200 mL of 115 M acetic acid was mixed with 10mL of 115 M NaO

20.0 mL of .115 M acetic acid was mixed with 10mL of .115 M NaOH and 10 mL of H20. Assume the reaction goes to completion. How many moles of acetate ions are formed and the number of moles of excess acetic acid left over? Also what is the molar concentrations for those two. The solution volumes are additive. Thank you so much I can\'t figure this out and I\'ve been looking at it for 2 hours. I need it for my lab report.

Solution

Moles of Acetic Acid = 0.02 x 0.115 = 0.0023 moles

moles of NaOH = 0.01 x 0.115 = 0.00115 moles

CH3COOH + NaOH -----> CH3COONa + H2O

Moles of Acetate ion formed = 0.00115 moles

Moles of Acetic Acid left = 0.0023 - 0.00115 = 0.00115 moles

Conc. of Acetate ion = 0.00115 / (0.02 + 0.01 + 0.01) = 0.02875 M

Conc of Acetic Acid left = 0.00115 / (0.02 + 0.01 + 0.01) = 0.02875 M