An aqueous solution contains 0.332 M ammonia ( NH 3 ). How many mL of 0.386 M nitric acid would have to be added to 250 mL of this solution in order to prepare a buffer with a pH of 9.180 ? Solution pH of basic buffer= 14 - ( pkb + log(NH4NO3/NH3) pKb of NH3 = 4.75 no of mol of NH3 present finally = 0.332*0.25 = 0.083-x mol no of mol of NH4NO3 = HNO3 = x M 9.18 = 14-(4.75+log(x/(0.083-x)) x = 0.0448 no of mol of HNO3 must add = 0.0448 mol volume of HNO3 must add =0.0448/0.386 = 0.116 L = 116 ml .