At 25oC, Kw = 1.00 x 10-14 = [H3O+] [OH-] from the equilibrium in water, 2 H2O ⇔ H3O+ + OH-
since [H3O+] = [OH-] , we could "let x = [H3O+] which equals [OH-], so 1.00 x 10-14 = x2; x = 1.00 x 10-7 M
Taisha L.
asked 12/16/22Calculate either [H3O+] or [OH−] for each of the solutions at 25 °C.
At 25oC, Kw = 1.00 x 10-14 = [H3O+] [OH-] from the equilibrium in water, 2 H2O ⇔ H3O+ + OH-
since [H3O+] = [OH-] , we could "let x = [H3O+] which equals [OH-], so 1.00 x 10-14 = x2; x = 1.00 x 10-7 M
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