Exercise 2: A diprotic acid (H₂A) is titrated with NaOH to determine its molar mass according to the following equation: H₂A (aq) + 2 NaOH(aq) 2H₂O(1) + Na₂A (aq) Several experiments are performed by weighing a sample of H₂A, dissolve in water, with addition of few drops of phenolphthalein (ph.ph) indicator. The following data are obtained. Mass of H₂A Volume (mL) of 0.1M NaOH Molar mass of H₂A Average molar mass 0.22 46.5 0.28 55.5 0.30 0.32 0.25 0.28 Hint: mole H₂A==xmole NaOH 11212² 62 63 52 57

Exercise 2: A diprotic acid (H₂A) is titrated with NaOH to determine its molar mass according to the following equation: H₂A (aq) + 2 NaOH(aq) 2H₂O(1) + Na₂A (aq) Several experiments are performed by weighing a sample of H₂A, dissolve in water, with addition of few drops of phenolphthalein (ph.ph) indicator. The following data are obtained. Mass of H₂A Volume (mL) of 0.1M NaOH Molar mass of H₂A Average molar mass 0.22 46.5 0.28 55.5 0.30 0.32 0.25 0.28 Hint: mole H₂A==xmole NaOH 11212² 62 63 52 57

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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Question

Exercise 2: A diprotic acid (H₂A) is titrated with NaOH to determine its molar mass
according to the following equation:
H₂A (aq) + 2 NaOH(aq)
2H₂O(1) + Na₂A (aq)
Several experiments are performed by weighing a sample of H₂A, dissolve in water, with
addition of few drops of phenolphthalein (ph.ph) indicator. The following data are
obtained.
Mass of H₂A Volume (mL) of 0.1M NaOH Molar mass of H₂A Average molar mass
46.5
55.5
0.22
0.28
0.30
0.32
0.25
0.28
Hint:
mole H₂A=-
Ul
<mole NaOH
62
63
52
57
Iodometry: Is a volumetric titration method, in which the appearance or disappearance of
elementary iodine indicates the end point

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What is the pH of an aqueous solution made by combining 37.69 mL of a 0.4289 M sodium acetate with 48.12 mL of a 0.3347 M solution of acetic acid to which 4.130 mL of a 0.0568 M solution of NaOH was added? Answer: 3.38 Check X
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