2 Suppose a standard strong-based solution was prepared by weighing out 2.514 g of NaOH in 500 mLof water. This solution was titrated against KHP to an end point established. 1.237 g of KHP dissolvedin 200 ml of water required 62.699 mL of the base solution.1. How many moles of NaOH were initially weighed out to prepare the "standard base solution?"12. What is the apparent molarity of the "standard base," using the mass of NaOH weighed out and thevolume of water used?3. How many moles of KHP were weighed out?4. How many moles of NaOH are needed to react completely with the KHP that was weighed out?5. What is the molarity of the "standard base" using the information obtained from KHP titration?6. Explain the difference in the answers to question 2 and question 5. Which is the "correct" value for theconcentration of the standard NaOH solution? Why?

Solution: moles of NaOH = 2.514/40 = 0.06285

Suppose a standard strong-based solution was prepared by weighing out 2.514 g of NaOH in 500 mL of water. This solution was titrated against KHP to an end point established. 1.237 g of KHP dissolved in 200 ml of water required 62.699 mL of the base solution. 1. How many moles of NaOH were initially weighed out to prepare the "standard base solution?"

Suppose a standard strong-based solution was prepared by weighing out 2.514 g of NaOH in 500 mL of water. This solution was titrated against KHP to an end point established. 1.237 g of KHP dissolved in 200 ml of water required 62.699 mL of the base solution. 1. How many moles of NaOH were initially weighed out to prepare the "standard base solution?"

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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