Question

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 25.0 mL of HNO3.

Answer 1

 pKb = - log 1.8 x 10^-5 = 4.7 
moles NH3 = 0.0750 L x 0.200 M=0.0150 
moles HNO3 = 0.0270 L x 0.500 M= 0.0135 
the net reaction is 
NH3 + H+ = NH4+ 
moles NH3 in excess = 0.0150 - 0.0135 =0.0015 
moles NH4+ formed = 0.0135 
total volume = 75.0 + 27.0 = 102.0 mL = 0.102 L 
[NH3]= 0.0015/ 0.102 L=0.0147 M 
[NH4+] = 0.0135/ 0.102 L = 0.132 M 

pOH = pKb + log [NH4+]/ [NH3]= 4.7 + log 0.132/ 0.0147=5.65 
pH = 14 - pOH = 14 - 5.65 =8.35 

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