Question

A 27.7 mL sample of 0.314 M ethylamine, C2H5NH2, is titrated with 0.325 M nitric acid. The pH before the addition of any nitric acid is .

Use the Tables link in the References for any equilibrium constants that are required.

kb: 4.3x10^-4

Answer 1

C2H5NH2 dissociates as:

C2H5NH2 +H2O     ----->     C2H5NH3+   +   OH-
0.314                   0         0
0.314-x                 x         x


Kb = [C2H5NH3+][OH-]/[C2H5NH2]
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 ((4.3*10^-4)*0.314) = 1.162*10^-2

since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Kb = x*x/(c-x)
4.3*10^-4 = x^2/(0.314-x)
1.35*10^-4 - 4.3*10^-4 *x = x^2
x^2 + 4.3*10^-4 *x-1.35*10^-4 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 4.3*10^-4
c = -1.35*10^-4

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 = 5.403*10^-4

roots are :
x = 1.141*10^-2 and x = -1.184*10^-2

since x can't be negative, the possible value of x is
x = 1.141*10^-2

So, [OH-] = x = 1.141*10^-2 M


use:
pOH = -log [OH-]
= -log (1.141*10^-2)
= 1.9428


use:
PH = 14 - pOH
= 14 - 1.9428
= 12.0572
Answer: 12.06