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9 February, 20:25

What is the molarity of a nitric acid solution if 43.13 mL 0.1000 M KOH solution is needed to neutralize 30.00 mL of the acid solution?

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  1. 9 February, 20:53
    0
    0.144M

    Explanation:

    First, let us write a balanced equation for the reaction. This is illustrated below:

    HNO3 + KOH - > KNO3 + H20

    From the equation,

    nA = 1

    nB = 1

    From the question given, we obtained the following:

    Ma = ?

    Va = 30.00mL

    Mb = 0.1000M

    Vb = 43.13 mL

    MaVa / MbVb = nA/nB

    Ma x 30 / 0.1 x 43.13 = 1

    Cross multiply to express in linear form

    Ma x 30 = 0.1 x 43.13

    Divide both side by 30

    Ma = (0.1 x 43.13) / 30 = 0.144M

    The molarity of the nitric acid is 0.144M
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