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Part 1: First Halfway point Ka, 2.34x10 Second Halfway Point: Ka, = 631×10 Moles of NaOH at the Equivalence Point First Equivalence Point Second Equivalence Point 1.1826 x 10 mol 1444x10 Moles of KHP at the Equivalence Point First Equivalence Point Same as the moles of NaOH, = 1.1826 10 mol Second Equivalence Point: Same as the moles of NaOH = 2.444 × 107 Molar mass of RHP =204.12 g/mol Mass of KHP used at first equivalence = 2.45 x 10'g Mass of KHP used at first equivalence = 4.99 x10 g Part 2: First Halfway point: Ka, 2.34x10 Second Halfway Point: K 631 10 Moles of NaOH at the Equivalence Point First Equivalence Point 1.1825 x 10 mol Second Equivalence Point 2.444 x 10 Moles of KHP at the Equivalence Point First Equivalence Point Same as the moles of NaOH, = 1.1826 x 10' mol Second Equivalence Point: Same as the moles of NaOH= 2444 × 10- Molar ma35 of KHP=204.22 g/mol Mass of KHP used at first equivalence = 1.1826 x 10' mol

Part 1: First Halfway point Ka, 2.34x10 Second Halfway Point: Ka, = 631×10 Moles of NaOH at the Equivalence Point First Equivalence Point Second Equivalence Point 1.1826 x 10 mol 1444x10 Moles of KHP at the Equivalence Point First Equivalence Point Same as the moles of NaOH, = 1.1826 10 mol Second Equivalence Point: Same as the moles of NaOH = 2.444 × 107 Molar mass of RHP =204.12 g/mol Mass of KHP used at first equivalence = 2.45 x 10'g Mass of KHP used at first equivalence = 4.99 x10 g Part 2: First Halfway point: Ka, 2.34x10 Second Halfway Point: K 631 10 Moles of NaOH at the Equivalence Point First Equivalence Point 1.1825 x 10 mol Second Equivalence Point 2.444 x 10 Moles of KHP at the Equivalence Point First Equivalence Point Same as the moles of NaOH, = 1.1826 x 10' mol Second Equivalence Point: Same as the moles of NaOH= 2444 × 10- Molar ma35 of KHP=204.22 g/mol Mass of KHP used at first equivalence = 1.1826 x 10' mol

Introductory Chemistry For Today
Introductory Chemistry For Today
8th Edition
ISBN:9781285644561
Author:Seager
Publisher:Seager
Chapter9: Acids, Bases, And Salts
Section: Chapter Questions
Problem 9.76E
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1. Compare your pKa, Ka and molar mass from your experiment to the expected values for KHP. How well do they compare? What might account for the differences?

2. Based on your results in Part 1, is the method used reliable in determining pKa, Ka and molar mass of a weak acid? Explain.

3. Discuss your thinking on how you determined the identity of your unknown. Were there two (or more) acids that were difficult for you to distinguish between? If so, how did you makeyour final decision?

Part 1: First Halfway point Ka, 2.34x10 Second Halfway Point: Ka, = 631×10 Moles of NaOH at the Equivalence Point First Equivalence Point Second Equivalence Point 1.1826 x 10 mol 1444x10 Moles of KHP at the Equivalence Point First Equivalence Point Same as the moles of NaOH, = 1.1826 10 mol Second Equivalence Point: Same as the moles of NaOH = 2.444 × 107 Molar mass of RHP =204.12 g/mol Mass of KHP used at first equivalence = 2.45 x 10'g Mass of KHP used at first equivalence = 4.99 x10 g Part 2: First Halfway point: Ka, 2.34x10 Second Halfway Point:
Transcribed Image Text:Part 1: First Halfway point Ka, 2.34x10 Second Halfway Point: Ka, = 631×10 Moles of NaOH at the Equivalence Point First Equivalence Point Second Equivalence Point 1.1826 x 10 mol 1444x10 Moles of KHP at the Equivalence Point First Equivalence Point Same as the moles of NaOH, = 1.1826 10 mol Second Equivalence Point: Same as the moles of NaOH = 2.444 × 107 Molar mass of RHP =204.12 g/mol Mass of KHP used at first equivalence = 2.45 x 10'g Mass of KHP used at first equivalence = 4.99 x10 g Part 2: First Halfway point: Ka, 2.34x10 Second Halfway Point:
K 631 10 Moles of NaOH at the Equivalence Point First Equivalence Point 1.1825 x 10 mol Second Equivalence Point 2.444 x 10 Moles of KHP at the Equivalence Point First Equivalence Point Same as the moles of NaOH, = 1.1826 x 10' mol Second Equivalence Point: Same as the moles of NaOH= 2444 × 10- Molar ma35 of KHP=204.22 g/mol Mass of KHP used at first equivalence = 1.1826 x 10' mol
Transcribed Image Text:K 631 10 Moles of NaOH at the Equivalence Point First Equivalence Point 1.1825 x 10 mol Second Equivalence Point 2.444 x 10 Moles of KHP at the Equivalence Point First Equivalence Point Same as the moles of NaOH, = 1.1826 x 10' mol Second Equivalence Point: Same as the moles of NaOH= 2444 × 10- Molar ma35 of KHP=204.22 g/mol Mass of KHP used at first equivalence = 1.1826 x 10' mol
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