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=...............
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.
Your answer is partially correct in terms of identifying the reaction equation and recognizing the formation of a buffer solution. However, it lacks the necessary calculations and explanation to determine the pH of the solution after the addition of KOH.
To calculate the pH of the solution, we need to consider the stoichiometry of the reaction and the equilibrium expression of the weak acid (acetic acid) dissociation.
Step 1: Calculate the moles of acetic acid initially present.
Moles of acetic acid = concentration (mol/L) x volume (L)
Moles of acetic acid = 0.39 mol/L x 0.024 L (24 mL converted to L)
Step 2: Determine the amount of KOH reacted.
Since KOH is a strong base and reacts in a 1:1 ratio with acetic acid, the moles of KOH reacted will be equal to the moles of acetic acid initially present.
Step 3: Calculate the moles of acetic acid remaining after the reaction.
Moles of acetic acid remaining = Moles of acetic acid initially present - Moles of acetic acid reacted
Step 4: Determine the concentration of acetic acid after the reaction.
Concentration of acetic acid after reaction = Moles of acetic acid remaining / Volume of the solution after the reaction (total volume - volume of KOH added).
Step 5: Use the equilibrium expression of the weak acid dissociation (Ka) to calculate the pH of the solution.
Ka = [H+][A-] / [HA]
[A-] = concentration of acetate ions (concentration of acetic acid after the reaction)
Using the calculated concentration of acetic acid and Ka value given (1.8 x 10^-5), you can now solve for [H+] and then convert it to pH using the equation pH = -log[H+].
Please let me know if you would like further assistance with the calculations.