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The 3-step ionization of phosphoric acid: H3PO4 = H + H2PO4 H2PO4 = H + HPO4² HPO4? = H* + PO4 Kal = 7.11 × 10-3 Ka2 Ka3 = 7.11 × 10-13 2- 6.32 x 10-8 %3D Available Aqueous Solutions: 1.00 M H;PO4 1.00 М КН-РО4 1.00 M K2HPO4 1.00 M HCI 1.00 M NaOH Available Solids: KH PO4 K2HPO4 K3PO4

The 3-step ionization of phosphoric acid: H3PO4 = H + H2PO4 H2PO4 = H + HPO4² HPO4? = H* + PO4 Kal = 7.11 × 10-3 Ka2 Ka3 = 7.11 × 10-13 2- 6.32 x 10-8 %3D Available Aqueous Solutions: 1.00 M H;PO4 1.00 М КН-РО4 1.00 M K2HPO4 1.00 M HCI 1.00 M NaOH Available Solids: KH PO4 K2HPO4 K3PO4

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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Boxes 1-4

buffer B: 100.0 mL at pH = 7.000, [conj.acid] =0.500M

 

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
5b) Part 1: Fill in the values below from your calculations for 10.0 mL 1.00 M HCI added to 50.0 mL of Buffer B. Number of moles converted = mol conj. acid/base Total volume of mixture = L solution Moles and Volume: Include numbers only, with correct significant figures and leading zeros 5b) Part 2: Fill in the values in the equation below for 10.0 mL 1.00 M HCI added to 50.0 mL of Buffer B. Change in pH = pH final - pH initial = Answer Change in pH = - 7,000 = pH values: Include numbers only, with correct significant figures, no space between sign and number, include leading zeros
Transcribed Image Text:5b) Part 1: Fill in the values below from your calculations for 10.0 mL 1.00 M HCI added to 50.0 mL of Buffer B. Number of moles converted = mol conj. acid/base Total volume of mixture = L solution Moles and Volume: Include numbers only, with correct significant figures and leading zeros 5b) Part 2: Fill in the values in the equation below for 10.0 mL 1.00 M HCI added to 50.0 mL of Buffer B. Change in pH = pH final - pH initial = Answer Change in pH = - 7,000 = pH values: Include numbers only, with correct significant figures, no space between sign and number, include leading zeros
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