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.

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.

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current

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