Question

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 15.0 mL of HNO3.

Answer 1

Answer;

The pH IS 9.6

Step-by-step explanation;

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