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OH = 10 . Thus, expressing the concentration in terms of CaCO3

[ OH-] = 10-3 gmmol/L = 10-3 geq/L = 10-3 (CaCO3 / 2) g/L as CaCO3 = 0.05 g/L as CaCO3 = 50 mg/L as CaCO3

11.2.4 Precipitation Potential

Figure 11.1 shows a pipe that is almost completely blocked due to precipitation of CaCO3. Precipitation potential is another criterion for water stability, and application of this concept can help prevent situations like the one shown in this figure. Understanding this concept requires prerequisite knowledge of the charge balance.

All solutions are electrically neutral and negative charges must balance the positive charges. Thus, the balance of charges, where concentration must be expressed in terms of equivalents, is

[ CO2-Lq + [ HCO-]eq + [ OH ] eq = [ H+]eq + [ Ca2+]eq (11.41)

Expressing in terms of moles,

2 [ CO2-] + [ HCO-] + [ OH ] = [ H+] + 2 [ Ca2+] (11.42)

Now, the amount of calcium carbonate that precipitates is simply the equivalent of the calcium ion that precipitates, Cappt. Because the number of moles of Cappt is equal to the number of moles of the carbonate solid CaCO3ppt that precipitates,

FIGURE 11.1 A water distribution pipe almost completely blocked with precipitated calcium carbonate.

[CaCO3ppi] is the precipitation potential of calcium carbonate. Cappi, in turn, can be obtained from the original calcium, Ca2+ore, minus the calcium at equilibrium, Caf^..

To use Equation (11.44), [ Cab+ore ] must first be known. To determine [ Cafer ], the charge balance equation derived previously will be used. [ Caflfier ] is the [Ca] in the charge balance equation, thus

2 [ CO3-] + [ HCO-] + [ OH ] = [ H] + 2 [ Caf. ] (11.45)

Making the necessary substitutions of the carbonate system equilibria equations to Equation (11.45) and solving for [ Cafer ], produces

7CO37Ca[Ca^,] 7hco37h^Ca[caafer] 7oh7h[H ]

f V 7oh7hj f

7co, 7Ca 7hco 7hk27Ca

[ H+]2 —^L-}2 + 8 [ H+]f 2Ksp.CaCO [ H+] + [H+]2 Ksp,CaCo;

7oh7hJ V 7CO37Ca 7HCO37HK27Ca

The [H+] in the previous equations is the saturation pH of Equation (11.25). Now. finally. the precipitation potential [CaCO3ppt] is

[CaCO3pp(] = [Capp(] = [Cabe/ore] - [Caafter] (11.47)

11.2.5 Determination of Percent Blocking POTENTIAL OF PIPES

Let Volpipe be the volume of the pipe segment upon which the percent blocking potential is to be determined. The amount of volume precipitation potential in this volume after a time i is (100([ Cafe ] - [ Cafe. ])Volpipet )/poaCo3 td. where Poaoo3 is the mass density of the carbonate precipitate and td is the detention time of the pipe segment. Letting Qpipe be the rate of flow through the pipe. td = Volpipe/Qpipe. Substituting this expression for td. the percent blocking potential Pbiock after time t is

Volpipe

PCaCO3 Volpipe

Pcaco = 2.6 g/cc = 2600 g/L. Note that since the concentrations are expressed in gram moles per liter. volumes and rates should be expressed in liters and liters per unit time. respectively. pCaCO should be in g/L.

Example 11.9 Water of the following composition is obtained after a softening-recarbonation process: [Ca ] = 10—3 gmol/L. [HCO—] = 10-4 gmol/L. [COf] = 3.2(10—3) gmol/L. [ H2CO3] = 10-9 gmol/L. pH = 8.7. pHs = 10. temperature = 25°C. p = 5(10—3). Calculate the equilibrium calcium ion concentration precipitation potential. Express precipitation potential in gmols/L and mg/L.

Solution: The saturation pH is given as pHs = 10. The actual pH is 8.7. so the system is not saturated with calcium carbonate and no carbonate will precipitate; the precipitation potential is therefore zero. At equilibrium at the Langelier saturation pH.

the calcium ion concentration will remain the same at [Ca ] = 10 gmol/L. Ans

Example 11.10 Assume that the pH of the treated water in Example 11.9 was raised to cause the precipitation of the carbonate solid. The water is distributed through a distribution main at a rate of 0.22 m /s. Determine the length of time it takes to clog a section of the distribution main 1 km in length, if the diameter is 0.42 m.

Solution:

pCaCO3 Volpipe

Therefore,

100 _ IQQO^B^2^) (100^ _ 16,373,410.18 s _ 189.51 days Ans

This example shows the importance of controlling the precipitation of CaCO3. Of course, in this particular situation, once the pipe is constricted due to blockage, the consumers would complain and the water utilities would then solve the problem; however, the problem situation may become too severe and the distribution pipe would have to be abandoned.

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