The salt formed by the reaction of the weak base aniline, C6H5NH2, with the strong acid nitric acid is anilinium nitrate, C6H5NH3NO3. What is the hydronium ion concentration of a 0.158 M solution of anilinium nitrate at 25∘C given that the value of Kb for aniline is 4.300×10−10?
use:
Ka = Kw/Kb
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Ka = (1.0*10^-14)/Kb
Ka = (1.0*10^-14)/4.3*10^-10
Ka = 2.326*10^-5
C6H5NH3+ + H2O -----> C6H5NH2 + H+
0.158 0 0
0.158-x x x
Ka = [H+][C6H5NH2]/[C6H5NH3+]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((2.326*10^-5)*0.158) = 1.917*10^-3
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
2.326*10^-5 = x^2/(0.158-x)
3.674*10^-6 - 2.326*10^-5 *x = x^2
x^2 + 2.326*10^-5 *x-3.674*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 2.326*10^-5
c = -3.674*10^-6
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 = 1.47*10^-5
roots are :
x = 1.905*10^-3 and x = -1.929*10^-3
since x can't be negative, the possible value of x is
x = 1.905*10^-3
So, [H+] = x = 1.905*10^-3 M
use:
pH = -log [H+]
= -log (1.905*10^-3)
= 2.72
Answer: 2.72
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