PH and equivalence point chem help!

Question

What is the pH at the equivalence point of a weak base-strong acid titration if 20.00 cm3 of NaOCl requires 28.30 cm3 of 0.50 M HCl? Ka = 3.0 × 10-8 for HOCl.

a. 4.03 b. 0.30 c. 3.18 d. 3.76

Answer 1

OCl- + H+ >> HClO

Moles HCl = 0.02830 L x 0.50 =0.0142 = moles OCl-

[HClO] = 0.0142 mol / 0.04830 L =  0.294 M

HClO <----> H+ + OCl-

3.0 x 10^-8 = x^2 / 0.294-x

x = [H+] = 9.39 x 10^-5 M

pH = 4.03

Answer 2

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...