Standardising HCl and the choice of indicator
A first-year diploma student is asked to standardise a hydrochloric acid solution using anhydrous sodium carbonate as the primary standard and then to use the acid to estimate the amount of sodium hydroxide in a drain-cleaning liquid. She weighs 0.530 g of dry Na2CO3, dissolves it in distilled water and makes the solution up to 100 mL. Using methyl orange indicator, she finds that 25.0 mL of this carbonate solution needs 20.0 mL of the acid for complete neutralisation. A classmate repeats the titration using phenolphthalein and reports that only 10.0 mL of the same acid is needed for the same 25.0 mL of carbonate solution, and argues that the student's value must be wrong. The teacher asks the student to explain the difference and then to calculate how much NaOH is present in 25.0 mL of the drain cleaner if 25.0 mL of the cleaner requires 31.0 mL of the same acid.
1. Calculate the molarity of the sodium carbonate solution and of the hydrochloric acid from the methyl orange titration.
Read the worked answer
Moles of Na2CO3 in the 100 mL flask = 0.530/106 = 0.0050 mol, so molarity = 0.0050/0.100 = 0.050 M. In 25.0 mL: 0.050 × 0.0250 = 1.25 × 10⁻³ mol. Reaction: Na2CO3 + 2HCl → 2NaCl + H2O + CO2, so HCl needed = 2 × 1.25 × 10⁻³ = 2.50 × 10⁻³ mol. This is contained in 20.0 mL, so [HCl] = 2.50 × 10⁻³ / 0.0200 = 0.125 M.
2. Explain why the classmate obtained 10.0 mL with phenolphthalein, and decide whose value is wrong, if either.
Read the worked answer
With phenolphthalein the colour change occurs at the first equivalence point, where carbonate is converted only to hydrogen carbonate: Na2CO3 + HCl → NaHCO3 + NaCl. That consumes 1.25 × 10⁻³ mol HCl, i.e. 1.25×10⁻³/0.125 = 0.0100 L = 10.0 mL. The classmate's 10.0 mL is correct for the phenolphthalein endpoint, not evidence that the 20.0 mL value is wrong. Methyl orange changes in acid, so it marks complete neutralisation to CO2 and needs twice the acid. Indicator choice defines the stoichiometry; the student should report which indicator was used.
3. Calculate the mass of NaOH in 25.0 mL of the drain cleaner and express it in g/L. State one limitation of the result.
Read the worked answer
Moles HCl in 31.0 mL = 0.125 × 0.0310 = 3.875 × 10⁻³ mol. NaOH + HCl → NaCl + H2O, so NaOH = 3.875 × 10⁻³ mol. Mass = 3.875 × 10⁻³ × 40 = 0.155 g in 25.0 mL. Concentration = 0.155/0.0250 = 6.2 g/L. This is the NaOH equivalent; if the cleaner contains other bases or carbonates, the result overestimates actual NaOH.