If 20 ml of 0.1 M solution of sodium sesquicarbonate (Na₂CO₃·NaHCO₃) is titrated against 0.05 M - HCl, using (i) phenolphthalein and (ii) methyl orange as indicators, what difference in titre values (

Question

If 20 ml of 0.1 M solution of sodium sesquicarbonate (Na₂CO₃·NaHCO₃) is titrated against 0.05 M - HCl, using (i) phenolphthalein and (ii) methyl orange as indicators, what difference in titre values (in ml) would be recorded?

TestHub · Accepted Answer

Salt taken = 20×0.1 = 2 mmol, containing 2 mmol CO₃²⁻ + 2 mmol HCO₃⁻. Phenolphthalein endpoint neutralizes only CO₃²⁻ → HCO₃⁻: H⁺ = 2 mmol → 2/0.05 = 40 ml. Methyl orange endpoint neutralizes fully: H⁺ = 2×2 + 2 = 6 mmol → 120 ml. Difference = 120-40 = 80 ml. Ans 0080.

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