Presentation is loading. Please wait.

Presentation is loading. Please wait.

Stabilization Water Stabilization. Water Stabilization As in water softening, when the concentrations of CaCO 3 and Mg(OH) 2 exceed their solubilities,

Similar presentations


Presentation on theme: "Stabilization Water Stabilization. Water Stabilization As in water softening, when the concentrations of CaCO 3 and Mg(OH) 2 exceed their solubilities,"— Presentation transcript:

1 Stabilization Water Stabilization

2 Water Stabilization As in water softening, when the concentrations of CaCO 3 and Mg(OH) 2 exceed their solubilities, the solids may continue to precipitate. This condition can cause scale to form, a solid that deposits due to precipitation of ions in solution. To prevent scale formation, the water must be stabilized.

3 Because of the universal presence of carbon dioxide, any water body is affected by the reaction products of carbon dioxide and water. The species produced from this reaction form the carbonate system equilibria. As discussed later, the stability or instability of water can be gaged using these equilibria. H2OH2O CO 2 If the pH is high, stabilization may be accomplished using one of several acids or using CO2, a process called recarbonation. If the pH is low, stabilization may be accomplished using lime or some other bases.

4 1. Carbonate Equilibria The carbonate equilibria is a function of : the ionic strength of water, activity coefficient, and the effective concentrations of the ionic species.

5 Carbonate Equilibria: Calcium is one of the major cations that can form scales as a result of the instability of water. Calcium plays an important role in the carbonate equilibria. We will therefore express the carbonate equilibria in terms of the interaction of the calcium ion and the carbonate species which are the reaction products of carbon dioxide and water.

6 Carbonate Equilibria: Since the equilibria occur in water, the dissociation of the water molecule must also be involved. Using calcium as the cation, the equilibrium equations of the equilibria along with the respective equilibrium constants at 25 º C are as follows (Rich, 1963):

7 Carbonate Equilibria:

8 The Ks are the values of the respective equilibrium constants. K sp.CaC03 is the equilibrium constant for the solubility of CaC0 3. The pair of braces, { }, are read as "the activity of,"

9 Carbonate Equilibria: the equilibrium constants are calculated using the activity. Activity: is a measure of the effectiveness of a given species in its participation in a reaction. It is an effective or active concentration and has units of concentrations.

10 Carbonate Equilibria: {sp} = γ [sp] Where sp represents any species involved in the equilibria. The pair of brackets, [], is read as "the concentration of,. γ is the activity coefficient.

11 1.1 Ionic Strength: As the particle ionizes, the number of particles increases. the activity coefficient is a function of the number of particles in solution. The number of particles is characterized by the ionic strength μ. 1.Carbonate Equilibria:

12 1.1 Ionic Strength: i is the index for the particular species and z is its charge. The concentrations are in gmmols/L In terms of the ionic strength. the activity coefficient is given by the DeBye- Huckellaw as follows (Snoeyink and Jenkins, 1980; Rich, 1963):

13 1.1 Ionic Strength: Langelier estimated as (TDS ) (TDS ) Russell estimated as (specific conductance ) (specific conductance )

14 Example 1

15 The pH of a solution is 7. Calculate the hydrogen ion concentration? Solution: pH = -log 10 {H + } 7 = -log 10 {H + } {H + } = gmmols/L Ans.

16 Example 2

17 The concentration of carbonic acid was analyzed to be 0.2 gmmols/L. If the pH of the solution is 7, what is the concentration of the bicarbonate ion if the temperature is 25°C? Solution:

18 Example 3

19 A sample of water has the following composition: CO 2 = 22.0 mg/L, Ca 2+ = 80 mg/L, Mg 2+ = I2.0 mg/L, Na + = 46.0 mg/L, HCO 3 - = mg/L, and SO 4 2- = 216 mg/L. What is the ionic strength of the sample? Solution:

20 gmmols/LMol. MassMg/LIon Ca Mg Na HCO SO 4 2- μ = 1/2[ (2 2 ) (2 2 ) (1) (1) (2 2 )] = Ans.

21 Example 4

22 In Example 3, calculate the activity coefficient and the activity in mg/L of the bicarbonate ion. Solution: { HCO 3 - } = 0.86(0.0025) = mg/L Ans. {sp} = γ [sp]

23 1.2. Equilibrium Constant As a Function Of Temperature: The equilibrium constants given previously were at 25°C. To find the values of the equilibrium constants at other temperatures, the Van't Hoff equation is needed;

24 Where: T: is the absolute temperature. ΔHº: is the standard enthalpy change, where the standard enthalpy change has been adopted as the change at 25°C at one atmosphere of pressure. R: is the universal gas constant. The value of R depends upon the unit used for the other variables. Table 11.1 gives its various values and units

25

26 Table 11.2 shows values of interest in water stabilization. It is normally reported as enthalpy changes.

27 The enthalpy change is practically constant with temperature; thus ΔHº may be replaced by ΔHº 298 Doing this and integrating the Van't Hoff equation from K T1 to K T2 for the equilibrium constant K and from T 1 to T 2 for the temperature, This equation expresses the equilibrium constant as a function of temperature.

