resultant ph?

Question

If 0.780g of phenol is dissolved in enough water to make 500ml of solution, what is the resultant pH?
C6H5OH (aq) <--> C6H50 (aq) + H............Ka=1.3x10^-10

So here the solution..i understand the first half.

0.780g x 1mol/94.11g = 8.29 x 10^-3
8.29x10^-3 / 0.500L = 0.0166M

He goes on to write this equation..

(x^2) / (0.0166 - x) = 1.3 x 10^-10

I understand that the x^2 is b/c of C6H50 & H

Why is C6H5OH...(0.0166 - x) and not (0.0166 + x)?
is it b/c the product of the reaction always takes up something?

There's a diff reaction where: N2 + O2 <--> 2NO
ea. gas is 0.300 atm.

Since the 2NO is a product...it's 0.300 - 2x? Would it still be that if the equation was reversed to 2NO <--> N2 + O2 or would it instead be 0.300 + 2x since it is on the left side?

Answer 1

First answer :

phenol is a weak acid ; let x = moles/lL phenol that dissociate to estabilish equilibrium.
This will give us x moles/L H+ and x moles/l C6H5O- and will leave 0.0166 -x moles / L phenol

Ka = 1.3 x 10^-10 = (x^2) / 0.0166-x

x = 1.47 x 10^-6

pH = 5.83

Answer 2

It would change.