6. You are an environmental engineer with a major water utility. You receive the
ID: 862556 • Letter: 6
Question
6. You are an environmental engineer with a major water utility. You receive the following water quality analysis for a source water: Parameter Concentration (mg/L) Arsenic ND (i.e., non-detect) Barium 0 Bicarbonate 131 Cadmium ND Calcium 35.8 Chloride 7.1 Chromium ND Copper 0.1 Cyanide ND Fluoride 0.7 Lead ND Magnesium 9.9 Nitrate 2.2 pH 7.6 Selenium ND Sodium 4.6 Sulphate 26.4 Zinc ND a. Check whether or not the analysis is correct (within 5%) using the four most abundant cations and the three most abundant anions. b. What is the hardness of the water in mg/L as CaCO3?Explanation / Answer
Four major cations present in water are calcium, potassium, sodium and magnesium. Here potassium is not given.
And the three major anions are chloride, fluoride and sulfate.
According to the principle of electroneutrality, the total charge of an aqueous solution must be zero. Therefore, the number of positive charges must be equal to the number of negative charges.
This can be obtained by calculating millieuivalents.
A Milliequivalent (meq) is 1/1000 of an equivalent.
the analysis is accurate, then the sum of milliequivalents of cations and anions should be nearly equal.
An error of more than 5% in the cation-anion balance might imply that the analysis is not accurate.
However, if the laboratory did not test for one of the major cations or anions, then a correct balance cannot be calculated. Here firstly, the laboratory has not calculated potassium, so, the analysis is not correct.
Secondly,
Calculating no.of meq of calcium = 35.8*2/ 40*1000 = 1.8* 10-3 meq
no.of meq of magnesium = 9.9*2/ 24.3*1000 = .81* 10-3 meq
no.of meq of sodium = 4.6*1/ 22.9*1000 = 0.2* 10-3 meq
Total meq of cations= (1.8 *10-3) + (0.81 *10-3) + (0.2*10-3) = 2.81 meq
If total meq of anions also comes out to be same or almost same, then it is said that the analysis is correct.
no.of meq of chloride = 7.1*1/ 35.43*1000 = 0.2* 10-3 meq
no.of meq of fluoride = 0.7*1/ 18.99*1000 = 0.03* 10-3 meq
no.of meq of sulfate = 26.4*2/ 96*1000 = 0.55* 10-3 meq
Total meq of anions= (0.55 *10-3) + (0.03 *10-3) + (0.2*10-3) = 0..78 meq
So, the analysis is wrong.
ANS 2......hard water comprises a mixture of calcium and magnesium, together with bicarbonate, sulphate, chloride, etc. When its hardness is expressed as mg/l as CaCO3, it
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