What are the [H3O+] and pH of 0.062 M C2H5NH2?

Question

4.3 x 10^-4 is the Kb of C2H5NH2. I keep trying this problem and getting the wrong answer. I'm out of ideas on what I can do. Can anyone help me to solve this please!!

Answer 1

Set up an ICE table

.. .. .. . .. C2H5NH2 + H2O <=> C2H5NH3(+) +OH-
Initial .. .. .. .. C
React .. .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. .. x
At equil .. .. C-x .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. .. x

Kb= [C2H5NH3(+)][OH-] / [C2H5NH2] = x^2/(C-x)=>
x^2/(0.062-x)= 4.3*10^-4

Let's assume that 0.062 >> x so that 0.062-x=0.062.
Then the equation is simplified to
x^2/0.062= 4.3*10^-4 =>
x =SQRT(0.062*4.3*10^-4) = 0.0052. This is NOT << 0.062, thus we MUST solve the quadratic

x^2/(0.062-x)= 4.3*10^-4 =>
x^2= 4.3*0.062*10^-4-(4.3*10^-4)x =>
x^2 +(4.3*10^-4)x -2.67*10^-5 =0
x1= 0.0049567= 0.00496 M
x2= -0.0053867 <0 rejected

Kw= [H+][OH-] => [H+]=Kw/[OH-] = (10^-14)/0.00496 = 2*10^-12
and pH= -log(2*10^-12) = 11.70

Answer 2

C2h5nh2

Answer 3

In order to find the [OH-] (even though you don't specifically need it, but can use it to find pH, and then [H+]) you can use the formula

6.4 x 10^-4 was the Kb value from my textbook, so I don't know about your value, but I'll use yours in the calculation

[OH-]=sqroot(Kb x initial concentration)
[OH-]=sqroot(4.3 x 10^-4 x .062)
[OH-]=5.2 x 10^-3 M

From [OH-] you can get pOH, and then pH
pOH= -log[OH-]
pOH=-log[5.2 x 10^-3]
pOH= 2.28
pH + pOH= 14
14-2.28=pH
pH=11.72

From the pH, you can get [H+] by using the formula
pH=-log[H+]
11.72=-log[H+]
-11.72=log[H+]
[H+]=inv log(-11.72)
[H+]=1.9 x 10^-12 M

Hope this helps! Sorry if I made any mistakes