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ACID BASE TITRATION (PART B)-REPORT SHEET CHEMISTRY 1108 SECTION PURPOSE: 0.20mL DATA: UNKNOWN NUMBER 18 Volume of acetic acid used in every run 25.00mL Concentration of NaOH solution 0.3085M TITRATION DATA Run #1 Run #2 Final reading of burette (NaOH) 21.60mL 42.30ml Initial reading of burette (NaOH) 21.60 ml 42.30 ml Volume of NaOH used in Stration 20.70 mL 20-20 mL Endpoint Observations (relative colour and # of light pink light pink/ drops in final addition) Colour Purple colour (7dmps) (5 drops) 21.40 ML light pink NAME: Karla Shampon BENCH NO: 8A (3 chips) Run # 3 462.50ML Added mure NaOH (was at 325-ML) 450.00 ML +12.50 ml = 62.50M

RESULTS AND CALCULATIONS: other pairs of runs. Record all results to the correct number of significant figures, with the Show a detailed calculation for one pair of runs only below, show only the answers for the proper units. 1. Calculate the % difference between your runs. Answer 1&2 2 & 3 3 & 1 2. Balanced Equation Using only your best runs from above, perform the following calculation. As above, show a detailed calculation for only one run; just show the results for the other run(s). 3. Using the provided molarity of the NaOH solution, calculate the % acetic acid (by mass in each of your samples. Note: The total mass of each 25.00 mL sample is given by: Total Mass = density x volume. Assume the density of the acetic acid solution is 1.000 g/ml. 4. Calculate the average % acetic acid (your best runs only). Answer