Double titration of a carbonated NaOH/Na2CO3 powder
A forensic-style practical receives 1.500 g of a white powder known to contain sodium hydroxide, sodium carbonate, and sodium chloride only. The powder has been stored in a loosely capped bottle in a damp store room, so some hydroxide may have carbonated. The analyst dissolves the whole sample in freshly boiled, cooled distilled water and makes the volume to 250.0 mL. A 25.00 mL aliquot is titrated with standardized 0.1000 M HCl. With phenolphthalein, the colour disappears after 14.20 mL. Methyl orange is then added, and a further 8.40 mL is required to reach the second endpoint. A second analyst argues that the first volume cannot be used to calculate hydroxide because phenolphthalein also responds to carbonate, and suggests precipitating carbonate with BaCl2 first. The laboratory supervisor needs the composition and a judgement on the method.
1. Calculate the mass percentages of NaOH, Na2CO3, and NaCl in the sample. Show the stoichiometric reasoning for the two endpoints.
Read the worked answer
Aliquot: n(HCl to phenolphthalein) = 0.1000 × 0.01420 = 1.420 mmol; n(HCl to methyl orange) = 0.1000 × 0.00840 = 0.840 mmol.
At the first endpoint: OH- + H+ → H2O and CO3^2- + H+ → HCO3-. At the second: HCO3- + H+ → H2CO3. Thus n(CO3^2-) = 0.840 mmol and n(OH-) = 1.420 - 0.840 = 0.580 mmol in 25.00 mL.
Scale by 10: Na2CO3 = 8.40 mmol × 105.99 = 0.890 g; NaOH = 5.80 mmol × 40.00 = 0.232 g; NaCl = 1.500 - 1.122 = 0.378 g.
Percentages: NaOH 15.5%, Na2CO3 59.3%, NaCl 25.2%.
2. Evaluate the second analyst's objection. When would using the simple difference V_ph - V_mo give a wrong or impossible result?
Read the worked answer
The objection is only partly right. Phenolphthalein does respond to carbonate, but only to its first protonation. The second volume measures that same carbonate. Therefore V_ph = OH + CO3 and V_mo = CO3, so subtraction gives OH.
If NaHCO3 were present, V_mo would exceed V_ph because bicarbonate would add to the second volume; then the simple formulas fail and V_ph - V_mo could be negative if interpreted as hydroxide.
BaCl2 precipitation can give an independent hydroxide value, but the BaCO3 must be removed completely and not titrated. For this OH/CO3 sample, the double titration is adequate; the BaCl2 method is a useful check rather than a necessity.
3. Suppose the powder had also contained sodium bicarbonate. How would the two titration volumes change, and why could the reported NaOH be wrongly low or even negative?
Read the worked answer
With bicarbonate present, the first volume is still OH + CO3, but the second volume becomes CO3 + HCO3. Hence V_mo > V_ph.
If the analyst still uses n(OH) = V_ph - V_mo, the result is negative, which is physically impossible. Even a small bicarbonate impurity makes the reported NaOH too low because the extra bicarbonate consumes acid in the second step and is misattributed.
The correct response is to test for bicarbonate (for example with BaCl2 or a separate aliquot) and use a three-component calculation. The supervisor should not accept a negative hydroxide result as an analytical mystery; it is evidence that the assumed model is wrong.