Problem 1 Class Text Problem 11.5. Draw a milliequivalent-per-liter bar graph for the following water analysis. alkalinity = 200 mg/L sulfate ion = 58 mg/L chloride ion=36 mg/L calcium hardness = 185 mg/L magnesium hardness = 50 mg /L sodium ion = 23 mg/L potassium ion = 20 mg/L pH = 7.7 Except make calcium hardness = 170 mg/L and alkalinity = 185 mg/L. Note that all hardness and alkalinity concentrations are as CaCO3, and all alkalinity can be assumed to exist as bicarbonate, because pH is in the natural water range of 6.5-9.5.
Problem 1 Class Text Problem 11.5. Draw a milliequivalent-per-liter bar graph for the following water analysis. alkalinity = 200 mg/L sulfate ion = 58 mg/L chloride ion=36 mg/L calcium hardness = 185 mg/L magnesium hardness = 50 mg /L sodium ion = 23 mg/L potassium ion = 20 mg/L pH = 7.7 Except make calcium hardness = 170 mg/L and alkalinity = 185 mg/L. Note that all hardness and alkalinity concentrations are as CaCO3, and all alkalinity can be assumed to exist as bicarbonate, because pH is in the natural water range of 6.5-9.5.
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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Transcribed Image Text:Problem 1
Class Text Problem 11.5. Draw a
milliequivalent-per-liter bar graph for the following water analysis.
alkalinity = 200 mg/L
sulfate ion = 58 mg/L
chloride ion = 36 mg/L
calcium hardness = 185 mg/L
magnesium hardness = 50 mg/L
sodium ion = 23 mg/L
potassium ion = 20 mg/L
pH = 7.7
Except make calcium hardness = 170 mg/L and alkalinity = 185 mg/L. Note that all hardness and alkalinity
concentrations are as CaCO3, and all alkalinity can be assumed to exist as bicarbonate, because pH is in the natural
water range of 6.5-9.5.
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