Assume an average minimum daily requirement for calcium is 1,150. Mg. What perce

Question

Assume an average minimum daily requirement for calcium is 1,150. Mg. What percentage of your daily requirements could be met by drinking 1.0 L of water that has a water hardness value of 290.17 mg/L? The average volume (mL) of EDTA used during this experiment was determined to be 2.9 mL.

What is the average concentration (moles per liter) of Ca2+ ions present in the water sample? The average concentration of Ca2+ ions present in a sample of water was determined to be 3.5 x 10-4 mol/L. What is the average water hardness (ppm) for this sample of water?

Explanation / Answer

an average minimum daily requirement for calcium is 1,150 mg.

Water has a water hardness value of 290.17 mg/L so Our body consumes 290.17mg of Ca2+ per 1 L of water.

percentage of our daily requirements by 1L water = 290.17/1150 *100 = 25.23%

Average concentration of Ca2+ = 290.17*10-3 / 40 = 7.25*10-3 M

If the average concentration of Ca2+ ions present in a sample of water was determined to be 3.5 x 10-4 mol/L,

then 1 L of water contain 3.5 x 10-4*40=0.014 g = 14mg of Ca2+ in 1 L.

Thus 106 mL od water will have 103*14mg of Ca2+ = 14 g of Ca2+ . Thus Hardness of water is 14 ppm.

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