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Result/conclusion Using the image attached please write • molar mass, pKa, % error • Sources of error • Sequence followed Please please please answer as fast as possible

Result/conclusion Using the image attached please write • molar mass, pKa, % error • Sources of error • Sequence followed Please please please answer as fast as possible

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Result/conclusion Using the image attached please write • molar mass, pKa, % error • Sources of error • Sequence followed Please please please answer as fast as possible
unknown acid #: 28 Mass of unknown acid: 0.300g Molarity of NaOH: 0.1002g Volume of NaOH at equivialence Point: 19.95 mL volume of NaOH at buffer Point: 9.97 PKa IPH) at puffer Point: 3.22 Molar mass: 150.75 g/mol. eqpt = equilvalent point☆ Calculations PHeqpt = 7.98 bP= boiling point ☆ ☆ mass = 0.300g☆ Vegpt =19.95mL Vop = Veapt -19.95 -9.97 2. 2 pka.bp (PH 6p) = 10.09348) (9.97) +2.289 = 3.22 bP Mbase = 0.1002 M Vbase = 19.95 mL x IL =0.01995L 1000mL mol pase =0.1002M x 19.95mL x. 12 = = 0.00199 mol 1000mL molar mass = mass 0.3009 =150.75 glmol mol = 0.00199m0 mol = mass - =0.300g molar mass = 0.00199 mol 150.0glmol
Transcribed Image Text:unknown acid #: 28 Mass of unknown acid: 0.300g Molarity of NaOH: 0.1002g Volume of NaOH at equivialence Point: 19.95 mL volume of NaOH at buffer Point: 9.97 PKa IPH) at puffer Point: 3.22 Molar mass: 150.75 g/mol. eqpt = equilvalent point☆ Calculations PHeqpt = 7.98 bP= boiling point ☆ ☆ mass = 0.300g☆ Vegpt =19.95mL Vop = Veapt -19.95 -9.97 2. 2 pka.bp (PH 6p) = 10.09348) (9.97) +2.289 = 3.22 bP Mbase = 0.1002 M Vbase = 19.95 mL x IL =0.01995L 1000mL mol pase =0.1002M x 19.95mL x. 12 = = 0.00199 mol 1000mL molar mass = mass 0.3009 =150.75 glmol mol = 0.00199m0 mol = mass - =0.300g molar mass = 0.00199 mol 150.0glmol
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