In a buffer pre-lab, we were asked to complete a table for 10 test tubes. In tube 1, you had 0.20 M of acetic acid only and were asked to find the pH. I think the answer is 2.72 (using a ICE table). In tube 2, you have 90 mL of Acetic Acid and 10 mL of Acetate soluion. HOw do you get the concentrations of Acetic Acid and Base concentration?

To find the concentrations of acetic acid and acetate in tube 2, you need to use the concept of dilution and the principles of solution stoichiometry.

Here's how you can calculate the concentrations:

1. Start with the given volumes of acetic acid and acetate solution in tube 2: 90 mL and 10 mL, respectively.

2. Determine the initial moles of acetic acid and acetate by multiplying their respective volumes by their concentrations. Let's assume the initial concentration of acetic acid is x M (which is what we want to find) and the concentration of the acetate solution is y M (which is given).

Initial moles of acetic acid = x M * 90 mL = 90x mmol
Initial moles of acetate = y M * 10 mL = 10y mmol

3. Assume that the acetic acid and acetate completely react to form a buffer solution. This means that the initial moles of acetic acid are equal to the final moles of acetic acid and acetate in the equilibrium state.

Final moles of acetic acid = 90x mmol
Final moles of acetate = 10y mmol

4. Use the ionization equation of acetic acid to form an equilibrium expression:

CH3COOH ⇌ CH3COO- + H+

The equilibrium constant expression for this reaction is:
Ka = ([CH3COO-][H+]) / [CH3COOH]

Since you want to find the concentrations of acetic acid and acetate, let's define the concentrations of acetic acid as [CH3COOH], acetate as [CH3COO-], and the concentration of H+ as [H+].

5. Assuming the initial concentration of acetic acid is x M and it ionizes completely, the equilibrium concentration of acetate will be equal to x M. The equilibrium concentration of H+ can be assumed to be negligible compared to the initial concentration of acetic acid.

Thus, the equilibrium concentrations of the reactants can be written as follows:
[CH3COOH] = x M
[CH3COO-] = x M
[H+] ≈ 0 M

6. Substituting these values into the equilibrium constant expression, we get:
Ka = (x * x) / x

Simplifying the equation, we have:
Ka = x

7. The equilibrium constant for acetic acid is approximately 1.8 * 10^-5 (at 25°C). Therefore, the concentration of acetic acid (x) can be calculated as follows:
x = Ka = 1.8 * 10^-5 M

8. As the concentration of acetic acid is now known, the acetate concentration can be determined as well, since they are equal in this buffer solution:
[CH3COO-] = x = 1.8 * 10^-5 M

Hence, in tube 2, the concentration of acetic acid (CH3COOH) is approximately 1.8 * 10^-5 M, and the concentration of acetate (CH3COO-) is also approximately 1.8 * 10^-5 M.