If 0.120mol KOH dissolved in 840mL of water what are [OH-] and [H3O+]?

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2007-05-06 09:52:05 · 2 answers · asked by tim_childress 1 in Science & Mathematics ➔ Chemistry

Molarity KOH =0.120 moles / 0.840 L = 0.143 M

[OH-] = 0.143 M

[ H+ ] [ OH-] = 1.00 x 10^-14

[ H+ ] = 1.00 x 10^-14 / 0.143 = 6.99 x 10^-14

2007-05-06 11:36:58 · answer #1 · answered by Anonymous · 0⤊ 0⤋

[OH-] = 0.12/0.84.

Use Kw to calculate [H+].

2007-05-06 17:35:32 · answer #2 · answered by Gervald F 7 · 0⤊ 0⤋