Question

This is for a titration experiment starting with 25 ml of HCL. I've answered two but am struggling with the others.

2. Find the equivalence point on the graph. What is the equivalence volume of NaOH at this point?

The pH at the equivalence point is 8.77 and the volume of NaOH at this point is 40.00 mL of NaOH

3. a. Calculate the unknown molarity of the diluted acetic acid from the volumes of acid and base at the equivalence point and the molarity of the NaOH M_a × V_a = M_b × V_b.

b. Once you find the molarity of your diluted solution use that to calculate the molarity of the original solution using the equation M₁ × V₁ = M₂ × V₂ a second time.

4. In experiment 1, you were able to calculate the concentration of the HCl solution using the initial pH. Would this same approach work with the acetic acid? Why or why not?

Answer 1

4. Same approch does not work for acetic acid. Because acetic acid is weak acid and it is not dissociated completely as ions. We can calculate the concentration of acetic acid from initial pH and Ka of acetic acid. But it is different approach.