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21 March, 05:53

The ph of a finished water from a water treatment process is 10.74. What amount of. 02 n sulfuric acid, in milliliters, is required to neutralize 1 l of finished water assuming alkalinity is zero

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  1. 21 March, 08:15
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    The pH of water is given to be 10.74.

    This means = pOH = 14 - pH = 3.26

    pOH = - log[OH-] = 3.26

    [OH-] = 0.00055 M / N

    The concentration of H2SO4 is given to be = 0.02 N

    volume of water = 1 L

    So moles of OH - = number of equivalents of OH-

    = concentration X volume (L) = 0.00055 X 1 = 0.00055

    So number of equivalents of H2SO4 requried = 0.00055

    Hence volume of 0.02 N H2SO4 required = Moles / Molarity

    = 0.00055/0.02 = 0.0275 L = 27.5 mL
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