Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Bigweight on December 03, 2019, 04:36:04 AM
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Hello everyone, I have to calculate the concentration of Diethylammine (Kb 8.6 10^-4) only knowing the PH of the solution, which is 8.2
How do I do it? I tried to do
C= [OH]^2/Kb
But the result doesn't make much sense..
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That equation is an approximation that applies only when the degree of ionisation is small.
Suppose the degree of ionisation is x. Then the concentration of free base is C(1-x) and [BH+] = [OH-] = Cx.
Can you work out what x and C are?
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That equation is an approximation that applies only when the degree of ionisation is small.
Suppose the degree of ionisation is x. Then the concentration of free base is C(1-x) and [BH+] = [OH-] = Cx.
Can you work out what x and C are?
I did the approximation for the reason that I can't work out X and C.
All I have is the PH of the solution and of course the Kb of Diethylammine
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Please, show your equation with numbers.
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Please, show your equation with numbers.
Ok then:
PH starts at 8.2 meaning POH is 5.8
This means that [OH] is 10^-5.8
So the concentration of Diethylammine is
[D] = (10^-5.8 )^2/8.6 10^-4
Which is 2.9 10^-17 and it makes no sense
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I did the approximation for the reason that I can't work out X and C.
All I have is the PH of the solution and of course the Kb of Diethylammine
You have (modifying your equation above)
C(1-x) = [OH-]2/Kb
and
Cx = [OH-]
So you have two equations and two unknowns
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This result is wrong - calculation error.
Solve more exact equation Kb=[OH-]2/(c-[OH-])
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C= [OH]^2/Kb
That's just an approximation. Each time you use an approximated equation you should check, if you are allowed to do so - that is, whether the approximation used when deriving holds.
Do you know how this equation is derived? What approximation is being used? Does it hold?