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March 29, 2017

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I wanted to calculate the average equivalent weight of an unknown acid for an acid-base titration experiment. The mass of the unknown that I had to obtain is 0.2 g and the equivalents of the acid is .001046 L x N. But is this equation right? Does the equivalent mass equal to the mass of the solid acid divided by the number of equivalents of acid? If so, isn't the average eq. wt. too high?

Thank you so much for any help.

  • Chemistry - ,

    liters x normality x equivalent weight = grams.

  • Chemistry - ,

    wait, liters x normality = grams because I wanted to find the equivalent wt

  • Chemistry - ,

    Wait, im not making sense. I know liters x normality gives you the equivalents of an acid. But are you sure it's not the grams I obtained divided by the equivalents of the acid.

  • Chemistry - ,

    Of course it is but that's what I wrote in a different form.
    liters x normality x eq wt = grams.
    liters x normality = # equiv so
    #eq x eq wt = grams and
    eq wt = grams/#equivalents

    I like
    liters x normality x equivalent weight = grams BECAUSE it solve one equation instead of three and get the same answer.

  • Chemistry - ,

    ok, thank you very much for your time and patience. I understand, im sorry if I bothered you too much. But thanks a lot.

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