Question

A student carries out a titration of 15.0 mL of 0.100M of acetic acid, CH3COOH, (Ka = 1.8 x 10-5) with 0.100M NaOH.
The reaction can be represented by the following generic equation below:
HA + OH- 🡪 A- + H2O
How would you calculate the pH of a solution after addition of the following volumes of NaOH?
1. 0 mL
2. 10 mL
3. 15 mL
4. 20 mL
5. 25 mL

Answer 1

So sorry if I'm late. I'll get back to elaborate more. However you must find the new concentrations of each by making a RICE Table and work backward from the Ka value. Recall that CH3COOH is a weak acid and NaOH is a strong base, so you have a weak acid-strong base titration. where NaOH is the titrant and CH3COOH is the analyte.



Helpful to draw a diagram of the titration at times, so that you could visualize how much of each dissociates. When there is 0 mL, CH3COOH dissociates little, but you would calculate the pH as you normally would; RICE Table, find [H+] from Ka, and use the 5% rule. Then do -log([H+].

Now, for when NaOH is added, remember that the strong base will neutralize the weak acid. So, you first need to find the mols of OH- with 10 mL of 0.1M NaOH, and find the mols of CH3COOH with 15.0 mL of 0.1m CH3COOH. Then, there will be a neutralization reaction, and you must find which has the excess by subtracting mols of NaOH from mols of CH3COOH. For whatever mols you get, you now need to convert back to concentration by dividing it by 10 mL, and now you can do the same Ka process. Keep doing this for 15, 20, and 25 mL. At some point, you'll end up with zero mols, and that means you are at the equivalence point!