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14 March, 08:22

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8*10-5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 17.0 mL of HNO3.

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  1. 14 March, 11:11
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    Given:

    Volume of NH3 = 75 mL

    Concentration of NH3 = 0.200 M

    Volume of HNO3 = 17 mL

    Concentration of HNO3 = 0.500 M

    Kb of NH3 = 1.8x10^-5

    Set-up a balanced equation:

    NH3 + HNO3 = = = > NH4NO3

    pH = 7 - 0.5*pKb - 0.5log C

    Solve for the pH using the given data.
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