Order Of Removal

Suppose that there are four gram-equivalents of Ca2+, one gram-equivalent of Mg2+, and 2.5 gram-equivalents of HCO—. Suppose further that the decision has been made to use lime. The question is which reaction takes precedence, Equation (10.15) or Equation (10.16)? Assuming the former takes precedence, the bicarbonate species present will be composed of 2.5 gram-equivalent of Ca(HCO3)2 and none of the Mg(HCO3)2. The amount of lime needed to remove this bicarbonate is 2.5(CaO/2) = 70 g. Assuming the latter takes precedence, the bicarbonate species will now be composed of one gram-equivalent of Mg(HCO3)2 and 1.5 gram-equivalent of Ca(HCO3)2. The corresponding amount of lime needed to remove the bicarbonates is 1.0(2CaO/ 2) + 1.5(CaO/2) = 98 g. These two simple calculations produce two very different results, so knowledge of the order of precedence of the reactions is very important. The order of precedence can be judged from the solubility of the precipitates, namely: CaCO3 and Mg(OH)2.

The Ksp for CaCO3 is 5(10—9) at 25°C, while that for Mg(OH)2 is 9(10—12), also at 25°C. Consider calcium carbonate, first. Before the solid is formed, the product of the concentrations of the calcium ions and the carbonate ions must be, at least, equal to 5(10—9). Let x be the concentration of the calcium ion; thus, x will also be the concentration of the carbonate ion. Therefore, x2 = 5(10-9) ^ x = 0.000071 gmole/L

Now, consider the magnesium hydroxide. Again, before the solid is formed, the product of the concentrations of the magnesium ions and the hydroxide ions must

be, at least, equal to 9(10 ). In magnesium hydroxide, there are two hydroxide ions to one of the magnesium ion. Thus, letting y be the concentration of the magnesium ions, the concentration of the hydroxide ions is 2y. Therefore, y (y2) = 9(10-12) ^ y = 0.00021 gmole/L

Now, comparing the two answers, CaCO3 is less soluble than Mg(OH)2, because it only needs 0.000071 gmole/L to precipitate it compared to 0.00021 gmole/L for the case of Mg(OH)2. Therefore, calcium carbonate will precipitate first and Equation (10.15) takes precedence over Equation (10.16). In this example, the answer would be 70 g instead of 98 g.


For the removal of the hardness of calcium and magnesium, the effect of carbon dioxide on the amounts of sludge produced and chemicals used were not considered; but because the type of raw water treated is exposed to the atmosphere, CO2 always dissolves in it. Therefore, in addition to the chemical reactions that have been portrayed, the precipitant chemicals must also react with carbon dioxide. This means that more lime than indicated in previous reactions would be needed. The reaction that portrays the consumption of lime by carbon dioxide is as follows:

As shown in Equation (10.36), solids of CaCO3 are produced. Let MCO^ be the mass of carbon dioxide in the raw water. Therefore, letting MCO be the carbonate solids produced from the dissolved carbon dioxide, mcaCO3cO2 = CCaQO3 (M co2 ) = 2.27 (Mqo2 ) (10.37)

Also, from Equation (10.36), the lime needed, MCaOCO , to react with the carbon dioxide in the raw water is 2


Refer to Figure 10.2. As shown, the lowest solubility of Ca2+ in equilibrium with CaCO3 is at a pH of approximately 9.3. This represents the optimum pH for adding lime or soda ash to precipitate calcium as CaCO3. Raising the pH to about 11 practically dissolves all the CaCO3. This is illustrated in the reaction below. The anion species responsible for raising the pH to 11 is the OH-. Thus, the chemical reaction is

sp CaOH+

This means that, in order to precipitate the calcium carbonate optimally, the pH should be maintained at the vicinity of 9.3 and that it should not be raised to pH 11. As shown in the chemical reaction, the calcium ion is converted into the complex CaOH+, which has a relatively high Ksp of 10-149, indicating the calcium carbonate is practically dissolved when the pH is raised to 11.

Also, from the figure, the lowest solubility of Mg2+ in equilibrium with Mg(OH)2 occurs at about pH 10.4; raising the pH above this value does not affect the solubility, since the solubility is already zero at pH 10.4. At this value of pH, the solubility of Ca2+ in equilibrium with CaCO3 is not affected very much. Also, lowering the pH to below 9.8 practically dissolves the Mg(OH)2. Thus, these two limiting conditions leave only a very small window of pH 9.8 to 10.4 for an optimal concurrent precipitation of Ca2+ and Mg2+. Above pH 10.4, the precipitation of CaCO3 suffers and below pH 9.8, the precipitation of Mg(OH) suffers.

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