The inverse log, or -log, as indicated by p, can be reversed. This converts pH back into [H3O+¹],POH back into [OH-1], and pKa back into Ka.1. Calculate the [H3O+¹] when the pH = 6.666[H3O+¹] = 10-PH = 10-6.666 =2. Calculate the [OH-1] when the pOH = 8.888[OH-1] = 10-POH = 10-8.888 =3. Calculate the Ka when the pka = 7.777K₂ = 10-pka = 10-7.777 =A. 1.67 x 10-8B. 2.158 x 10-7C. 1.29 x 10-9

(1) The pH of the solution = 6.666 [H3O+]=?

The inverse log, or -log, as indicated by p, can be reversed. This converts pH back into [H3O+¹], POH back into [OH-¹], and pKa back into Ka. 1. Calculate the [H3O+¹] when the pH = 6.666 [H3O+1] = 10-PH = 10-6.666 = 2. Calculate the [OH-1] when the pOH = 8.888 [OH-1] = 10-POH = 10-8.888 = 3. Calculate the Ka when the pka = 7.777 Ka = 10-pka = 10-7.777 : = A. 1.67 x 10-8 B. 2.158 x 10-7 C. 1.29 x 10-9

The inverse log, or -log, as indicated by p, can be reversed. This converts pH back into [H3O+¹], POH back into [OH-¹], and pKa back into Ka. 1. Calculate the [H3O+¹] when the pH = 6.666 [H3O+1] = 10-PH = 10-6.666 = 2. Calculate the [OH-1] when the pOH = 8.888 [OH-1] = 10-POH = 10-8.888 = 3. Calculate the Ka when the pka = 7.777 Ka = 10-pka = 10-7.777 : = A. 1.67 x 10-8 B. 2.158 x 10-7 C. 1.29 x 10-9

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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