Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.200 M HClO(aq) with 0.200 M KOH(aq). The ionization constant for HClO can be found here.

A. Before any addition of KOH
B. After addition of 25.0 mL of KOH
C. After 30.0 mL of KOH
D. After 50.0 mL of KOH
E. After 60.0 mL of KOH

Can you explain what you don't understand here?

The first thing you do is calculate where the equivalence point is. A is just calculation of pH of a weak acid. Then you have all of the calculations leading up to the eq. pt, all of the calculations after the eq. pt. and finally the eq. pt pH.

To calculate the pH for each case in the titration, we need to determine the number of moles of acid (HClO) and base (KOH) present at each point in the titration and then use the appropriate equations to find the pH.

Here's how you can calculate the pH for each case:

A. Before any addition of KOH:
Since no KOH has been added yet, the solution only contains HClO. We can assume it completely dissociates, so the concentration of HClO is equal to its initial molar concentration. This means the number of moles of HClO is:

moles of HClO = concentration of HClO × volume of HClO

moles of HClO = 0.200 M × 50.0 mL (convert mL to L by dividing by 1000)

Next, use the equation for pH:

pH = -log[H+]

Since HClO is a strong acid, it completely dissociates into H+ and ClO-. Therefore, the concentration of H+ is equal to the initial concentration of HClO. Take the negative logarithm of the H+ concentration to find the pH.

B. After addition of 25.0 mL of KOH:
To determine the pH after adding 25.0 mL of KOH, we need to calculate the moles of KOH that reacted with HClO. From the balanced chemical equation, we know that the stoichiometric ratio between HClO and KOH is 1:1.

moles of KOH = concentration of KOH × volume of KOH

moles of KOH = 0.200 M × 25.0 mL (convert mL to L by dividing by 1000)

Since 25.0 mL is half of the initial volume of HClO, half of the moles of HClO have reacted with KOH. Subtract half of the initial moles of HClO from the moles of HClO to find the remaining moles.

moles of HClO remaining = moles of HClO - (moles of KOH / 2)

Now, calculate the concentration of HClO remaining by dividing the moles of HClO remaining by the total new volume (50.0 mL + 25.0 mL).

concentration of remaining HClO = (moles of HClO remaining) / (total new volume)

Finally, find the pH using the concentration of remaining HClO as we did in part A.

C. After 30.0 mL of KOH:
Using the same method as part B, calculate the moles of KOH that reacted, find the remaining moles of HClO, calculate the concentration of remaining HClO, and then find the pH.

D. After 50.0 mL of KOH:
After adding 50.0 mL of KOH, the reaction is complete and all the HClO has reacted. Therefore, the concentration of HClO is zero, and the pH is 7 (neutral).

E. After 60.0 mL of KOH:
Since 60.0 mL is more than the total volume of the original HClO solution, there is an excess of KOH. Therefore, the pH will be determined by the excess KOH. Calculate the moles of KOH that reacted, and then use the remaining moles to calculate the concentration of OH-. Finally, use the pOH equation to find the pH. Remember to subtract the pOH from 14 to obtain the pH.

Note: It's important to consult the ionization constant for HClO to determine if it behaves as a strong or weak acid and modify the calculations accordingly.

By following these steps, you can calculate the pH at each stage of the titration.