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29 October, 22:27

If 15.98 mL of 0.1080 M KOH solution reacts with 52.00 mL of HC2H302,

what is the molarity of the acid solution? (Don't forget significant figures)

KOH + HC2H302 - KC2H302 + H2O

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Answers (1)
  1. 29 October, 22:49
    0
    0.0332 M

    Explanation:

    Step 1:

    The balanced equation for the reaction.

    KOH + HC2H302 - > KC2H302 + H2O

    From the balanced equation above,

    The mole ratio of the acid (nA) = 1

    The mole ratio of the base (nB) = 1

    Step 2:

    Data obtained from the question.

    Volume of base (Vb) = 15.98 mL

    Molarity of base (Mb) = 0.1080 M

    Volume of acid (Va) = 52.00 mL

    Molarity of acid (Ma) = ?

    Step 3:

    Determination of the molarity of the acid.

    Using the formula:

    MaVa/MbVb = nA/nB

    The molarity of the acid can be obtained as follow

    MaVa/MbVb = nA/nB

    Ma x 52 / 0.1080 x 15.98 = 1

    Cross multiply to express in linear form

    Ma x 52 = 0.1080 x 15.98

    Divide both side by 52

    Ma = (0.1080 x 15.98) / 52

    Ma = 0.0332 M

    Therefore, the molarity of the acid solution is 0.0332 M
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