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1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M HNO₂ (Ka = 4.0 x 10-4) by 0.100 M NaOH 100.0 mL of 0.100 M HC3H5O₂ (Ka = 1.3 x 10-5) by 0.100 M NaOH 100.0 mL of 0.100 M HCI by 0.100 M NaOH 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 NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI

1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M HNO₂ (Ka = 4.0 x 10-4) by 0.100 M NaOH 100.0 mL of 0.100 M HC3H5O₂ (Ka = 1.3 x 10-5) by 0.100 M NaOH 100.0 mL of 0.100 M HCI by 0.100 M NaOH 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 NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI

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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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M HNO₂ (Ka = 4.0 x 10-4) by 0.100 M NaOH 100.0 mL of 0.100 M HC3H5O₂ (Ka = 1.3 x 10-5) by 0.100 M NaOH 100.0 mL of 0.100 M HCI by 0.100 M NaOH 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 NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI 2) Rank the following titrations in order of increasing pH at the equivalence point of the titration (1= lowest pH and 5 = highest pH). 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 KOH by 0.100 M HCI 100.0 mL of 0.100 M HC3H50₂ (Ka = 1.3 x 10-5) by 0.100 M NaOH 100.0 mL of 0.100 M HNO₂ (Ka = 4.0 x 10-4) by 0.100 M NaOH 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI
Transcribed Image Text:1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M HNO₂ (Ka = 4.0 x 10-4) by 0.100 M NaOH 100.0 mL of 0.100 M HC3H5O₂ (Ka = 1.3 x 10-5) by 0.100 M NaOH 100.0 mL of 0.100 M HCI by 0.100 M NaOH 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 NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI 2) Rank the following titrations in order of increasing pH at the equivalence point of the titration (1= lowest pH and 5 = highest pH). 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 KOH by 0.100 M HCI 100.0 mL of 0.100 M HC3H50₂ (Ka = 1.3 x 10-5) by 0.100 M NaOH 100.0 mL of 0.100 M HNO₂ (Ka = 4.0 x 10-4) by 0.100 M NaOH 100.0 mL of 0.100 M NH3 (Kb = 1.8 x1 0-5) by 0.100 M HCI
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