Washing Soda Purity from Two Endpoints and a Burette Bubble
In a school laboratory, a student is asked to determine the percentage purity of a sample of washing soda (Na2CO3·10H2O, molar mass 286 g mol−1) using a standard 0.100 M HCl solution. The student weighs 2.50 g of the impure sample, dissolves it in distilled water, and makes the solution up to 250.0 mL. A 25.0 mL aliquot is titrated with the HCl. With phenolphthalein, the colour just disappears after 18.60 mL of acid. The student stops at this first endpoint and plans to report the purity. A lab partner insists that the titration should be continued with methyl orange; the total volume to the second endpoint is 37.10 mL. The burette had an air bubble in the tip before the start, and the bubble disappeared during the first few millilitres of titration. The student asks whether the first-endpoint result is reliable, what the two endpoint volumes say about the sample, and how the air bubble affects the calculation.
1. Compare the observed first and second endpoint volumes with those expected for pure Na2CO3·10H2O. What do the volumes reveal about the sample’s hydration and about the presence of NaOH or NaHCO3? Show the calculation.
Read the worked answer
For pure Na2CO3·10H2O: n(sample) = 2.50 g / 286 g mol−1 = 8.74 × 10−3 mol total, so the 25.0 mL aliquot contains 8.74 × 10−4 mol Na2CO3.
First endpoint (phenolphthalein): CO3^2− + H+ → HCO3^−, so 1 mol HCl per mol Na2CO3. Expected V1 = 8.74 × 10−4 mol / 0.100 mol L−1 = 8.74 × 10−3 L = 8.74 mL.
Second endpoint total: HCO3^− + H+ → H2CO3, so 2 mol HCl per mol Na2CO3. Expected V2 = 17.48 mL.
Observed V1 = 18.60 mL and V2 = 37.10 mL, both about 2.13 times the pure-decahydrate values. Also, V2 ≈ 2V1 (37.10 mL vs 2 × 18.60 = 37.20 mL).
For a NaOH/Na2CO3 mixture, V2 < 2V1 because NaOH is titrated only at the first endpoint. For a NaHCO3/Na2CO3 mixture, V2 > 2V1 because original HCO3^− is titrated only at the second endpoint. The observed near-equality means the acid-consuming base is essentially carbonate only; NaOH and NaHCO3 are not present in significant amounts.
The effective molar mass is 2.50 g / (0.100 × 0.03710 / 2 × 10) mol = 134.8 g mol−1, between anhydrous Na2CO3 (106) and the decahydrate (286). The sample is partially dehydrated/effloresced washing soda, not pure Na2CO3·10H2O.
2. The air bubble: in which direction does it bias the calculated moles of HCl and the reported purity? Would it also bias the conclusion about hydration from V2 − V1? Explain.
Read the worked answer
The bubble sits below the stopcock before the titration. When acid is run out, some acid fills the space previously occupied by the bubble instead of reaching the flask. Therefore the burette reading overstates the volume actually delivered to the conical flask.
If the recorded volume is used in n(HCl) = cV, the calculated moles of HCl are too high. With the 2:1 HCl:Na2CO3 stoichiometry, the calculated mass of Na2CO3 and the reported percentage purity are also too high.
The hydration conclusion uses V2 − V1, not an absolute volume. The bubble is expelled early in the titration, so both endpoint readings carry essentially the same offset. The difference V2 − V1 therefore cancels the bubble error and remains reliable for deciding whether the sample is carbonate-only and for estimating its effective molar mass. A small residual error is possible if the bubble is not completely expelled before the first endpoint, but that is not the main problem here.
3. Using V2 = 37.10 mL, calculate the percentage of Na2CO3 (anhydrous, 106 g mol−1) in the sample. Then explain why reporting the result as percentage Na2CO3·10H2O would be misleading.
Read the worked answer
Total HCl used to the second endpoint: n(HCl) = 0.100 mol L−1 × 0.03710 L = 3.710 × 10−3 mol.
Na2CO3 + 2HCl → 2NaCl + CO2 + H2O, so n(Na2CO3) in the 25.0 mL aliquot = 3.710 × 10−3 / 2 = 1.855 × 10−3 mol.
In the 250.0 mL flask: n(Na2CO3) = 1.855 × 10−2 mol.
Mass as anhydrous Na2CO3 = 1.855 × 10−2 mol × 106 g mol−1 = 1.966 g.
Percentage Na2CO3 = (1.966 g / 2.50 g) × 100 = 78.6%.
If the same moles were expressed as Na2CO3·10H2O, the mass would be 1.855 × 10−2 × 286 = 5.305 g, which exceeds the 2.50 g sample. That would give an impossible purity above 100%. The sample has lost part of its water of crystallisation, so the honest report is either the anhydrous Na2CO3 content or the actual hydration state; calling it pure decahydrate is misleading.