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1 February, 18:04

A solution of malonic acid, H2C3H2O4, was standardized by titration with 0.1000 M NaOH solution. If 21.09 mL of the NaOH solution were required to neutralize completely 12.30 mL of the malonic acid solution, what is the molarity of the malonic acid solution

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  1. 1 February, 18:17
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    1000cm³ = 1mL = 1dm³

    C₃H₄O₄ + 2NaOH → Na₂C₃H₂O₄ + 2H₂O

    0.10M = 1000cm³

    x M = 21.09cm³

    xM = (21.09 * 0.10) / 1000

    x M = 0.002109 moles

    From equation of the reaction,

    1 mole of C₃H₄O₄ reacts with 2 moles of NaOH

    1 mole of C₃H₄O₄ = 2 moles of NaOH

    y mole of C₃H₄O₄ = 0.002109 moles of NaOH

    y = (0.002109 * 1) / 2

    y = 0.0010545 moles

    0.0010545 moles is present in 12.30cm³

    0.0010545 mole = 12.30cm³

    z mole = 1000cm³

    z = (0.0010545 * 1000) / 12.30

    Z = 0.0857M

    The molarity of malonic acid is 0.0857M
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