Solution pKa Theoretical Type (Acidic, basic, or neutral) | pH 0.1 M NaCl 7.00 0.1 M HAC 4.74 0.1 M NH4Cl 9.30 0.1 M H2CO3 6.35 0.1 M NaHCO3 10.33 [Products] HAC [H+] [AC] H+ + Ac K₁ = [Reactants] [HAC] [AC] -logK= -log[H+]- log [HAC] [AC] pka = pH -log [HAC] [AC] pH = pKa + log (1) [HAC] Because Ac and HAC are in the same solution, the concentration ratio is a mole ratio so the Henderson-Hassebach equation can be rewritten as n Ac pH = pK+Log n HAC (2)

Solution pKa Theoretical Type (Acidic, basic, or neutral) | pH 0.1 M NaCl 7.00 0.1 M HAC 4.74 0.1 M NH4Cl 9.30 0.1 M H2CO3 6.35 0.1 M NaHCO3 10.33 [Products] HAC [H+] [AC] H+ + Ac K₁ = [Reactants] [HAC] [AC] -logK= -log[H+]- log [HAC] [AC] pka = pH -log [HAC] [AC] pH = pKa + log (1) [HAC] Because Ac and HAC are in the same solution, the concentration ratio is a mole ratio so the Henderson-Hassebach equation can be rewritten as n Ac pH = pK+Log n HAC (2)

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter16: Acids And Bases

Section: Chapter Questions

Problem 108CP: . Consider 0.25 M solutions of the following salts: NaCl. RbOC1, KI, Ba(ClO4),, and NH4NO3. For each...

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