Titration Calculation?? PLEASE HELP!!

Question

Suppose our unknown acid was acetic acid, CH3COOH (Ka = 1.8 x 10-5) and that your initial aqueous solution contained 0.100 mol of acetic acid in total volume of 100 mL.

To this acetic acid solution you begin adding a 1.0 mol/L NaOH solution. Let's see what happens after adding different amounts of NaOH.

Suppose you have added a total of 20.0 mL of NaOH solution since the beginning of the titration. How many mol of acetate ions (CH3COO–) are present in the reaction mixture at this point?

 

i created an ice table and came toward the calculation of .0000022 mols..... am i right? and if not could you please fully explain it. thank you so much :)

Answer 1

Ch3COOH + NaOH ==> CH3COONa + H2O

Initially you have 0.100 mol of CH3COOH

You then add 20 ml of 1.0 M NaOH = 0.02 L x 1 mol/L = 0.02 moles NaOH.

This will more than neutralize ALL the CH3COOH, leaving 0.1 moles acetate and excess NaOH and no acetic acid.