1)
CH3NH2 dissociates as:
CH3NH2 +H2O -----> CH3NH3+ + OH-
2.93*10^-2 0 0
2.93*10^-2-x x x
Kb = [CH3NH3+][OH-]/[CH3NH2]
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 ((4.47*10^-4)*2.93*10^-2) = 3.619*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)
4.47*10^-4 = x^2/(2.93*10^-2-x)
1.31*10^-5 - 4.47*10^-4 *x = x^2
x^2 + 4.47*10^-4 *x-1.31*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 4.47*10^-4
c = -1.31*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 = 5.259*10^-5
roots are :
x = 3.402*10^-3 and x = -3.849*10^-3
since x can't be negative, the possible value of x is
x = 3.402*10^-3
So, [OH-] = x = 3.402*10^-3 M
use:
pOH = -log [OH-]
= -log (3.402*10^-3)
= 2.4682
use:
PH = 14 - pOH
= 14 - 2.4682
= 11.53
2)
[CH3NH2] = 2.93*10^-2 - x
= 2.93*10^-2 - 3.402*10^-3
= 0.0259 M
Answer: 0.0259 M
3)
[CH3NH3+] = x
= 3.402*10^-3
Answer: 3.40*10^-3 M
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