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)

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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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
Transcribed Image Text: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)
Transcribed Image Text:[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)
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