You took 25.00 mL of unknown water solution and titrated it with EDTA. The EDTA solution molarity was 0.01 M, and it took 16.50 mL to reach the end point. With the blank it took only 0.98 mL to reach end point. A) How many milliliters of EDTA solution were used to titrate hardness that actually came from the unknown? B) How many moles of EDTA reacted with hardness-causing ions from the unknown sample? C) How many moles of hardness-causing ions were present in the unknown sample?
You took 25.00 mL of unknown water solution and titrated it with EDTA. The EDTA solution molarity was 0.01 M, and it took 16.50 mL to reach the end point. With the blank it took only 0.98 mL to reach end point. A) How many milliliters of EDTA solution were used to titrate hardness that actually came from the unknown? B) How many moles of EDTA reacted with hardness-causing ions from the unknown sample? C) How many moles of hardness-causing ions were present in the unknown sample?
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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You took 25.00 mL of unknown water solution and titrated it with EDTA. The EDTA solution molarity was 0.01 M, and it took 16.50 mL to reach the end point. With the blank it took only 0.98 mL to reach end point.
A) How many milliliters of EDTA solution were used to titrate hardness that actually came from the unknown?
B) How many moles of EDTA reacted with hardness-causing ions from the unknown sample?
C) How many moles of hardness-causing ions were present in the unknown sample?
D) Assuming that the total hardness of water is due to CaCO3, how many grams CaCO3 does it correspond to?
E) What is the total hardness of the unknown water in ppm CaCO3?
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