1. Titration What is the concentration in moles/liter of a vinegar (solution of acetic acid in water)?

2. Titration Titration – experimental method of determining the concentration of a solution Purpose: To determine the concentration of a solution of commercial vinegar (solution of acetic dissolved in water) by reaction with known amount of sodium hydroxide Link to acetic acid NaOH reaction Link to acid base titration animation Chang

3. Link to titration technique video Link to reading meniscus Link to measuring volume

4. Prelab Question #1: Vinegar Concentration on label = 5.0 % mass/volume – calculate molarity % = part/ 100 5.0 % m/v = 5.0 g CH 3 COOH/ 100 mL Solution Molarity = moles CH 3 COOH/ L of Solution Molar Mass C 2 H 4 O 2 : C 2 : 24.02 + H 4 : 4.04 + O 2 : 32.00 = 60.06 g/mole

5. Prelab Question #1: Vinegar Concentration on label = 5.0 % mass/volume – calculate molarity Molar Mass C 2 H 4 O 2 : C 2 : 24.02 + H 4 : 4.04 + O 2 : 32.00 = 60.06 g/mole ( 1 mole C 2 H 4 O 2 60.06 g C 2 H 4 O 2 ) =0.083 moles 5.0 g C 2 H 4 O 2 Molarity = 0.083 moles C 2 H 4 O 2 0.100 L = 0.83 M Use 0.83 M as accepted value in final lab % error calculations

6. Prelab #2: Write the balanced equation for the reaction of acetic acid CH 3 COOH and sodium hydroxide, NaOH Important hints: This is an acid-base reaction; What is an acid? What is a base? Which is which? NaOH is an ionic base – don’t forget to separate it into ions before reacting it. CH 3 COOH + NaOH → CH 3 COOH + Na + + OH -1 → H2OH2O + CH 3 COO -1 + Na +1

7. 3A) Acetic acid, sodium hydroxide, sodium acetate and water are all clear and colorless. What color is the Phenolpthlalein indicator in acidic solution and what color is it in basic solution?

8. 3B) What is the endpoint of a titration? What change will you observe in the solution when the endpoint is reached? At endpoint : moles acid = moles of base Start Endpoint Past Endpoint

10. H+H+ HInd H+H+ H+H+ OH -1

11. H+H+ HInd H+H+ H2OH2O

12. H+H+ H+H+ H2OH2O OH -1

13. H+H+ HInd H2OH2O H2OH2O

14. H+H+ H2OH2O H2OH2O OH -1

15. HInd H2OH2O H2OH2O H2OH2O

16. H2OH2O H2OH2O H2OH2O OH -1

17. H2OH2O H2OH2O H2OH2O H2OH2OInd -1

18. Titration Equation for acid-base rxn with 1:1 Stoichiometry M A V A = M B V B M A = Molarity of acid M B = Molarity of Base V A = Volume of acid V B = Volume of Base Concept: Moles acid x Liter = Moles Base x Liter Liter HA HA HA NaOH NaOH NaOH H 2 O H 2 O H 2 O NaA NaA NaA

19. Example Calculation of ∆V base Initial Volume = 15.72 mLFinal Volume = 26.75 mL ∆V base = V f – V i = 26.75 – 15.72 mL = 11.03 mL

20. Sample Data Table and Calculation Acid Base M A = ? M B = 0.50 M Volume Final: 10.05 mL Volume Final: 26.75 mL - Volume Initial: 0.00 mL - Volume Initial: 15.72 mL V A = 10.05 mL V B = 11.03 mL M A V A = M B V B M A = M B V B VAVA = (0.50 M) (11.03 mL) 10.05 mL =.55 M