1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 =highest pH).2 200.0 mL of 0.100 M (C₂H5)2NH (Kb = 1.3 x 10-³) by 0.100 M HCI4 100.0 mL of 0.100 M HCI by 0.100 M NaOH3 100.0 mL of 0.100 M KOH by 0.100 M HCI15100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI200.0 mL of 0.100 M HC₂H30₂ (Ka = 1.8 x 10-5) by 0.100 M NaOHConsider the major species present at the halfway point. From this, you should be able to deduce the order of the pHwith only minimal calculationsSubmit Answer You have entered that answer before Incorrect. Tries 5/45 Previous Tries2) Rank the following titrations in order of increasing pH at the equivalence point of the titration (1= lowest pH and5 highest pH).2 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI3200.0 mL of 0.100 M HC₂H30₂ (Ka = 1.8 x 10-5) by 0.100 M NaOH5 100.0 mL of 0.100 M KOH by 0.100 M HCI41100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH200.0 mL of 0.100 M (C₂H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCI

Answered: Image /qna-images/answer/0599f511-3687-4830-998f-108071953dad.jpg

) Rank the following titrations in order of increasing pH at the halfway point to equ ighest pH). 2 200.0 mL of 0.100 M (C₂H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCI 100.0 mL of 0.100 M HCI by 0.100 M NaOH 3 100.0 mL of 0.100 M KOH by 0.100 M HCI 4 1 5 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI 200.0 mL of 0.100 M HC₂H302 (Ka = 1.8 x 10-5) by 0.100 M NaOH Consider the major species present at the halfway point. From this, you should be ab vith only minimal calculations Submit Answer You have entered that answer before Incorrect. Tries 5/45 Previou ) Rank the following titrations in order of increasing pH at the equivalence point of = highest pH). 2 € 100.0 mL of 0.100 M NH3 (Kb: = 1.8 x1 0-5) by 0.100 M HCI 3 € 200.0 mL of 0.100 M HC₂H302 (Ka = 1.8 x 10-5) by 0.100 M NaOH 5 100.0 mL of 0.100 M KOH by 0.100 M HCI 4 € 100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH 1 € 200.0 mL of 0.100 M (C₂H5)2NH (K₁ = 1.3 x 10-³) by 0.100 M HCI

) Rank the following titrations in order of increasing pH at the halfway point to equ ighest pH). 2 200.0 mL of 0.100 M (C₂H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCI 100.0 mL of 0.100 M HCI by 0.100 M NaOH 3 100.0 mL of 0.100 M KOH by 0.100 M HCI 4 1 5 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI 200.0 mL of 0.100 M HC₂H302 (Ka = 1.8 x 10-5) by 0.100 M NaOH Consider the major species present at the halfway point. From this, you should be ab vith only minimal calculations Submit Answer You have entered that answer before Incorrect. Tries 5/45 Previou ) Rank the following titrations in order of increasing pH at the equivalence point of = highest pH). 2 € 100.0 mL of 0.100 M NH3 (Kb: = 1.8 x1 0-5) by 0.100 M HCI 3 € 200.0 mL of 0.100 M HC₂H302 (Ka = 1.8 x 10-5) by 0.100 M NaOH 5 100.0 mL of 0.100 M KOH by 0.100 M HCI 4 € 100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH 1 € 200.0 mL of 0.100 M (C₂H5)2NH (K₁ = 1.3 x 10-³) by 0.100 M HCI

Chemistry

10th Edition

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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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