Calculate [H3O+] and [OH−] for each of the following solutions.
A) pH= 8.60
B) pH= 2.89
A) pH= 8.60
-log[H^+] = 8.6
log[H^+] = -8.6
[H^+] = 10^-8.6 = 2.5*10^-9 M
Kw = [H^+][OH^-]
[OH^-] = Kw/[H^+]
= 10^-14/2.5*10^-9 = 4*10^-6 M
B) pH= 2.89
-log[H^+] = 2.89
log[H^+] = -2.89
[H^+] = 10^-2.89 = 0.0013M
Kw = [H^+][OH^-]
[OH^-] = Kw/[H^+]
= 10^-14/0.0013 = 7.7*10^-12M
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