Two Water Samples
A cadet is asked to compare two water samples before they are used in the engine room. Sample X is fresh water from a shore supply; a 100 cm³ portion requires 2.4 cm³ of 0.010 mol/dm³ silver nitrate solution to reach the endpoint in a chloride titration. Sample Y is from the ship's evaporator; the same test on 100 cm³ requires 0.6 cm³ of the same solution. The cadet is told the evaporator outlet should not exceed 10 mg/dm³ chloride expressed as Cl⁻ (Ar: Cl = 35.5). The cadet must decide which sample is acceptable and explain the chemistry in a way the engine-room staff can follow.
1. Calculate the chloride concentration of each sample in mg/dm³ and decide which is acceptable.
Read the worked answer
The reaction is Ag⁺ + Cl⁻ → AgCl(s), a 1:1 ratio.
Sample X: moles AgNO₃ = 0.010 × (2.4 ÷ 1000) = 2.4 × 10⁻⁵ mol. So Cl⁻ = 2.4 × 10⁻⁵ mol in 100 cm³. Mass = 2.4 × 10⁻⁵ × 35.5 = 8.52 × 10⁻⁴ g = 0.852 mg in 100 cm³. Per dm³ (×10): 8.52 mg/dm³.
Sample Y: moles = 0.010 × (0.6 ÷ 1000) = 6.0 × 10⁻⁶ mol. Mass = 6.0 × 10⁻⁶ × 35.5 = 2.13 × 10⁻⁴ g = 0.213 mg in 100 cm³ → 2.13 mg/dm³.
Decision: both are below 10 mg/dm³, so both pass; Sample Y is the cleaner at about a quarter of the chloride. If the limit were applied strictly to the evaporator outlet only, Y is clearly acceptable and X is also within limit.
2. Explain why the result is multiplied by ten, and what error would arise if a cadet forgot this step.
Read the worked answer
The titration was run on 100 cm³, but the limit is expressed per dm³ (1,000 cm³). Since 1,000 ÷ 100 = 10, the mass found in the 100 cm³ portion must be multiplied by 10 to give mg/dm³.
If forgotten: Sample X would be reported as 0.852 mg/dm³ instead of 8.52 mg/dm³, and Sample Y as 0.213 instead of 2.13. Every result would be ten times too low.
Why that is dangerous: a genuinely failing sample at, say, 45 mg/dm³ would be reported as 4.5 mg/dm³ and passed. The error is systematic, so it would never show up as an odd result - every reading would simply look reassuringly clean. The safeguard is to write the units at each step (mg per 100 cm³ → mg per dm³) so the conversion cannot be skipped silently.
3. The engine-room staff ask why silver nitrate is used rather than simply measuring the water's appearance. Give the reasoning.
Read the worked answer
Chloride in water is colourless and invisible. Clear-looking water can still carry well over the limit, so appearance tells you nothing about chloride content.
Silver nitrate works because silver ions react specifically and quantitatively with chloride ions to form silver chloride, a pale precipitate:
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
The endpoint is reached when all chloride has reacted and the first excess of silver ions is detected (commonly with a indicator such as potassium chromate, which gives a colour change). Because the reaction is 1:1, the volume of standard silver nitrate used is directly proportional to the chloride present.
In short: the titration converts an invisible contaminant into a measurable volume. Appearance-based checks cannot do this, and chloride is exactly the kind of contaminant that damages boiler and engine surfaces while looking perfectly harmless.