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.
I am hoping that these answers
have satisfied your query and it will be able to help you in your endeavors, and
if you would like, feel free to ask another question.
