Question

Calculate the pH during the titration of 10.00 mL of 0.400 M hypochlorous acid with 0.500 M NaOH.

The Ka for HOCl is 3.0 x 10-8 M.

What is the pH when half the acid has been neutralized?
'I got 7.82 by using the HH formula but its wrong, help.

Answer 1

Show your work so I can see what you did.

Answer 2

I think you converted Ka to pKa wrong.

Answer 3

Ill post my work

Answer 4

Actually can you help me find the pH at the equilibrium point, my work:
HOCl OH-
I .004 .004
C -.004 -.004
E 0 0

.01*.400=.004
.008*.5=.004

I don't know what to do next...

Answer 5

I think you're on the wrong track.
I posted the work for that at your other post. I left the z for you to do but it is
HOCl+ NaOH ==> NaOCl + H2O

You have 10 x 0.4 = 4 millimols HClO initially. The equivalence point will come at mL x M = mL x M =
10 x 0.4 = mL x 0.5
mL NaOH = 10 x 0.4/0.5 = 8 mL of the base; therefore, (NaClO) at the equivalence point is 4 mmols/18 mL = ?. Plug that in for z and solve for x = OH^= and convert to pH.