0.4 g mixture of NaOH, Na₂CO₃ and some inert impurities was first titrated with N/10 HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5 mL of same HCl was required for the next end point. The weight percentage of Na₂CO₃ in the mixture is ________. (Rounded off to the nearest integer)

Solution

Chemistry ma’am · 4 steps

Read the methyl orange volume as the carbonate's second half: it is the clean measure of Na₂CO₃.

1. Phenolphthalein stage. Phenolphthalein turns colourless near pH 8.3. By then all the NaOH is neutralised, but Na₂CO₃ has only gone halfway, to NaHCO₃. So the 17.5 mL covers all the NaOH plus one proton's worth of carbonate.
2. Methyl orange stage. Methyl orange then needs extra acid to turn NaHCO₃ into CO₂. Each NaHCO₃ came from one Na₂CO₃, so the 1.5 mL counts moles of Na₂CO₃ one to one.

   n(Na₂CO₃) = n(NaHCO₃) = (1.5 × 0.1)/1000 = 1.5 × 10⁻⁴ mol

3. Mass and percentage. Mass of Na₂CO₃ is 1.5 × 10⁻⁴ × 106 = 0.0159 g. Out of 0.4 g, that is 0.0159/0.4 × 100 = 3.975%. As a check, NaOH used 17.5 - 1.5 = 16 mL, so it makes up 16% of the mixture.
4. So. The weight percentage of Na₂CO₃ is 3.975%, which rounds to 4.

The common slip: 8. Treating the methyl orange reading as only half the carbonate, so taking 2 × 1.5 mL as moles of Na₂CO₃, doubles the mass and gives 8, not 4.

The idea to keep. In a double-indicator titration, carbonate reacts in two steps. The methyl orange volume alone gives moles of Na₂CO₃; the phenolphthalein volume minus it gives NaOH.

Answer from the NTA final answer key. Solution by PrepWiser.