OK except for two things.
#1. The equation you say to use, although it will work ok, is not the H-H equation.
#2. It may bae clear what the answer is but you don't give it.
#3. An extemporaneous remark. I don't know how this answer is to be used. In some venues it would be just right. But for a class assignment I think it is too wordy.
Please judge my answer:
Question:
24 mL of 0.39 mol/L acetic acid is titrated with a standardized 0.33 mol/L KOH solution. Calculate the pH of the solution after 17 mL of the KOH solution has been added. Assume the Ka of acetic acid is 1.8 x 10-5.
Answer:
The equation for this reaction proves itself to be CH3COOH + NaOH ===> CH3COONa + HOH The mol of CH3COOH initially would be equal to M x L =? The mol of NaOH initially would be equal to M x L = ? I realize that it is imperative to look at how much has been reacted, how much CH3COOH has remained unaffected, or unreacted and how much of the salt has been created, or formed. I find that there exists salt and a surplus of CH3COOH, which reveals a buffer solution. Using the Hasselbaclch equation and calculating pH, the answer becomes clear. Using the equation gives Ka=[H^+][A^-]/[HA]. Plugging in the values gives Ka=...............
1 answer