions: NH4+ + OH- → NH3 + H2O; NH3 + HCl → NH4Cl; HCl + NaOH → NaCl + H2O. Total HCl = 0.2000 L × 1.00 mol/L = 0.2000 mol. Excess HCl = 0.0224 L × 0.500 mol/L = 0.01120 mol. Ammonia nitrogen = 0.2000 - 0.01120 = 0.1888 mol. Mass of N = 0.1888 × 14.01 = 2.645 g. Percentage N = 2.645/17.65 × 100 = 14.99%. The specification is 14.5-15.5%, so this sample is within specification, very close to the 15.0% label.
2. Evaluate the quality manager's proposal to release on the first result. What do the top and bottom samples indicate, and what should be done before release?
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Do not release on one result. The top sample at 13.9% is below specification, while the bottom sample at 15.1% is inside it. That pattern is a stratification signal, not ordinary random error: the first sample may be near the average but does not prove homogeneity. Without an agitator, the customer could draw low-nitrogen material from the top. Before release, circulate or agitate if possible; take a stratified sampling plan covering top, middle and bottom at several points; test composites and individual samples; require all results within 14.5-15.5%, or homogenise/blend until they are. If the tank cannot be homogenised, release only the conforming portion and rework the rest. The waiting lorry is a commercial cost, not a quality excuse. The quality manager's approach confuses a passing sample with a passing batch.
3. If the homogenised batch is found to be 14.8% N, calculate how much 21.0% N material must be added to bring the whole batch to 15.0% N, assuming no mass loss.
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Initial nitrogen = 12,000 × 0.148 = 1,776 kg. Let x kg of 21.0% N material be added. Final nitrogen = 1,776 + 0.21x; final mass = 12,000 + x. Require (1,776 + 0.21x)/(12,000 + x) = 0.15. Then 1,776 + 0.21x = 1,800 + 0.15x, so 0.06x = 24 and x = 400 kg. Check: final mass 12,400 kg; final N = 1,776 + 84 = 1,860 kg; 1,860/12,400 = 15.0%. Add 400 kg only after checking compatibility and mixing thoroughly; otherwise the calculation is correct but the batch may still be inhomogeneous.