Determine the pH of a solution of 12.00 mL of 0.2 M acetic acid with 10 mL of a 0.025M NaOH? (Ka = 1.8E–5)

I've calculated my excess moles of acetic to be .00215. Total volume being .022L.

Where do I go from here? Thanks!

Use the Henderson-Hasselbalch equation.

You have (acid). Look to see how much of the sodium acetate was formed and calculate concn of that. Plug in HH equation and pH comes out as the answer.

4.18

I don't think so.

0.00215 is right and 0.022 is right. mols sodium acetate should be 0.00025
If you will post your set up I'll find the error.

To determine the pH of the solution, you need to calculate the concentration of the remaining acetic acid after the neutralization reaction with NaOH.

Since acetic acid (CH3COOH) and sodium hydroxide (NaOH) react in a 1:1 molar ratio, the moles of NaOH used also represents the moles of acetic acid neutralized. In your case, the excess moles of acetic acid are 0.00215.

Now, to calculate the concentration of the remaining acetic acid, you need to subtract the moles of acetic acid neutralized from the initial moles of acetic acid.

Initial moles of acetic acid = (Volume of acetic acid solution in liters) x (Concentration of acetic acid in M)

Initial moles of acetic acid = (0.012 L) x (0.2 M) = 0.0024 moles

Remaining moles of acetic acid = Initial moles of acetic acid - Moles of acetic acid neutralized

Remaining moles of acetic acid = 0.0024 moles - 0.00215 moles = 0.00025 moles

Next, calculate the concentration of acetic acid in the remaining solution by dividing the remaining moles by the total volume of the solution.

Concentration of acetic acid = Remaining moles of acetic acid / Total volume of solution in liters

Concentration of acetic acid = 0.00025 moles / 0.022 L ≈ 0.01136 M

Now that you have the concentration of acetic acid in the solution, you can use the expression for the dissociation of acetic acid, using the equilibrium constant (Ka), to find the pH.

Ka = [H+][CH3COO-] / [CH3COOH]

[H+] is the concentration of hydrogen ions (protons)

To begin, set up the dissociation equation for acetic acid:

CH3COOH ⇌ H+ + CH3COO-

Initially, the concentration of H+ is 0, as no acid has dissociated yet.

At equilibrium, the concentration of H+ will be equal to the concentration of CH3COO-, since the ratio of H+ to CH3COO- is 1:1.

Using the equilibrium constant (Ka), you have:

Ka = [H+][CH3COO-] / [CH3COOH]

Since [H+] = [CH3COO-], you can rewrite the equation as:

Ka = [H+]^2 / [CH3COOH]

Rearranging the equation, you get:

[H+]^2 = Ka x [CH3COOH]

Taking the square root of both sides gives you:

[H+] = sqrt(Ka x [CH3COOH])

Now, substitute the values you have:

[H+] = sqrt((1.8E–5) x (0.01136 M))

[H+] = sqrt(2.0448E–7 M)

[H+] ≈ 1.43E–4 M

Finally, calculate the pH using the concentration of hydrogen ions:

pH = -log[H+]

pH = -log(1.43E–4)

pH ≈ 3.85

Therefore, the pH of the solution is approximately 3.85.