Question 9: 30.0 mL of a 0.20 M solution of the weak base methylamine is titrated with 0.10 M nitric acid (HNO3). The Kb of methylamine can be found in Appendix I. Calculate the pH at each volume of base addition as listed below and report your answer to two decimal places The initial pH (after 0.0 mL of nitric acid addition) is 9.94 × The pH after 10.0 mL of nitric acid addition is 9.74 x The pH after 30.0 mL of nitric acid addition is 8.34 x The pH after 60.0 mL of nitric acid addition is 11.34 x The pH after 65.0 mL of nitric acid addition is 2.28 ✓

Principles of Modern Chemistry
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Chapter15: Acid–base Equilibria
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Question 9:
30.0 mL of a 0.20 M solution of the weak base methylamine is titrated with 0.10 M nitric acid
(HNO3). The Kb of methylamine can be found in Appendix I. Calculate the pH at each volume of
base addition as listed below and report your answer to two decimal places
The initial pH (after 0.0 mL of nitric acid addition) is 9.94 ×
The pH after 10.0 mL of nitric acid addition is 9.74 x
The pH after 30.0 mL of nitric acid addition is 8.34 x
The pH after 60.0 mL of nitric acid addition is 11.34 x
The pH after 65.0 mL of nitric acid addition is 2.28 ✓
Transcribed Image Text:Question 9: 30.0 mL of a 0.20 M solution of the weak base methylamine is titrated with 0.10 M nitric acid (HNO3). The Kb of methylamine can be found in Appendix I. Calculate the pH at each volume of base addition as listed below and report your answer to two decimal places The initial pH (after 0.0 mL of nitric acid addition) is 9.94 × The pH after 10.0 mL of nitric acid addition is 9.74 x The pH after 30.0 mL of nitric acid addition is 8.34 x The pH after 60.0 mL of nitric acid addition is 11.34 x The pH after 65.0 mL of nitric acid addition is 2.28 ✓
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