A Drain Cleaner Review: Titration Data, Indicator Errors and an Unidentified Alkali
A student buys a bottle of liquid drain cleaner whose label reads 'contains 5.0% w/v sodium hydroxide'. Using a pipette she transfers 25.0 cm3 of the cleaner to a 250.0 cm3 volumetric flask and makes it up to the mark with distilled water. She then titrates 25.0 cm3 portions of the diluted solution with 0.100 mol dm−3 hydrochloric acid, using methyl orange as indicator. Her recorded titres are 21.10 cm3, 22.40 cm3, 22.50 cm3 and 24.90 cm3. She averages all four values, calculates the concentration of alkali in the original liquid, converts it to grams per decimetre, and posts an online review saying the product contains much less alkali than the label claims. The manufacturer replies that the active ingredient is potassium hydroxide, not sodium hydroxide, and asks her to withdraw the review. [Mr: NaOH = 40.0; KOH = 56.0]
1. Decide which titre values should be used and calculate the mean titre, justifying each rejection.
Read the worked answer
Which values to keep
Titres: 21.10, 22.40, 22.50, 24.90 cm3
- 24.90 cm3 - reject. It is 2.4–2.5 cm3 above the two closest values, far outside the 0.10–0.20 cm3 agreement expected of concordant titres. The likely cause is running past the endpoint, so the recorded volume includes excess acid.
- 21.10 cm3 - reject. It is about 1.3 cm3 below the pair. This is characteristic of a rough first titration, where the endpoint is approached too quickly or the burette has just been filled and the jet not yet cleared of air.
- 22.40 and 22.50 cm3 - keep. They agree within 0.10 cm3 and are the concordant pair.
Mean titre
(22.40 + 22.50) ÷ 2 = 22.45 cm3
Averaging all four would give 22.725 cm3, about 1.2% too high - small, but the whole point of discarding non-concordant values is to remove systematic error of exactly this kind. If fewer than two concordant titres had been obtained, the correct action would be to repeat the titration rather than to average everything.
2. Using the correct mean titre, calculate the mass concentration of the alkali in the original cleaner and assess the label claim on the assumption that the alkali is sodium hydroxide.
Read the worked answer
Step 1 - moles of acid used
n(HCl) = 0.100 × (22.45 ÷ 1000) = 2.245 × 10−3 mol
Step 2 - moles of alkali in the 25.0 cm3 portion titrated
NaOH + HCl → NaCl + H2O, 1 : 1, so n(NaOH) = 2.245 × 10−3 mol
Step 3 - scale up through the dilution
The 25.0 cm3 portion came from a 250.0 cm3 flask, a tenfold dilution, so the original 25.0 cm3 of cleaner contained
2.245 × 10−3 × 10 = 2.245 × 10−2 mol
Step 4 - concentration and mass concentration
c = 2.245 × 10−2 ÷ 0.0250 = 0.898 mol dm−3
Mass concentration = 0.898 × 40.0 = 35.9 g dm−3, i.e. about 3.6% w/v
Assessment
3.6% is roughly 72% of the claimed 5.0% w/v, so on the sodium hydroxide assumption the cleaner is under strength and the student's review would be defensible.
3. Explain how the manufacturer's reply changes the conclusion, and describe a test that would settle the identity of the alkali.
Read the worked answer
Effect of the reply
The titration measures moles of alkali, not mass. To convert moles to grams, the correct molar mass must be used. If the alkali is KOH (Mr = 56.0):
Mass concentration = 0.898 × 56.0 = 50.3 g dm−3, i.e. 5.0% w/v
That agrees with the label almost exactly. So the review should be withdrawn: the student measured correctly but then attached the result to the wrong formula. The lesson is that a titration gives the amount of substance; identifying which substance it is requires separate evidence.
Test to settle it - flame test
Clean a nichrome wire with concentrated hydrochloric acid, dip it into the undiluted cleaner, and hold it in a non-luminous flame.
- Sodium: persistent golden yellow flame.
- Potassium: lilac flame, best seen through cobalt blue glass, which filters out the yellow from any sodium impurity.
Whichever colour appears identifies the cation and therefore which molar mass should be used. Until that is done, no claim about percentage strength is safe.