At the equivalence point, all of the base has reacted:
Na(1+) + OCl(1-) + H(1+)
Na(1+) + HOCl
First find out the concentration of your NaOCl.
Vol 1 * Conc 1 = Vol 2 * Conc 2
(cm3=mL)
20 (mL) * Conc 1 = 28.30 (mL) * 0.50 (mol/L) * 1/1000 (L/mL)
Conc 1 = 0.0007075 mol/mL = 0.7075 mol/L
If you had 20 mL NaOCl, you have 20 mL * 0.0007075 mol/mL = 0.01415 mol NaOCl. Everything is a 1:1 ratio, so 0.01415 mol NaOCl = 0.01415 mol HOCl.
You added 28.30 mL of HCl, which makes the total volume 20 mL +28.30 mL = 48.30 mL. This makes the new molarity of the HOCl = 0.01415 mol / 48.30 mL * 1000 mL/L = 0.292 mol/L
Then you write out your equilibrium equation for HOCl in the presence of water:
HOCl +H2O
OCl(1-) + H3O(+)
Ka = [OCl] [H3O] / [HOCl]
(There are charges included with the ions as shown in the equation, but it looked to complicated typed out like that)
The you set up at table: Remember ICE
HOCl + H2O
OCl(1-) + H3O(+)
Initial 0.292 - 0 0
Change -x - +x +x
--------------------------------------...
Equilibrium 0.292-x - x x
Plug these values into your Ka Euqation and set equal to your Ka values:
Ka = [ x]*[ x]/[0.292-x]=3.0 *10^-8
Multiply it out, use the quadratic and all that nonsense. After determining x=9.36*10^-5 , you know that [H3O(1+)]= 9.36*10^-5 mol/L
pH = -log[H3O(1+)], so pH=-log (9.36*10^-5), so pH= 4.03 at equivalence point.
Don't you have a textbook that gives example of how to do this? This is pretty basic, there aren't any major tricks to this question...