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25 October, 03:06

In a titration of 50 ML of 0.20m NaOH is used to neutralize 10.0 ML of HCL. Calculate the molarity of the acid

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  1. 25 October, 03:31
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    1M

    Explanation:

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

    NaOH + HCl - > NaCl + H2O

    From the equation above,

    nA (mole of acid) = 1

    nB (mole of the base) = 1

    Data obtained from the question include:

    Vb (volume of base) = 50mL

    Mb (Molarity of base) = 0.2M

    Va (volume of acid) = 10mL

    Ma (Molarity of acid) = ?

    Using MaVa/MbVb = nA/nB, we can calculate the molarity of the acid as follow:

    Ma x 10/0.2 x 50 = 1

    Cross multiply to express in linear form

    Ma x 10 = 0.2 x 50

    Divide both side by 10

    Ma = (0.2 x 50) / 10

    Ma = 1M

    The molarity of the acid is 1M
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