The Effloresced Carbonate: Standardising HCl with a Reagent That Is Not 100% Pure
A first-year practical group must submit 250.0 mL of exactly 0.0500 M sodium carbonate solution for the next day's titration. The only reagent available is a bottle labelled 'Na2CO3·10H2O, assay 92.0% as Na2CO3'; the remainder is water lost by efflorescence. The balance reads to ±0.001 g and the standard flask is class A. Student A calculates the mass needed by assuming pure anhydrous Na2CO3, weighs 1.325 g, dissolves it with two rinses and makes up to the mark. Student B wants to dry the salt overnight at 110 °C, but the oven is broken. The class will use this solution to standardise about 0.1 M HCl, using methyl orange and 20.0 mL pipetted portions. Student A's three concordant titres average 21.60 mL of the carbonate solution. A demonstrator remarks that the balance tolerance is irrelevant and the flask tolerance is 'a rounding error', but that the result is still wrong. The report must state the true concentration of both solutions and identify the dominant error.
1. Calculate the mass of the 92.0% material actually required, the concentration Student A really prepared, and the relative error.
Read the worked answer
Amount required: n = 0.2500 L × 0.0500 mol L⁻¹ = 1.250 × 10⁻² mol. M(Na2CO3) = 2(23.0) + 12.0 + 3(16.0) = 106.0 g mol⁻¹, so pure mass = 1.250 × 10⁻² × 106.0 = 1.325 g. Because the solid is only 92.0% Na2CO3 by mass, the mass to weigh is 1.325/0.920 = 1.440 g - not 1.325 g. Student A actually introduced n = 1.325 × 0.920/106.0 = 1.219/106.0 = 1.150 × 10⁻² mol, so c = 1.150 × 10⁻²/0.2500 = 0.0460 M. Relative error = (0.0460 − 0.0500)/0.0500 = −8.0%. Judgement: the balance contributes 0.001/1.325 = 0.08% and the flask about 0.15/250 = 0.06%, together ~0.1%. The 8% assay error is roughly eighty times larger, so the solution is wrong regardless of how carefully it is weighed or made up to the mark.
2. Using the 21.60 mL titre, compute the HCl concentration both with the nominal 0.0500 M value and with the true value. State the direction and size of the error.
Read the worked answer
Reaction: Na2CO3 + 2HCl → 2NaCl + H2O + CO2, so n(HCl) = 2n(Na2CO3). Nominal: n(Na2CO3) = 0.0500 × 0.02160 = 1.080 × 10⁻³ mol; n(HCl) = 2.160 × 10⁻³ mol; c(HCl) = 2.160 × 10⁻³/0.02000 = 0.1080 M. True: n(Na2CO3) = 0.0460 × 0.02160 = 9.94 × 10⁻⁴ mol; n(HCl) = 1.987 × 10⁻³ mol; c(HCl) = 0.0994 M. The nominal figure overestimates the acid by (0.1080 − 0.0994)/0.0994 = +8.7%, exactly the factor 0.0500/0.0460 = 1.087. Direction matters: because the assumed carbonate concentration was too high, each titre is converted into too many moles of HCl, so the acid is reported as stronger than it is. A systematic multiplicative error of this kind is not random; repeating the titration improves precision but cannot remove it.
3. The demonstrator calls the flask tolerance 'a rounding error'. Quantify this and recommend, with reasons, what the group should do in the time available.
Read the worked answer
Flask: 0.15 mL in 250.0 mL = 0.06%. Balance: 0.001 g in 1.325 g = 0.075%. Pipette and endpoint reading add perhaps 0.1%. All together they are about 1% at worst, against 8% from the assay - the remark is fair. The key point is that efflorescence water does not neutralise acid; it is inert diluent, so treating the material as 92.0% Na2CO3 for the mole calculation is legitimate provided the assay is used. The error is entirely in Student A's assumption of 100% purity. Recommendation: do not buy a four-decimal balance. Either (i) weigh out the assay-corrected mass, 1.440 g, and label the flask with the calculated concentration, or (ii) standardise the HCl directly against a reliable primary standard such as freshly dried oxalic acid dihydrate or borax. Drying the carbonate is the ideal fix, but with the oven out of order the assay correction is the only defensible option.