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27 August, 22:42

The titration of 25.0 mL of an unknown concentration H2SO4 solution requires 83.6 mL of 0.12 M LiOH solution for complete neutralization. What is molarity (M) of the H2SO4 solution

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  1. 28 August, 01:56
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    0.20M

    Explanation:

    Step 1:

    Data obtained from the question. This include the following:

    Volume of acid, H2SO4 (Va) = 25mL

    Molarity of acid, H2SO4 (Ma) =.?

    Volume of base, LiOH (Vb) = 83.6mL

    Molarity of base, LiOH (Mb) = 0.12 M

    Step 2:

    The balanced equation for the reaction. This is given below:

    H2SO4 + 2LiOH - > Li2SO4 + 2H2O

    From the balanced equation above,

    Mole ratio of the acid, H2SO4 (nA) = 1

    Mole ratio of the base, LiOH (nB) = 2

    Step 3:

    Determination of the molarity of the acid, H2SO4. This can be obtain as follow:

    MaVa/MbVb = nA/nB

    Ma x 25 / 0.12 x 83.6 = 1/2

    Cross multiply

    Ma x 25 x 2 = 0.12 x 83.6

    Divide both side by 25 x 2

    Ma = (0.12 x 83.6) / (25 x 2)

    Ma = 0.20M

    Therefore, the molarity of the acid, H2SO4 is 0.20M.
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