Two-Endpoint Titration of an Adulterated Washing Soda Sample
A first-year student is given a 2.50 g sample of commercial washing soda suspected to be mixed with baking soda and common salt. She dissolves it completely and makes the solution up to 250.0 mL, then titrates 25.0 mL portions with 0.100 M HCl. With phenolphthalein the first permanent faint pink disappears after 10.0 mL of acid. She then adds methyl orange and continues titrating; the orange-red end point is complete only after a total of 24.5 mL of acid has been delivered from the burette. A classmate titrates a fresh portion using methyl orange only, sees the end point at 24.5 mL, and announces that the whole sample is pure Na2CO3. The bottle carries no purity figure. (M: Na2CO3 = 106, NaHCO3 = 84, NaCl = 58.5.)
1. Calculate the mass percentage of Na2CO3, NaHCO3 and NaCl in the sample, showing the stoichiometry you rely on.
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Stoichiometry of the two stages. Stage 1 (phenolphthalein, pH ~8.3): CO3^2- + H+ -> HCO3^- Stage 2 (methyl orange, pH ~4): HCO3^- + H+ -> H2CO3 (for the carbonate part) and the original HCO3^- + H+ -> H2CO3.
Per 25.0 mL aliquot. Acid to phenolphthalein = 10.0 x 0.100 = 1.00 mmol, so moles of CO3^2- = 1.00 mmol (1:1). Total acid = 24.5 x 0.100 = 2.45 mmol. The second stage uses 2.45 - 1.00 = 1.45 mmol. Of this, 1.00 mmol is needed to convert the HCO3^- just formed from the carbonate, leaving 1.45 - 1.00 = 0.45 mmol of original HCO3^-.
Scale up to the 250.0 mL flask (factor 10). n(Na2CO3) = 10.0 mmol -> 10.0e-3 x 106 = 1.060 g n(NaHCO3) = 4.5 mmol -> 4.5e-3 x 84 = 0.378 g Alkali mass = 1.438 g, so NaCl (by difference) = 2.50 - 1.438 = 1.062 g
Percentages. Na2CO3 = 1.060/2.50 x 100 = 42.4% NaHCO3 = 0.378/2.50 x 100 = 15.1% NaCl = 1.062/2.50 x 100 = 42.5% The three add to 100.0%, a useful internal check.
2. The classmate's methyl-orange-only calculation is wrong. Show what number he would report, explain why it is physically impossible, and state the general reason a single end point cannot solve this mixture.
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His number. He treats all 2.45 mmol of HCl per aliquot as neutralising carbonate: n(Na2CO3) = 2.45 mmol per aliquot, so 24.5 mmol in 250 mL, i.e. 24.5e-3 x 106 = 2.597 g in a 2.50 g sample - a purity of 103.9%.
Why it is impossible. It exceeds 100% of the sample mass, which no composition can do. The error is stoichiometric: utilising the whole acid for CO3^2- assumes each carbonate takes only one proton, but a full neutralisation (methyl orange) consumes two protons per carbonate.
The general reason. A single end point gives one measurable number - the total proton-accepting capacity in equivalents - and there are two unknowns, n(CO3^2-) and n(HCO3^-). One equation, two unknowns; the system is underdetermined. The phenolphthalein end point provides the second, independent equation, because it counts only the first protonation of carbonate.
Diagnostic habit. Any purity calculation above 100% is a signal that the equivalence factor is wrong, not that the supplier cheated. Note also that if the sample were pure Na2CO3, the methyl orange volume would be exactly twice the phenolphthalein volume; here 24.5 mL > 2 x 10.0 mL, which is itself proof that HCO3^- is present independently. If instead V(MO) had been less than 2 x V(Ph), the sample would contain NaOH, not NaHCO3 - a different case entirely.
3. Suppose she overshoots the phenolphthalein end point by 0.5 mL but records the total volume correctly. Quantify the resulting error and explain why the answer is not simply 'everything is overestimated'.
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Recalculate with V1 = 10.5 mL instead of 10.0 mL, V_total still 24.5 mL. CO3^2- = 10.5 x 0.100 = 1.05 mmol per aliquot (true value 1.00), an error of +0.05 mmol per aliquot = +0.50 mmol on the full 250 mL. Mass error = 0.50e-3 x 106 = 0.053 g, i.e. about +2.1 percentage points on a 42.4% figure. HCO3^- = V_total - 2V1 = 2.45 - 2.10 = 0.35 mmol (true 0.45), an error of -0.10 mmol per aliquot = -1.0 mmol overall = -0.084 g, about -3.4 percentage points.
Why the errors have opposite signs. The second stage uses V_total - V1, so inflating V1 simultaneously inflates the carbonate term and deflates the bicarbonate term. The total alkalinity (V_total, measured separately from the start of the titration) is untouched, so the sum of the two errors largely cancels in equivalents: +0.05 mmol of carbonate consumes exactly the extra 0.10 mmol... precisely, one extra CO3^2- assigned replaces one HCO3^- and one extra proton.
Practical lesson. Errors at the first end point distort the split between carbonate and bicarbonate far more than they distort the total; errors in V_total distort the total. To control the split, the faint-pink end point must be taken at the first permanent tinge - a drop of overshoot here is worth roughly 1% of the whole sample per 0.1 mL, so a 0.5 mL overshoot is a badly sloppy titration.