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27 January, 11:05

A sample of pure oxalic acid (H2C2O4.2H2O) weighs 0.2000 g and requires 30.12 ml of KOH solution for

complete neutralization. Calculate the molarity of the KOH solution. (0.105 M)

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  1. 27 January, 12:37
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    The molarity of KOH is 0.1055 M

    calculation

    Step 1: write the equation for reaction between H₂C₂O₄.2H₂O and KOH

    H₂C₂O₄.2H₂O + 2 KOH → K₂C₂O₄ + 4 H₂O

    step 2: find the moles of H₂C₂O₄.2H₂O

    moles = mass: molar mass

    from periodic table the molar mass H₂C₂O₄.2H₂O = (1 x2) + (12 x2) + (16 x4) + 2 (18) = 126 g/mol

    = 0.2000 g : 126 g/mol = 0.00159 moles

    step 3: use the mole ratio to calculate the moles of KOH

    H₂C₂O₄.2H₂O : KOH is 1:2

    therefore the moles of KOH = 0.00159 x 2 = 0.00318 moles

    step 4: find molarity of KOH

    molarity = moles/volume in liters

    volume in liters = 30.12/1000=0.03012 L

    molarity is therefore = 0.00318/0.03012 = 0.1055 M
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