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.

Show your work so I can see what you did.

I think you converted Ka to pKa wrong.

Ill post my work

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...

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.

To calculate the pH when half the acid has been neutralized during the titration, we need to make some assumptions and use a balanced chemical equation for the reaction.

First, let's write the balanced chemical equation for the reaction between hypochlorous acid (HOCl) and sodium hydroxide (NaOH):

HOCl + NaOH --> NaOCl + H2O

From the balanced equation, we can see that one mole of HOCl reacts with one mole of NaOH to form one mole of NaOCl and one mole of water.

Given that we have 10.00 mL of 0.400 M HOCl, we can calculate the initial number of moles of HOCl:

moles of HOCl = concentration (M) x volume (L)
= 0.400 M x 0.01000 L
= 0.00400 moles

Since one mole of HOCl reacts with one mole of NaOH, at the halfway point of the reaction, 0.00200 moles of HOCl will react with 0.00200 moles of NaOH. The remaining 0.00200 moles of NaOH will be in excess.

To find the concentration of HOCl and NaOCl at the halfway point, we need to subtract the moles of HOCl that reacted from the initial moles of HOCl:

moles of HOCl remaining = initial moles of HOCl - moles of HOCl reacted
= 0.00400 moles - 0.00200 moles
= 0.00200 moles

Using the volume of the solution (10.00 mL) and the remaining number of moles of HOCl (0.00200 moles), let's calculate the concentration of HOCl at the halfway point:

HOCl concentration = moles of HOCl remaining / volume of solution (L)
= 0.00200 moles / 0.01000 L
= 0.200 M

Now, we need to use the equilibrium expression for the dissociation of HOCl:

Ka = [H+][OCl-] / [HOCl]

Given that the Ka for HOCl is 3.0 x 10^(-8) M, and assuming x is the concentration of H+ ions, the concentrations of H+ and OCl- can be approximated as x during the dissociation of HOCl.

Therefore, the equilibrium expression becomes:

3.0 x 10^(-8) M = x^2 / 0.200 M

Since the dissociation is not complete, we can assume that x is much smaller than 0.200 M, which simplifies the equation to:

3.0 x 10^(-8) M = x^2 / 0.200 M

Rearranging the equation:

x^2 = (3.0 x 10^(-8) M) * (0.200 M)
x^2 = 6.0 x 10^(-9)
x = sqrt(6.0 x 10^(-9))
x ≈ 2.45 x 10^(-4) M

Therefore, the concentration of H+ ions at the halfway point is approximately 2.45 x 10^(-4) M.

Finally, we can calculate the pH using the equation:

pH = -log[H+]
pH = -log(2.45 x 10^(-4))
pH ≈ 3.61

So, the pH when half the acid has been neutralized is approximately 3.61 (rounding to two decimal places).

Note: The Henderson-Hasselbalch equation (HH formula) is typically used for weak acid and conjugate base solutions that are not undergoing a titration. In this case, since we are dealing with a reaction and using the equilibrium expression, we need to follow the steps outlined above.