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 15.0 ml of hno3.

The pH IS 9.6

Moles NH3 initially present = 0.0750 L X 0.2 mol/L = 0.015 mol NH3  

Moles HNO3 added = 0.015 L X 0.500 mol/L = 7.5X10^-3 mol HNO3 added  

NH3 + HNO3 --> NH4+ + NO3-  

So, after the addition, the solution contains 7.50X10^-3 mol NH4+ and 9.5X10^-3 mol NH3. The concentrations are:  

[NH4+] = 7.5X10^-3 mol / 0.090 L = 0.0833 M  

[NH3] = 1,5X10^-2mol / 0.090 L = 0.1667 M  

The equilbirium involved is:  

NH3 + H2O <--> NH4+ + OH-  

Kb = [NH4+][OH-]/[NH3] = 1.8X10^-5  

1.8X10^-5 = (0.0833)[OH-]/0.1667  

[OH-] = 3.602 X10^-5  

pOH = 4.44

pH = 14.00 - pOH

     = 9.557 or 9.6

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