A 100 ml mixture of Na₂CO₃ and NaHCO₃ is titrated against 1 M HCl. If V₁ L and V₂ L are consumed when phenolphthalein and methyl orange are used as indicators, respectively, in two separate titrations

Question

A 100 ml mixture of Na₂CO₃ and NaHCO₃ is titrated against 1 M HCl. If V₁ L and V₂ L are consumed when phenolphthalein and methyl orange are used as indicators, respectively, in two separate titrations, which of the following is true for molarities in the original solution?

TestHub · Accepted Answer

Phenolphthalein neutralizes CO₃²⁻ to HCO₃⁻ only: mol Na₂CO₃ in 100 ml = V₁ mol (1 M HCl), so molarity of Na₂CO₃ = 10V₁, making (a) false. Methyl orange gives total: V₂ = 2n(CO₃²⁻) + n(HCO₃⁻), so n(NaHCO₃) = V₂ - 2V₁ mol in 100 ml, i.e. molarity of NaHCO₃ = 10(V₂ - 2V₁), option (b).

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