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23 December, 04:48

A 10.8ml sample of sulfuric acid titrated with 80.0 ml of 0.200 m mg solution. What is the concentration of the sample given the acid-base reaction shown below

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  1. 23 December, 04:59
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    1.48 M

    Explanation:

    Step 1:

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

    Mg + H2SO4 - > MgSO4 + H2

    Step 2:

    Determination of the number of mole of Mg in 80.0 mL of 0.200 M Mg solution. This is illustrated below:

    Molarity of Mg = 0.200 M

    Volume of solution = 80 mL = 80/1000 = 0.08L

    Mole of Mg = ?

    Molarity = mole / Volume

    0.2 = mole / 0.08

    Mole = 0.2 x 0.08

    Mole of Mg = 0.016 mole.

    Step 3:

    Determination of the number of mole of H2SO4 that reacted. This is illustrated below:

    Mg + H2SO4 - > MgSO4 + H2

    From the balanced equation above,

    1 mole of Mg reacted with 1 mole of H2SO4.

    Therefore, 0.016 mole of Mg will also react with 0.016 mole of H2SO4.

    Step 4:

    Determination of the concentration of the acid.

    Mole of H2SO4 = 0.016 mole.

    Volume of acid solution = 10.8 mL = 10.8/1000 = 0.0108 L

    Molarity = ?

    Molarity = mole / Volume

    Molarity = 0.016/0.0108

    Molarity of the acid = 1.48 M

    Therefore, the concentration of acid is 1.48 M
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