How many milliliters of 0.202 M KOH should be added to 500.0 mL of 0.0233 M

tartaric acid (C4H6O6; FW 150.087) to adjust the pH to 2.75? Ka1 = 1.0 x 10-13 and Ka2 = 4.6 x10-5.

I ended up with KOH mL being about 1.5mL. The equiv. point would be around 57mL. I see the answer is to be around 20mL. I have no idea how this comes about.

To solve this problem, we need to apply the concept of acid-base titration and the equilibrium equations of the given acids.

1. Write the balanced equation for the reaction between tartaric acid (C4H6O6) and KOH:
C4H6O6 + y KOH -> x K+ + z H2O + C4H4O6−

2. Identify the two equilibria involving the tartaric acid (C4H6O6):
Ka1 = [C4H4O6−][H+]/[C4H6O6]
Ka2 = [C4H4O6−][H+]/[C4H6O6−]

3. Use the given Ka values:
Ka1 = 1.0 x 10^-13
Ka2 = 4.6 x 10^-5

4. Set up an ICE (Initial, Change, Equilibrium) table for the reaction involving tartaric acid:
Initial: [C4H6O6] = 0.0233 M, [H+] = 10^(-pH) = 10^(-2.75) M, [C4H4O6−] = 0 M
Change: -x, -x, +x
Equilibrium: [C4H6O6 - x], [H+ - x], [C4H4O6− + x]

5. Since the molar ratio between C4H6O6 and H+ is 1:1, we can assume that the change in concentration of H+ (x) will be approximately equal to the change in concentration of C4H6O6 (x).

6. Substitute the values in the Ka1 expression and solve for x:
Ka1 = (x^2) / (0.0233 - x)
Since Ka1 is very small compared to 0.0233, we can assume 0.0233 - x ≈ 0.0233:
1.0 x 10^-13 = (x^2) / (0.0233)
x ≈ √(1.0 x 10^-13 * 0.0233)

7. Calculate the concentration of KOH using the balanced equation and stoichiometry:
C4H6O6 + y KOH -> x K+ + z H2O + C4H4O6−
The molar ratio between C4H6O6 and KOH is 1:1, so the concentration of KOH = x = √(1.0 x 10^-13 * 0.0233).

Now, let's calculate the KOH volume.

8. Calculate the moles of KOH:
moles of KOH = concentration of KOH * volume of KOH (in liters)
volume of KOH = moles of KOH / concentration of KOH
volume of KOH = moles of KOH / 0.202 M

9. Convert the volume of KOH to milliliters:
volume of KOH (in mL) = volume of KOH (in L) * 1000

Using these steps, you can calculate the volume of KOH needed to adjust the pH to 2.75.