Titration of Acetic Acid Chemistry 1105

Titration  Commercial White Vinegar  Determination of Molar Concentration Of Acetic Acid

Materials  Acetic Acid, app 5 % per volume Commercial White Vinegar  NaOH, 0.5 M Strong Base Strong Base Will Burn Skin Will Burn Skin Rinse Thoroughly Rinse Thoroughly

Instructions  Titration Assembly Burette Clamp Burette Clamp Ring Stand Ring Stand Burette Burette  Volume Reading Top to Bottom Top to Bottom Subdivisions (tenths of a mL) Subdivisions (tenths of a mL) School of Agronomy and Horticulture. The University of Queensland. Brisbane, Australia

Procedure  Titration Right?  But All Reagents and Products Are Colorless?  Need An Indicator (In)

Information  There are three ways to determine the equivalence point: (acid-base indicator, pH thermometric)  We will be working with end point method which uses an indicator as a visual representation of the equivalence point

Indicators  Indicator (In) Phenolphthalein Phenolphthalein Laxative Laxative  In General: HIn + H 2 O ↔ H 3 O + + In - Adding Base Changes Equilibrium To Product Side Due To Acid Consumption Adding Base Changes Equilibrium To Product Side Due To Acid Consumption Specifically: HIn (Colorless) → In - (Purple) Specifically: HIn (Colorless) → In - (Purple)

Indicator  Colorless  purple  Remain Pale Purple After Swirling  The longer it takes for color to disappear, the closer to the end point you are getting

Procedure  Weigh a clean and dry 250 mL flask and record the value  Obtain approx. 50 mL of vinegar in a 100 mL beaker  Using graduate cylinder take 15 ml and add it to the flask  Reweight the flask and record the value  To this add 10 mL of water (remember use only 15 mL of vinegar)

Procedure  Using a 200 ml beaker acquire ≈ 100 mL of NaOH  Rinse burette with NaOH ( 5 mls)  Barrel, stopcock and delivery tube  Drain this and fill burette Initial volume reading must be between 0-10 mL Initial volume reading must be between 0-10 mL  Add 3-5 Drops Phenolphthalein Into Erlenmeyer flask (vinegar + water) Run this experiment 3 times

Reminders  Again Remember Molarity = Moles Solute / Liters Solution Molarity = Moles Solute / Liters Solution M = mol / L M = mol / L Mathematical Manipulation Mathematical Manipulation  Therefore…

Calculations  [NaOH] (M) * Vol NaOH (L) = mol NaOH 0.50 mol/L * L = mol NaOH 0.50 mol/L * L = mol NaOH  Since The Stoichiometry Is 1 to 1 1 CH 3 COOH + 1 NaOH → H 2 O + Na + - OOCCH 3 1 CH 3 COOH + 1 NaOH → H 2 O + Na + - OOCCH 3 The Amount Of Moles NaOH Is Equal To The Amount Of Moles CH 3 COOH The Amount Of Moles NaOH Is Equal To The Amount Of Moles CH 3 COOH

Calculations  To Calculate The Molar Concentration Of Acetic Acid In Commercial White Vinegar: Simply divide the amount of moles CH 3 COOH by the liters of white vinegar Simply divide the amount of moles CH 3 COOH by the liters of white vinegar mol / L = 1.0 mol/L CH 3 COOH mol / L = 1.0 mol/L CH 3 COOH  After Titrations, Average Your [CH 3 COOH] Results

Experiment  Goggles Required  After The 1’st Experiment Empty waste Empty waste Rinse Erlenmeyer with DI wash-bottle and drain Rinse Erlenmeyer with DI wash-bottle and drain Don’t worry about excess water in Erlenmeyer Don’t worry about excess water in Erlenmeyer Does not change the moles of Acetic acid Does not change the moles of Acetic acid

Experiment  Fill Erlenmeyer With Another 10.0 mL White Vinegar and repeat  After 3’rd Titration Empty and rinse Erlenmeyer Empty and rinse Erlenmeyer Empty and rinse burette Empty and rinse burette Store burettes with stopcock open Store burettes with stopcock open Empty waste beaker contents into lab waste container Empty waste beaker contents into lab waste container Wipe down bench with wet paper towel Wipe down bench with wet paper towel Put all your equipment back into the drawers Put all your equipment back into the drawers

Lab Report – My Example Titration1 Titration 2 Titration 3 Known Quantities [NaOH] (M) mol/Lmol/Lmol/L Volume Vinegar LLL Experimental Values Burette (i) mL mL Burette (f) mL mL Calculations Vol NaOH (L) LL L mols NaOH mol mol mols CH 3 COOH mol mol [CH 3 COOH] (M) mol/L mol/L Avg. [CH 3 COOH] (M) SignificantFigures mol/L mol/L

Calculations example  M=mole/L of NaOH =.1M  15 ml used  *.1 =.0015 moles NaOH  Vi = 1.5ml Vf = 26.5  Volume used = = 25 ml  CH3COOH NaOH  NaOOCCH3 + H2O  1 to 1 so moles of NaOH = moles AcA .0015mol= 0.025L * [AcA]  Moles= 0.06 moles AcA  Molar AcA = 0.06mol/1L =0.06M AcA