J.R. S. answered 03/22/22
Ph.D. in Biochemistry--University Professor--Chemistry Tutor
pH = -log [H3O+]
2.96 = - log [H3O+]
[H3O+] = 1x10-2.96
[H3O+] = 1.09648x10-3 M (and btw, 5 decimal places is ridiculous)
J.R. S. answered 03/22/22
Ph.D. in Biochemistry--University Professor--Chemistry Tutor
pH = -log [H3O+]
2.96 = - log [H3O+]
[H3O+] = 1x10-2.96
[H3O+] = 1.09648x10-3 M (and btw, 5 decimal places is ridiculous)
Joseph G. answered 03/22/22
Senior Chemistry Student
pH = -log[H3O+]
You can rearrange this equation algebraically to solve for [H3O+]:
-pH = log[H3O+]
10^-pH = 10^(log[H3O+])
**log rule: 10^(logX) = X
So, 10^-pH = [H3O+]
In other words, you can solve for the H3O+ concentration (same as H+) by raising 10 to the power of -pH.
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