Question

Data Table 2: Water Hardness

Average Volume of EDTA Used (mL)	Concentration Ca2+ Ions Per Liter of Water (mol/L)	Water Hardness (ppm CaCO3)
3.33 mL	0.00333 mol/L	333.2 ppm

Exercise 1 - Questions

1. Based on the analysis of your local water, would you classify its hardness as soft, moderate, hard, or very hard? Explain your answer.

2. Approximately how much calcium would you ingest by drinking eight 8-oz glasses of your local water? Hint: 1 oz (fluid ounce) = 29.57 mL.

3. Assume an average minimum daily requirement for calcium is 1,150 mg. Calculate what percentage of your daily requirements could be met by drinking 1.0 L of your local water.

Answer 1

1. Ans. Water hardness = 333.2 ppm = 333.2 mg/L

CaCO3 (in mg/L)	Type
0-60	Soft
61-120	Moderately hard
121-180	Hard
>180	Very hard

Water containing CaCO₃ of above 180 mg/L are classified as very hard.

Therefore, the local water is very hard.

b. Ans.  Water hardness = 333.2 ppm = 333.2 mg/L

$333.2mg/L = \frac{333.2mg}{1000mL}$

$333.2mg/L = 0.3332mg/mL$

That is. 1 mL of water has 0.3332mg of CaCO₃

$1 oz = 29.57 mL$

$8 oz =8\times 29.57 mL$

$8 oz = 236.56 mL$

1 mL water contains 0.3332 mg of CaCO₃

236.56 mL water will contain

$= 0.3332mg/mL\times 236.56mL$

$= 78.82mg$ of CaCO₃

Molar mass of CaCO₃ = 100 g/mol = 100mg/mmol

100mg of CaCO₃ has 40mg of Ca

Therefore, 78.82 mg of CaCO₃ will have

$= \frac{40}{100}\times 78.82 mg$

$= 31.53 mg$

Drinking 8 oz local water would mean ingesting 31.53 mg of calcium

_{3 Ans.} Water hardness = 333.2 ppm = 333.2 mg/L

1L water has 333.2mg of CaCO₃

100 mg of CaCO₃ has 40 mg of Ca

333.2 mg of CaCO₃ will have

$= \frac{40}{100}\times 333.2 mg$

$= 133.28 mg$

Percentage met by drinking 1L local water is

$= \frac{133.28 mg}{1150mg}\times 100$

$= 11.59$%

Therefore, 11.59% of calcium requirements would be met by drinking 1L of the local water.