A75.0-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.

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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