Question
Consider a titration of 100.0 mL of 0.100 M H2NNH2 (Kb=3.0X10^-6) by 0.200 M HNO3. Calculate the pH at the equivalence point for this titration. PLEASE SHOW ALL WORK
Explanation / Answer
at equivalence point we have only salt ( acid and base gets neutralised) NH2NH2 + H2O + HNO3 N2H5NO3 + H2O , here N2H5NO3 is salt of weak base and strong acid , it dissociates to give H+ , HA ---> H+ + A- ( where HA is base hydrazene nitrate) , Ka = Kw/Kb = 10^ -14/3 x10^ -6 = 3.3 x10^ -9 , HA moles = ( base moles = acid moles = ( 100 x0.1/1000) = 0.01 , vol of base = 100 ml , vol of acid = 50 ml ( since 50 x0.2 M acid = 100 x0.1 moles of base) [HA] = 0.01 x1000/150 = 0.066 M, at equi [HA] = 0.066-x , [H+]=[A-] = x , Kb = [H+][A-]/[HA] , 3.3 x10^ -9 = x^2/(0.06-x) , x = 1.4 x10^ -5 = [H+] , pH = -log[H+] = -log( 1.4 x10^ -5) = 4.854