A source water for a potable water system contains 185 mg/L (as CaCO3) of carbon

Question

A source water for a potable water system contains 185 mg/L (as CaCO3) of carbonate hardness (CH) due to calcium, 40 mg/L (as CaCO3) of carbonate hardness due to magnesium, 225 mg/L (as CaCO3) of total hardness (TH), 10 mg/L of H2CO3 (as CaCO3) and 230 mg/L of HCO3- (as CaCO3). If a target of 90 mg/L of total hardness (TH) is to be achieved, how much lime (as CaCO3) must be added? (report your answer to nearest 1 mg/L and do not forget to add 20 mg/L to theoretical value to promote a reasonable reaction rate; do not enter units)

Explanation / Answer

Total hardness as of CaCO3 equivalent = 225 mg/L

Hence the CaCO3 equivalent of lime(Ca(OH)2) required to make the hardness 0 mg/L = 225mg/L

But we need to achieve a target of 90 mg/L of total hardness

Hence the CaCO3 equivalent of lime(Ca(OH)2) required to achieve the hardness 90 mg/L = 225mg/L - 90 mg/L

= 135 mg/L

Also we need to add 20 mg/L to theoretical value to promote a reasonable reaction rate.

Hence the total CaCO3 equivalent of lime(Ca(OH)2) required =  135 mg/L + 20 mg/L

= 155 mg/L eqv. of CaCO3 (answer)

If we convert this equivalent of CaCO3 as the weight of lime(Ca(OH)2), then

155 mg/L eqv. of CaCO3 = W*50 / E

where W = weight of Ca(OH)2

E = equivalent weight of Ca(OH)2 = 37 g

=> 155 = W*50 / 37

=> W = 114 mg/L of Ca(OH)2

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