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Draw the titration curve that would result if 35 mL of a 0.2 M CH3NH2 (kb = 4.60 x 10-4) were titrated with 0.20 M HCl(ag). Calculate the pH for each of the following cases (a) Initial pH (b) Volume of HCI(ag): 17.5 mL (c) Volume of HCI(ag): 34.9 mL (d) Volume of HCI(ag): 35.0 mL (e) Volume of HCl(ag): 35.1 mL

Draw the titration curve that would result if 35 mL of a 0.2 M CH3NH2 (kb = 4.60 x 10-4) were titrated with 0.20 M HCl(ag). Calculate the pH for each of the following cases (a) Initial pH (b) Volume of HCI(ag): 17.5 mL (c) Volume of HCI(ag): 34.9 mL (d) Volume of HCI(ag): 35.0 mL (e) Volume of HCl(ag): 35.1 mL

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Draw the titration curve that would result if 35 mL of a 0.2 M CH3NH2 (kb = 4.60 x 10-4) were titrated with 0.20 M HCl(ag). Calculate the pH for each of the following cases (a) Initial pH (b) Volume of HCl(ag): 17.5 mL (c) Volume of HCl(ag): 34.9 mL (d) Volume of HCI(ag): 35.0 mL (e) Volume of HCl(ag): 35.1 mL
Transcribed Image Text:Draw the titration curve that would result if 35 mL of a 0.2 M CH3NH2 (kb = 4.60 x 10-4) were titrated with 0.20 M HCl(ag). Calculate the pH for each of the following cases (a) Initial pH (b) Volume of HCl(ag): 17.5 mL (c) Volume of HCl(ag): 34.9 mL (d) Volume of HCI(ag): 35.0 mL (e) Volume of HCl(ag): 35.1 mL
Expert Solution
Step 1

 

                [CH3NH2] = 0.2 M

Volume of CH3NH2 = 35 mL

                           Kb = 4.60 * 10-4

millimoles of CH3NH2 = 35 mL * 0.2 M

                                                         = 7.0 = milliequivalents

Initial pH

For a weak base, [

[OH-] = Kb* [Base]

    [OH-] = 4.6 * 10-4 * [0.2]            = 92 * 10-6            = 9,59 * 10-3  M

 

 

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