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5) ApH resulting from adding 10.0 mL 1.00 M HCI a) What would the ApH be of 50.0 mL of pure water if 10.0 mL 1.00 M HC1 is added?

5) ApH resulting from adding 10.0 mL 1.00 M HCI a) What would the ApH be of 50.0 mL of pure water if 10.0 mL 1.00 M HC1 is added?

Chemistry
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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5) ApH resulting from adding 10.0 mL 1.00 M HCI a) What would the ApH be of 50.0 mL of pure water if 10.0 mL 1.00 M HCl is added? Answer: Buffer Problem Set – Page 6 of 7 b) What would the ApH be of 50.0 mL of Buffer B if 10.0 mL 1.00 M HCl is added?
Transcribed Image Text:5) ApH resulting from adding 10.0 mL 1.00 M HCI a) What would the ApH be of 50.0 mL of pure water if 10.0 mL 1.00 M HCl is added? Answer: Buffer Problem Set – Page 6 of 7 b) What would the ApH be of 50.0 mL of Buffer B if 10.0 mL 1.00 M HCl is added?
The 3-step ionization of phosphoric acid: Kal = 7.11 × 10-3 H;PO4 = H* + H2PO4 H-PO4 = H + HPO4² HPO4² = H + PO4 2- 6.32 x 10-8 Ka2 = Ka3 = 7.11 × 10-13 2- 3- Available Aqueous Solutions: 1.00 M HC1 1.00 M H3PO4 1.00 M KH2PO4 1.00 M K2HPO4 1.00 M NaOH Available Solids: KH2PO4 K2HPO4 K3PO4
Transcribed Image Text:The 3-step ionization of phosphoric acid: Kal = 7.11 × 10-3 H;PO4 = H* + H2PO4 H-PO4 = H + HPO4² HPO4² = H + PO4 2- 6.32 x 10-8 Ka2 = Ka3 = 7.11 × 10-13 2- 3- Available Aqueous Solutions: 1.00 M HC1 1.00 M H3PO4 1.00 M KH2PO4 1.00 M K2HPO4 1.00 M NaOH Available Solids: KH2PO4 K2HPO4 K3PO4
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10 ml of 1.00 M HCl is added to 50ml of water.

 

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