(CH3)2NH dissociates as:
(CH3)2NH +H2O -----> (CH3)2NH2+ + OH-
1.98*10^-2 0 0
1.98*10^-2-x x x
Kb = [(CH3)2NH2+][OH-]/[(CH3)2NH]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((5.9*10^-4)*1.98*10^-2) = 3.418*10^-3
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Kb = x*x/(c-x)
5.9*10^-4 = x^2/(1.98*10^-2-x)
1.168*10^-5 - 5.9*10^-4 *x = x^2
x^2 + 5.9*10^-4 *x-1.168*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 5.9*10^-4
c = -1.168*10^-5
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 4.708*10^-5
roots are :
x = 3.136*10^-3 and x = -3.726*10^-3
since x can't be negative, the possible value of x is
x = 3.136*10^-3
So, [OH-] = x = 3.136*10^-3 M
use:
pOH = -log [OH-]
= -log (3.136*10^-3)
= 2.50
use:
PH = 14 - pOH
= 14 - 2.50
= 11.50
[(CH3)2NH] = 0.0198 - x
= 0.0198 - 3.136*10^-3 M
= 0.0167 M
[(CH3)2NH] = x
= 3.136*10^-3 M
Answer:
PH = 11.50
[(CH3)2NH] = 0.0167 M
[(CH3)2NH] = 3.14*10^-3 M
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