PH and Ka need help?

Question

Calculate the pH of a 3.3 10-5 M phenol (Ka = 1.6 10-10).

Answer 1

Let's write phenol for the shake of simplicity as HA (where A is C6H5O)

.. .. .. .. .. .. .. HA <=> H+ +A-
initial .. .. .. .. C
Dissociate.. .. x
Produce .. .. .. .. .. .. ..x .. .. x
At equil ,, ,, ,, C-x .. .. x .. .. x

Ka = [H+][A-]/ [HA] =x^2/(C-x) =>

Let's assume that C>>x and thus C-x=C. Then the equation is simplified to
Ka=x^2/C => x= squareroot (Ka*C) =>
x = SQRT( (1.6*10^-10)*(3.3*10^-5) ) = 7.27 *10^-8

if the value of x that we calculated wasn't << C then we would have to solve the quadratic :x^2+Kax-KaC=0

The problem now is that the H+ coming from phenol are in the order of magnitude of that coming from water. So we have to take into account both equillibrium reactions.
One way to solve this is:

.. .. .. .. .. .. .. HA <=> H+ +A-.
initial .. .. .. .. C
Dissociate.. .. x
Produce .. .. .. .. .. .. ..x .. .. x
At equil ,, ,, ,, C-x .. .x+y .. .. x

and
.. .. .. .. .. .. H2O<=> H+ +OH-
Dissociate .. y
Produce .. .. .. .. .. .. .. y .. .. y
At Equil. .. .. .. .. .. .. .. x+y .. y

Note that [H2O] is never written in the equilirium constant expressions in such cases since it is considered constant thus I didn't put anything in the ICE table.

So now you have a system of two equations:
Ka= [H+][A-]/ [HA] => x(x+y)/(3.3*10^-5-x) =1.6*10^-10
Kw =[H+][OH-] => (x+y)y =10^-14

Solving that we get 3 sets of solutions out of which only one is acceptable (the others have at least one negative value)
x=4.27*10^-8
y=8.09*10^-8

so pH= -log[H+]= -log(x+y) = -log((4.27+8.09)*10^-8) = 6.91

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Answer 2

take the square root of 3.3 10-5 times 1.6 10-10
you get 7.27 10-8
now take -log of that # and u get 7.14 that is a pH
hope this helps

Answer 3

the pH is very close to 7(between 6.5 and 6.99)