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
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![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
K₂ = 10-pka = 10-7.777 =
A. 1.67 x 10-8
B. 2.158 x 10-7
C. 1.29 x 10-9](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90a9e6d8-18c9-4369-a7ca-3ea93a2f0515%2F3d1f13e4-2f3c-4c51-8ae1-184aac6b836e%2F8xod8v6_processed.png&w=3840&q=75)
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