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 17.0 mL of HNO3.

Given:

Volume of NH3 = 75 mL

Concentration of NH3 = 0.200 M

Volume of HNO3 = 17 mL

Concentration of HNO3 = 0.500 M

Kb of NH3 = 1.8x10^-5

Set-up a balanced equation:

NH3 + HNO3 = = = > NH4NO3

pH = 7 - 0.5*pKb - 0.5log C

Solve for the pH using the given data.