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23 March, 02:25

A volume of 80.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 13.7 mL of 1.50 M H2SO4 was needed? The equation is

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  1. 23 March, 04:54
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    The equation is H2SO4 + 2KOH = ==> K2SO4 + 2H2O

    The molarity of base is 0.257 M

    Explanation:

    Using titration equation CAVA/CBVB = NA/NB

    Where NA is the number ov mole of acid = 1

    Using titration equation

    CAVA/CBVB = NA/NB

    Where NA is the number of mole of acid = 1

    NB is the number of mole of base = 2

    CA is the molarity of acid = 1.5M

    CB is the molarity of base = to be calculated

    VA is the volume of acid = 13.7 mL

    VB is the volume of base = 80 mL

    Substituting

    1.5*80/CB*13.7 = 1/2

    Therefore CB = 1.5*13.7*2/80*1

    CB = 0.257 M.
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Home » Chemistry » A volume of 80.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 13.7 mL of 1.50 M H2SO4 was needed? The equation is
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