Question

A 75.0-ml volume of 0.200M NH3 (Kb=1.8 x 10^-5) is titrated with 0.500M NHO3. Calculate the pH after the addition of 13.0mL of NH3.

Answer 1

The solution would be like this for this specific problem:

moles NH3 = 0.0750 L x 0.200 M=0.015
moles HNO3 = 0.0130 L x 0.500 M=0.0065

NH3 + H+ = NH4+
moles NH3 = 0.015 - 0.0065=0.0085
moles NH4+ = 0.0065
pKb = 4.74
pOH = 4.74 + log 0.0065/0.0085=4.62
pH = 9.38

So the pH after the addition of 13.0 mL of HNO_3 is 9.38.

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