pH problem?

Question

A 0.20 M sodium chlorobenzoate (NaC7H4ClO2) solution has a pH of 8.65. Calculate the pH of a 0.12 M chlorobenzoic acid (HC7H4ClO2) solution.

This might sound kind of stupid, but does anyone know how I would turn 0.00005988 = (y squared/(0.12 - y)) into Ax2 + Bx + C form?

Answer 1

chlorobenzoic acid is a weak monoprotic acid. So it is of the general form HA which we can use to avoid typing the long formula.
The salt will hydrolyze according to

.. .. .. .. .. .. .. A(-) + H2O <=> HA + OH-
initial .. .. .. .. 0.2
React .. .. .. ..x
Produce .. .. .. .. .. .. .. .. .. .. .. . x .. .. x
At Equil. .. 0.2-x .. .. .. .. .. .. .. .. x .. .. x

Kb= [HA][OH-]/ [A(-)]= x^2/(0.2-x)
but [H+]=10^-pH and [OH-]=Kw/[H+] = 10^-14 /10^-pH = 10^-14/10^-8.65= 10^-5.35 =4.47*10^-6 =x
substitute and find Kb
Then Ka=Kw/Kb, so you can find Ka
Finally
.. .. .. .. .. .. .. HA <=> H+ +A-
initial .. .. .. .. 0.12
Dissociate .. .. y
Produce .. .. .. .. .. .. .. y .. .. y
At equil .. .. ..0.12-y .. .y .. .. y

Ka= [H+][A-]/[HA] =y^2/(0.12-y)

solve for y (either the quadratic or do the assumption that y<<0.12 and 0.12-y=0.12.In the latter case the equation is simplified to Ka*0.12=y^2 => y=squareroot (0.12*Ka). If you find that y>0.001 then the assumption is not valid and you have to solve the quadratic to find y)

Finally pH=-logy

[Edit] Ka=y^2/(0.12-y) => Ka(0.12-y) =y^2 =>
Ka*0.12-Kay= y^2 =>
y^2+Ka y-0.12Ka=0