mt.1 Calculate the pH of a 0.154 M solution of ethylenediamine (H2NCH2CH2NH2). The pKa values for the acidic form ofethylenediamine (H3 NCH, CH, NH) are 6.848 (pKa1) and 9.928 (pKa2).pH =11.51Correct AnswerCalculate the concentration of each form of ethylenediamine in this solution at equilibrium.4.12 x10-3[H2NCH2CH2NH3] ==Incorrect Answer[HNCH2CH2NH3] =7.05 x10-8Correct AnswerMM

Step 1:The concentration of ethylene diamine = 0.154 M pKa1 = 6.848pKa2 = 9.928pKb1 = 14 - pKa2 = 14…

Answered: Calculate the pH of a 0.154 M solution…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter15: Complex Acid/base Systems

Section: Chapter Questions

Problem 15.16QAP

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The base ethylamine (CH3CH2NH2) has a Kb of. A closely related base, ethanolamine(HOCH2CH2NH2), has a Kb of 3.2105. (a) Which of the two bases is stronger? (b) Calculate the pH of a 0.10M solution of the strong base?
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Using the diagrams shown in Problem 10-37, which of the four acids is the weakest acid?
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