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A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the...

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution after the addition of 50.0 mL of HNO3. The Kb of NH3 is 1.8 × 10-5.

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Expert Solution

no of moles of NH3 = molarity * volume in L

                                  = 0.1*0.1 = 0.01 moles

no of moles of HNO3 = molarity * volume in L

                                    = 0.1*0.05 = 0.005 moles

                 NH3 + HNO3 ---------------> NH4NO3

I              0.01       0.005                         0

C          -0.005    -0.005                          0.005

E           0.005        0                              0.005

                   Kb = 1.8*10^-5

                    PKb = -logKb   = -log1.8*10^-5 = 4.75

             POH   = PKb + log[NH4NO3]/[NH3]

                       = 4.75 + log0.005/0.005

                      = 4.75 +0

                       = 4.75

          PH    = 14-POH

                    = 14-4.75 = 9.25

queen_honey_blossom answered 2 years ago
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