A 75.0 mL volume of 0.200 M NH3 (Kb=1.8x10^-5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 28.0 mL of HNO3.

Answer 1

mmols NH3 = mL x M = 75.0 x 0.200 = 15.0
mmols HNO3 = 28.0 x 0.500 = 14.0
.......NH3 + HNO3 ==> NH4NO3
I......15.0....0........0
add...........14.0.......
C.....-14.0..-14.0.....+14.0
E.......1.0.....0......+14.0

So you are 1.0 mmol away from the equivalence point. Basically you have a buffer problem. You have a small amount of NH3 left and you have (relatively) large amount NH4^+ formed. Use the Henderson-Hasselbalch equation. NH3 will be the base and NH4^+ will be the acid.

Answer 2

Thanks.