Two-Indicator Titration of a Drain Cleaner
A quality-control analyst receives 1.200 g of a commercial drain cleaner that contains sodium hydroxide, sodium carbonate and inert filler. The solid is dissolved and made up to 250.0 mL with distilled water. A 25.00 mL aliquot is titrated with standard 0.1000 M HCl. With phenolphthalein, the pink colour disappears after 11.60 mL of acid. Methyl orange is then added to the same flask, and a further 7.20 mL of the same acid is needed to reach the second end point. The analyst must report the percentage of NaOH and of Na2CO3 in the sample and must explain the two-stage titration to a trainee who suggests that phenolphthalein alone is enough. The trainee also asks whether the same two titres could quantify a solid mixture of NaOH and NaHCO3 instead of NaOH and Na2CO3. M(NaOH) = 40.00, M(Na2CO3) = 106.0 g mol⁻¹.
1. Calculate the mass percentage of NaOH, Na2CO3 and inert material in the sample.
Read the worked answer
What each end point measures. Phenolphthalein changes colour when OH⁻ is fully neutralised and CO3²⁻ has been converted to HCO3⁻. Methyl orange then measures the second proton: HCO3⁻ → H2CO3.
Aliquot: • CO3²⁻ = 7.20 mL × 0.1000 M = 0.720 mmol • OH⁻ = (11.60 − 7.20) mL × 0.1000 M = 0.440 mmol (Check: 0.440 + 0.720 = 1.160 mmol = the phenolphthalein titre.)
Scale to 250.0 mL (×10): OH⁻ = 4.40 mmol, CO3²⁻ = 7.20 mmol.
• NaOH: 4.40 × 10⁻³ × 40.00 = 0.1760 g → 14.7 % • Na2CO3: 7.20 × 10⁻³ × 106.0 = 0.7632 g → 63.6 % • Inert: 1.200 − (0.1760 + 0.7632) = 0.2608 g → 21.7 %
Sum check: 14.7 + 63.6 + 21.7 = 100.0 %.
2. What would the analyst report if phenolphthalein alone were used and all the acid were attributed to NaOH? Explain the size of the discrepancy.
Read the worked answer
Phenolphthalein titre = 11.60 mL → 1.160 mmol in the aliquot → 11.60 mmol in 250.0 mL.
Apparent NaOH = 11.60 × 10⁻³ × 40.00 = 0.464 g → 0.464/1.200 × 100 ≈ 38.7 %.
That is about 2.6 times the true 14.7 %, an absolute overestimate of ~24 percentage points.
Why: during the phenolphthalein step each carbonate ion consumes one proton and each hydroxide ion consumes one, so the two bases are counted together. Treating the whole titre as NaOH therefore counts carbonate as if it were hydroxide, and it ignores the second proton of carbonate entirely.
Judgement: a single-indicator result cannot be reported for a carbonate-containing base; a second end point (or a BaCl2 precipitation step) is required. The trainee's shortcut is only valid for carbonate-free samples.
3. The trainee asks whether the same two titres could quantify a solid mixture of NaOH and NaHCO3. What happens on dissolution, and how would the two volumes be interpreted?
Read the worked answer
On dissolution the two solids cannot coexist as such:
OH⁻ + HCO3⁻ → CO3²⁻ + H2O.
Let a = mmol of NaOH and b = mmol of NaHCO3 in the aliquot.
• If a > b: all HCO3⁻ is converted, leaving OH⁻ = a − b and CO3²⁻ = b. Then phenolphthalein corresponds to (a − b) + b = a, and methyl orange to b. • If b > a: all OH⁻ is consumed, leaving CO3²⁻ = a and HCO3⁻ = b − a. Phenolphthalein still corresponds to a, and methyl orange to a + (b − a) = b.
In both cases the phenolphthalein titre gives a and the second titre gives b, so the method does work - but for a different reason than in the NaOH/Na2CO3 case, where the first titre gives OH⁻ + CO3²⁻ and the second gives CO3²⁻ only.
Practical caveat: the interpretation is valid only if dissolution is complete and the acid–base reaction between the components has gone to completion before titration; a partly reacted or poorly mixed sample would give an intermediate pattern and an incorrect split.