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Titration of 0.01500 grams of diprotic acid required 36.22ml of 0.0120 M NaOH to reach the...

Titration of 0.01500 grams of diprotic acid required 36.22ml of 0.0120 M NaOH to reach the endpoint. What is the mass of a mole of the acid?

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Answer #1

From law of milli equivalence N1V1 = N2V2

                                          (Weight/Equivalent weight)1000 = N2V2    ------- (1)

N2 = Normality of NaOH = 0.0120N (because Normality and molarity is same for NaOH)

V2 = volume of NaOH = 36.22ml

weight of acid = 0.015grams

substitute in eq (1)       (0.015/EW)1000 = 0.012 * 36.22

EW = 34.5113

Equivalent weight = molecular weight / No of replaceable H+

Molecular weight = 34.5113 * 2    (since it is diprotic acid No of replaceable H+ are 2)

Molecular weight = 69.02 = weight of one mole of compound

mass of a mole of the acid = 69.02grams

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