Posted by Sarah on Sunday, April 6, 2008 at 11:04pm.
Kb = (SNH^+)(OH^-)/(SN)
= (x)(x) / 0.001 - x
= x^2 / 0.001 - x
You haven't included Kb. Look in the problem, I know you posted it, and this last equation becomes
(x)(x)/(.001-x) = Kb. After adding the Kb into the mix, this is a quadratic equation unless the 0.001-x can be simplified to 0.001.
(x)(x)/(.001-x) = Kb
(x^2)/(0.001 - x) = 1.0 x 10^-6
So, in the form of ax^2 +bx + c = 0
this is
x^2 + 0.001x + 1.0 x 10^-6?
See my later post to this question. But what you have written isn't correct.
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