Lab Report #6-2-3: 0.1 M NagPO4 solution, using the given pH data, write expression for equilibrium constant (Ka or Kb): O Ka = [HPO4²-] [H*] / [PO43•] Ka = [HPO42-] [OH"] / [PO4³-] %3D Kb = [HPO42] [OH"] / [PO43] %3D Kb = [HPO42-] [H*] / [PO43-] %3D

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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pH values
3.5
Solutions
H20 (unboiled)
H20 (boiled)
7.0
NaCl
7.0
NaC2H302
9.1
NHẠC1
4.5
NaHCO3
9.5
Na3PO4
11.9
Na2CO3
11.0
Transcribed Image Text:pH values 3.5 Solutions H20 (unboiled) H20 (boiled) 7.0 NaCl 7.0 NaC2H302 9.1 NHẠC1 4.5 NaHCO3 9.5 Na3PO4 11.9 Na2CO3 11.0
Lab Report #6-2-3:
0.1 M Na3PO4 solution, using the given pH data, write expression for equilibrium
constant (Ka or Kb):
O Ka = [HPO42] [H*] / [PO43°]
Ka = [HPO42] [OH"] / [PO43-]
Kb = [HPO42] [OH"] / [PO4³]
Kb = [HPO42] [H*] / [PO43]
Transcribed Image Text:Lab Report #6-2-3: 0.1 M Na3PO4 solution, using the given pH data, write expression for equilibrium constant (Ka or Kb): O Ka = [HPO42] [H*] / [PO43°] Ka = [HPO42] [OH"] / [PO43-] Kb = [HPO42] [OH"] / [PO4³] Kb = [HPO42] [H*] / [PO43]
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