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6 June, 13:46

You measure 2.34 g of K2Cr2O7 into a volumetric flask. You dilute the K2Cr2O7 with water to a volume of 250 mL. It takes 35.7 mL of this solution to titrate 17.8 mL of a solution of Na2C2O4in the presence of excess acid. What was the concentration of the original Na2C2O4 solution

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  1. 6 June, 13:52
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    0.191M of Na₂C₂O₄ is the concentration of the original Na₂C₂O₄ solution

    Explanation:

    The reaction of potassium dichromate, K₂Cr₂O₇ with sodium oxalate, Na₂C₂O₄ in the presence of acid H⁺ is:

    K₂Cr₂O₇ + 3Na₂C₂O₄ + 14H⁺ → 2Cr³⁺ + 6CO₂ + 7H₂O + 6Na⁺ + 2K⁺

    Thus, 1 mole of K₂Cr₂O₇ reacts with 3 moles of Na₂C₂O₄

    Moles of 2.34g of K₂Cr₂O₇ (Molar mass: 294.185g/mol):

    2.34g K₂Cr₂O₇ ₓ (1mol / 294.185g) = 7.954x10⁻³ moles K₂Cr₂O₇

    In 250mL = 0.250L:

    7.954x10⁻³ moles K₂Cr₂O₇ / 0.250L = 0.0318M K₂Cr₂O₇

    Moles in 35.7mL = 0.0357L of this solution are:

    0.0357L ₓ (0.0318mol / L) = 1.136x10⁻³ moles K₂Cr₂O₇ in solution. As 1 mole of K₂Cr₂O₇ reacts with 3 moles of Na₂C₂O₄, to titrate the moles of K₂Cr₂O₇ in solution you need:

    1.136x10⁻³ moles K₂Cr₂O₇ * (3 moles Na₂C₂O₄ / 1 mole K₂Cr₂O₇) =

    3.408x10⁻³ moles of Na₂C₂O₄

    In 17.8mL = 0.0178L:

    3.408x10⁻³ moles of Na₂C₂O₄ / 0.0178L =

    0.191M of Na₂C₂O₄ is the concentration of the original Na₂C₂O₄ solution
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