Given:
[OH-] = 5.7*10^-12 M
use:
[H+] = Kw/[OH-]
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
[H+] = (1.0*10^-14)/[OH-]
[H+] = (1.0*10^-14)/5.7*10^-12
[H+] = 1.8*10^-3 M
Answer: 1.8*10^-3 M
Revie Calculate [H3 Ogiven OH solution. in each aqueous Part A (OH=5.7x10-12 M Express your answer...
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