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Using the calculation attached as a guide, what would be the new pH after the addition of another 205 µL of 5M NaOH? Please show work. Thanks.

Using the calculation attached as a guide, what would be the new pH after the addition of another 205 µL of 5M NaOH? Please show work. Thanks.

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
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Using the calculation attached as a guide, what would be the new pH after the addition of another 205 µL of 5M NaOH? Please show work. Thanks.

Original solution: 50 mL of 0.05 M H,PO4 moles of H2PO4 = Concentration × Volume = 0. 05M × 50 × 10-3 L = 0. 0025 moles NaOH Added: 205 µL5M NaOH mmoles of NaOH = Concentration × Volume = 5M × 205 × 10-6 L = 0. 001025 moles ICE Table for the reaction of H,PO, with NaOH H,PO4 OH HPO,2 H20 + + --------- Initial 0.0025 mol 0.001025 mol Change - 0.001025 0.001025 + 0.001025 Equilibrium 0.001475 0.001025 0.001025 mol 205x10-6 + 50x10¬³ L [HPO,²-] 0.001025 mol = 0. 020416 M 0.050205 L [H2 PO,-] 0.001475 mol 0.001475 mol = 0. 029379 M 205x10-6 + 50x10-3 0.050205L pH can be calculated u sin g Hasselbach – Henderson equation pH = pK. + log ( HPO,²-] [H2 PO4 0.020416 pH = 7.2 + log 0.029379 PН —D 7.2 — 0. 1581 pH 7.04
Transcribed Image Text:Original solution: 50 mL of 0.05 M H,PO4 moles of H2PO4 = Concentration × Volume = 0. 05M × 50 × 10-3 L = 0. 0025 moles NaOH Added: 205 µL5M NaOH mmoles of NaOH = Concentration × Volume = 5M × 205 × 10-6 L = 0. 001025 moles ICE Table for the reaction of H,PO, with NaOH H,PO4 OH HPO,2 H20 + + --------- Initial 0.0025 mol 0.001025 mol Change - 0.001025 0.001025 + 0.001025 Equilibrium 0.001475 0.001025 0.001025 mol 205x10-6 + 50x10¬³ L [HPO,²-] 0.001025 mol = 0. 020416 M 0.050205 L [H2 PO,-] 0.001475 mol 0.001475 mol = 0. 029379 M 205x10-6 + 50x10-3 0.050205L pH can be calculated u sin g Hasselbach – Henderson equation pH = pK. + log ( HPO,²-] [H2 PO4 0.020416 pH = 7.2 + log 0.029379 PН —D 7.2 — 0. 1581 pH 7.04
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