A chemist titrates 230.0 mL of a 0.5410 M aniline (CH,NH,) solution with 0.4431 M HBr solution at 25 °C.Calculate the pH at equivalence. The p K, of aniline is 4.87.Round your answer to 2 decimal places.Note for advanced students: you may assume the total volume of the solution equals the initial volume plus thevolume of HBr solution added.

Given: volume of aniline = 230 mL molarity of aniline = 0.5410 M molarity of HBr = 0.4431 M pKb of…

A chemist titrates 230.0 mL of a 0.5410 M aniline (CH,NH,) solution with 0.4431 M HBr solution at 25 °C. 6. Calculate the pH at equivalence. The p K, of aniline is 4.87. b. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added.

A chemist titrates 230.0 mL of a 0.5410 M aniline (CH,NH,) solution with 0.4431 M HBr solution at 25 °C. 6. Calculate the pH at equivalence. The p K, of aniline is 4.87. b. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added.

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter15: Additional Aqueous Equilibria

Section: Chapter Questions

Problem 93QRT: When 40.00 mL of a weak monoprotic acid solution is titrated with 0.100-M NaOH, the equivalence...

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