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17 November, 21:55

I am having some trouble figuring out how to approach the following problem: 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 17.0mL of HNO3.

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  1. 18 November, 00:26
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    We look at the end of the day:

    n (HNO3) added = 0.500*17.0/1000 = 0.00850 mol

    n (NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol

    [NH3] left = 0.00650*1000 / (17.0+75.0) = 0.070652

    M [OH-] = Kb * [NH3] = 0.070652*1.8*10^ (-5) = 1.27174 x 10^ (-6)

    pOH = - log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10
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