Calculate the pH values and draw the titration curve for the titration of 500 mL of 0.010 M acetic acid (pKa 4.76) with 0.010 MKOH.

Calculate the pH of the solution after 490 mL of the titrant have been added.
Calculate the pH of the solution after 500 mL of the titrant have been added.
Calculate the pH of the solution after 510 mL of the titrant have been added.
Calculate the pH of the solution after 750 mL of the titrant have been added.

To calculate the pH values and draw the titration curve, we need to understand the reaction that occurs during the titration of acetic acid (CH3COOH) with potassium hydroxide (KOH), and how it affects the pH of the solution.

In this titration, acetic acid acts as a weak acid, while potassium hydroxide acts as a strong base. The reaction is as follows:
CH3COOH + KOH → CH3COOK + H2O

Initially, we have 500 mL of 0.010 M acetic acid. To find the moles of acetic acid present, we use the formula:
moles = concentration (M) × volume (L)
moles = 0.010 M × 0.500 L
moles = 0.005 mol

At the start of the titration, before any titrant is added, the acetic acid is in its acidic form and has not reacted with the base. Hence, the pH is determined by the ionization of acetic acid.
The equation for the ionization of acetic acid is:
CH3COOH ⇌ CH3COO- + H+

To determine the pH of an acidic solution, we need to calculate the concentration of H+ ions. In the case of acetic acid, since it is a weak acid, we can use the dissociation constant (Ka) and the Henderson-Hasselbalch equation to determine the pH.

The Henderson-Hasselbalch equation is given as:
pH = pKa + log([A-] / [HA]),
where pH is the negative logarithm of the H+ ion concentration, pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

For acetic acid, the pKa is 4.76, and initially, we have only acetic acid present, so [A-] = 0 and [HA] is equal to the initial concentration of acetic acid.

Now, let's calculate the pH values for the given scenarios:

1. After 490 mL of the titrant have been added:
In this case, we add 490 mL of 0.010 M KOH to the solution. This reaction is a one-to-one reaction, so the moles of acetic acid that reacted is also 0.005 mol. Hence, after 490 mL of titrant is added, 0.005 mol of acetic acid has reacted.

To calculate the remaining concentration of acetic acid, we subtract the moles of reacted acetic acid from the initial moles:
Remaining moles of acetic acid = initial moles - reacted moles
Remaining moles of acetic acid = 0.005 mol - 0.005 mol
Remaining moles of acetic acid = 0 mol

Since all the acetic acid has reacted, the solution now contains only the conjugate base (CH3COO-) and water. Therefore, to calculate the pH, we will use the concentration of the conjugate base and the water equilibrium equation:
[H+] = √(Ka × [A-] / [HA])

Since [HA] is now 0, the pH is determined solely by the concentration of the conjugate base, which is the same as the concentration of KOH added:
pH = -log([OH-])
pH = -log(0.010 M)
pH = 2

2. After 500 mL of the titrant have been added:
At this point, all the acetic acid has reacted, and the pH is determined by the concentration of the conjugate base, which is the concentration of KOH added.
pH = -log([OH-])
pH = -log(0.010 M)
pH = 2

3. After 510 mL of the titrant have been added:
Since we have already consumed all the acetic acid, the pH remains the same as in the previous step.
pH = 2

4. After 750 mL of the titrant have been added:
Since we have added a total volume of 750 mL (500 mL acetic acid + 250 mL KOH), the reaction has entered the excess base region. At this point, we have an excess of OH- ions, and the pH is determined by the concentration of OH- ions.
pOH = -log([OH-])
pOH = -log(0.010 M)
pOH = 2

To convert pOH to pH:
pH = 14 - pOH
pH = 14 - 2
pH = 12

Overall, the pH values are as follows:
- After 490 mL of titrant: pH = 2
- After 500 mL of titrant: pH = 2
- After 510 mL of titrant: pH = 2
- After 750 mL of titrant: pH = 12

To draw the titration curve, you can plot the pH on the y-axis and the volume of titrant (in mL or mL added) on the x-axis. The titration curve will show the change in pH as the volume of titrant increases.

This is at least an hour's worth of explaining. Surely you know how to to portions of this. How what you know how to do, explain what you don't understand, and I can help you through with the rest of it.