what is the pH of...?

Question

Calculate the pH at the following points in the titration of 50.00mL of a 0.0100M NaOC6H5 solution with HCl. The Ka of HOC6H5= 1.05x10^-10
(a) Midpoint, (when 1/2 base is neutralized)
(b) Equivalence Point, (after all base is neutralized)

Answer 1

By definition at the midpoint for the base you have pOH=pKb
but pOH=14-pH and pKb=14-pKa, so
14-pH=14-pKa => pH=pKa= -log(1.05*10^-10)= 9.98

Or if you want, for simplicity we use the Henderson-Haselbalch equation instead of an ICE table:
pH=pKa+log([conj.base]/[acid])
but [conj.base]=[acid]=0.5 *Cbase, where Cbase is the initial concetration of the base just after mixing.
So you get pH=pKa+log1 =pKa

For b we have to set up an ICE table for the dissociation of the acid. in order to do that you need to provide the concentration of HCl so that we can calculate the volume needed.So if Va mL HCl was requires you have

Cacid= MbVb/(Vb+Va)= 0.01*50/(50+Va)
The conversion factor from mL to L is simplified here, that's why I didn't convert the units.

.. .. .. .. .. .C6H5OH <=> H+ + C6H5O-
Initial .. .. .. Cacid
Dissoc. .. .. .. x
Produce .. . .. . .. . . . . .. x.. .. .. .. x
At equil. .. Cacid-x .. .. .. x .. .. .. .. x

Ka= x^2/(Cacid-x)

solve for x

and pH=-logx