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 mL

To find the pH at different volumes of KOH during the titration, we need to determine the amount of HClO4 remaining and calculate its concentration. Then we can use the concentration of HClO4 to find the pH.

First, let's determine the amount of HClO4 initially present in the solution. We start with 24mL of 0.050M HClO4, so the initial moles of HClO4 can be calculated as:

moles of HClO4 = volume (L) x concentration (mol/L)
moles of HClO4 = 0.024 L x 0.050 mol/L = 0.0012 mol

Now, let's consider the volume of KOH added and calculate the amount of HClO4 that reacts with the KOH.

For Vb = 0.1 mL:
The reaction between HClO4 and KOH is 1:1, so the moles of HClO4 that reacted with KOH can be calculated as:

moles of HClO4 reacted = moles of KOH added

Since we added 0.1 mL of 0.100 M KOH, the moles of KOH added can be calculated as:

moles of KOH added = volume (L) x concentration (mol/L)
moles of KOH added = 0.0001 L x 0.100 mol/L = 0.00001 mol

Therefore, the amount of HClO4 remaining after the reaction is:

moles of HClO4 remaining = initial moles - moles of HClO4 reacted
moles of HClO4 remaining = 0.0012 mol - 0.00001 mol = 0.00119 mol

To calculate the concentration of HClO4 remaining, we need to convert the moles to volume by dividing by the final volume of the solution.

For Vb = 0.1 mL, the final volume would be 24.1 mL:
concentration of HClO4 = moles of HClO4 remaining / final volume (L)
concentration of HClO4 = 0.00119 mol / 0.0241 L = 0.0492 M

Now, we can find the pH using the concentration of HClO4.

pH = -log[H+]

Since HClO4 is a strong acid, it dissociates completely and the concentration of H+ is equal to the concentration of HClO4. Therefore:

pH = -log(0.0492) = 1.31

Using this method, you can repeat the calculations for the other volumes of KOH (5 mL, 10 mL, 12.40 mL, 12.50 mL, 12.60 mL, 13 mL) to find the pH at each point during the titration.