In 12.21, how does Kw = [H3O+][OH-] manage to turn into [OH-] = Kw/[H3O+]?
Also, are we expected to memorize that Kw = [H3O+][OH-] = 1.0 x 10^-14?
12.21
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Mike Vinci 2B
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Re: 12.21
Hi Connie,
Kw is the ionic product of water and is a constant. I'm not sure if we're required to memorize it (it could be on the equations and constants sheet), however I'd double check with a moderator on this forum. In regards to your question, the equation of this formula Kw = [H3O+][OH-] can be solved with algebra by dividing both sides by [H3O+] to get [OH-]=Kw/[H3O+]. The remainder of the problem is a matter of plugging in given information which is 1.0x10^-14 for Kw(it's always this) and the molarity of [H30+] which is also given. By dividing Kw/[H30+], you will get the concentration or molarity of OH-.
Kw is the ionic product of water and is a constant. I'm not sure if we're required to memorize it (it could be on the equations and constants sheet), however I'd double check with a moderator on this forum. In regards to your question, the equation of this formula Kw = [H3O+][OH-] can be solved with algebra by dividing both sides by [H3O+] to get [OH-]=Kw/[H3O+]. The remainder of the problem is a matter of plugging in given information which is 1.0x10^-14 for Kw(it's always this) and the molarity of [H30+] which is also given. By dividing Kw/[H30+], you will get the concentration or molarity of OH-.
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