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.

Question

Kale Mahoney

Chemistry

4 August, 00:56

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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Kirby · Answer 1

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