A75.0-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. 

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