Consider the titrations of 24mL of 0.050M HClO4 with 0.100 M KOH. Find the pH at the following volumes of KOH; Vb: 0.1, 5, 10, 12.40, 12.50, 12.60, 13

To find the pH at different volumes of KOH during the titration of HClO4 with KOH, we need to understand the reaction that occurs during the titration.

The reaction between HClO4 (a strong acid) and KOH (a strong base) is a neutralization reaction that produces water and a salt. In this case, the salt produced is potassium perchlorate (KClO4).

The balanced chemical equation for the reaction is:
HClO4 + KOH -> KClO4 + H2O

Since HClO4 is a strong acid, it completely dissociates in water to form H+ ions. KOH also fully dissociates in water to produce OH- ions. During the titration, the OH- ions from KOH will react with the H+ ions from HClO4 to form water.

To calculate the pH at different volumes of KOH, we need to determine how much of the strong acid and strong base react with each other. This will depend on their stoichiometry, which is determined by the balanced chemical equation.

Given:
- Volume of HClO4: 24 mL
- Concentration of HClO4: 0.050 M
- Concentration of KOH: 0.100 M

Let's calculate the amount of HClO4 that reacts with KOH at each volume of KOH.

Volume of HClO4 (Va): 24 mL
Concentration of HClO4: 0.050 M

Amount of HClO4 (na) = Volume (Va) * Concentration (Ca)
na = 24 mL * 0.050 M
na = 1.2 mmol

Now, let's consider the volumes of KOH:

1. For Vb = 0.1 mL:
The volume of KOH is very small, so the reaction is not complete. We can neglect the OH- ions from KOH, and the pH will be determined by the H+ ions from HClO4. To calculate the pH, we need to determine the concentration of H+ ions remaining after reacting with a small amount of KOH. Since the reaction is not complete, the remaining H+ ions will determine the pH. We can use the Henderson-Hasselbalch equation to calculate the pH.

Henderson-Hasselbalch equation:
pH = -log10[H+]
[H+] = Concentration of H+ ions

In this case, the concentration of H+ ions will be na - nb, where nb is the amount of KOH that reacted with HClO4.

2. For Vb = 5 mL:
To calculate the pH, we need to determine the amount of KOH that reacted with HClO4. Since the reaction is not complete, we can assume that the reaction goes to completion with respect to the OH- ions from KOH but not with the H+ ions from HClO4. To calculate nb, we can use the stoichiometry of the balanced equation to determine the ratio of HClO4 to KOH.

3. For Vb = 10 mL:
We can repeat the same procedure as above, calculating the amount of KOH that reacted with HClO4.

4. For Vb = 12.40 mL:
Again, repeat the same procedure as above.

5. For Vb = 12.50 mL:
Continue the same procedure.

6. For Vb = 12.60 mL:
Carry out the same procedure as before.

7. For Vb = 13 mL and beyond:
At this point, we have reached the equivalence point where the stoichiometric amounts of HClO4 and KOH have reacted exactly. The pH will be determined by the resulting salt, which in this case is potassium perchlorate (KClO4). Since KClO4 is a strong salt, it will fully dissociate into K+ and ClO4- ions. The pH at this point will depend on the hydrolysis of ClO4- ions.