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.
Strychnine, C21H22N2)2(aq) is a weak base but a powerful poison. Calculate the pH of a 0.001 mol/L solution of strychnine. The Kb of strychnine is 1.0 x 10^-6
DrBob222, I am planning on obtaining a pHD in about nine years:P Can you give me some advice on how to be that 1% of the population who has a pHD?
chemistry - DrBob222, Sunday, April 6, 2008 at 12:38am
Start now!
Let's call strychnine SN.
SN + HOH ==> SNH^+ + OH^-
Kb = (SNH^+)(OH^-)/(SN)
(SNH^+) = x
(OH^-) = x
(SN) = 0.001 - x
Plug into Kb and solve for x = (OH^-)
pOH = - log(OH^-)
and pH = 14 - pOH
Kb = (SNH^+)(OH^-)/(SN)
= (x)(x) / 0.001 - x
= x^2 / 0.001 - x
How is the next step supposed to look?
3 answers
(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?
(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.
1 answer