28 1.3. ΔH 298 o, For Pertinent Chemical Reactions Of the Carbonate Eq.: According to Hess's low, if the chemical reaction can be written in steps, the enthalpy changes can be obtained as the sum of the steps;

29

30 The values of the ΔH 298 o s are obtained from the previous Table The values in the table indicate ΔH 298 o of formation having negative values. If the reaction is not a formation but a breakup such as;, The sign is positive.

31 Example 5

32

33

34 Hand-off Any questions?

35 2. Criteria For Water Stability at Normal Conditions: In the preceding discussions, a criterion for stability was established using the equilibrium constant K sp. At normal conditions, as especially in the water works industry, specialized forms of water stability criteria have been developed. These are saturation pH, Langelier index, and the precipitation potential of a given water.

36 2.1. Saturation pH and Langelier Index: Because pH is easily determined the determination of saturation pH is convenient method of determining the stability of water. If the condition is at equilibrium no precipitate or scale will form. If the pH of the sample is determined, this can be compared with the equilibrium pH to see if the water stable or not. Therefore, we now proceed to derive the equilibrium pH. Equilibrium pH is also called saturation pH.

37 In natural systems, the value of the pH is strongly influenced by the carbonate equilibria reactions. The CO 3 2- species of these reactions well pair with a cation, thus the equilibrium reactions into a dead end by forming a precipitate. For example, the complete carbonate equilibria reactions as follows:

38 c is the charge of the cation that pairs with CO 3 2- forming the precipitate Cation 2 (CO 3 ) c(s). We call the formation of this precipitate as the dead end of the carbonate equilibria, since the carbonate species in solution are removed by the precipitation.

39 In order to find the dead end cation, several cations can possibly pair with the carbonate. The pairing will be governed by the value of K sp. The cation with smallest K sp value is the one that can form a dead end for the carbonate equilibria reactions. however, of all the possible cations, Ca 2+ is the one that is found in great abundance in the nature compared to the rest. Thus, although all the other cations have much more smaller K sp s than the calcium, they are of no use as dead ends if they do not exist (see Table 11.3).

40

41 As will be shown later, the saturation pH may conveniently be expressed in terms of total alkalinity. The species which are the components of the carbonate equilibria, they also represent as components of the total alkalinity of the carbonate system equilibria. They may be added together to produce the value of the total alkalinity. A convenient common unit is the gram equivalent. Letting [A] geq represent the total alkalinity;

42 Equations to calculate the carbonate equilibria species:

43 Then the total alkalinity equation becomes; Let Solving for {H+},

44 thus, the saturation pH, pH s is The Langlier Index (or Saturation Index) (LI) is the difference between the actual pH and the saturation pH of the a solution, thus;

45 Example 6

46

47

48

49

50 Example 7

51

52 2.2. Determination of {Ca 2+ }: The activity of the calcium ion is affected by its complexation with anions. Ca 2+ forms complexes with the carbonate species, OH - and SO The complexation reactions are as follows: Ca anion = anion complex of calcium ion = represent complex of calcium ion.

53 Let the total concentration of the calcium species as determined by titration be [Ca T ]. Thus, the concentration of the calcium ion [Ca 2+ (aq) ] is Table 11.4 shows the equilibrium constants of the previous complexes at 25 o C.

54

55 By applying Hesss law and using solubility product relations the above equation becomes:

56 Example 8

57

58

59 2.3. Total Alkalinity As Calcium Carbonate: The unit of concentration that we used for alkalinity is equivalent per unit volume. But alkalinity also can be expressed in terms of CaCO 3. Expressing the alkalinity in terms of CaCO 3 is a sort of equivalence. depending upon the chemical reaction it is involved with, CaCO 3 can have more than one value for its equivalent mass.

60 If the reaction is CaCO 3 has an equivalent mass of CaCO 3 /2, because the number of reference species is 2. To illustrate the use of this concept, assume gmmol/L of the hydroxide ion and express this concentration in terms of CaCO 3. the pertinent reaction is [OH - ] = gmmol/L = geq/L = (CaCO 3 /2) g/L as CaCO 3 = 0.05 g/L as CaCO 3 = 50 mg/L as CaCO 3.

61 2.4. Precipitation Potential: Precipitation potential is another criterion for water stability, and application of this concept can help prevent situations like the one shown in this figure. 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

62 Expressing in terms of moles, the amount of calcium carbonate that precipitates is simply the equivalent of the calcium ion that precipitates, Ca ppt. Because the number of moles of Ca ppt is equal to the number of moles of the carbonate solid CaCO 3ppt that precipitates, [CaCO 3ppt ] = [Ca ppt ]

63 [CaCO 3ppt ] is the precipitation potential of calcium carbonate, Ca ppt, in turn, can be obtained from the original calcium, Ca 2+ before, minus the calcium at equilibrium, Ca 2+ after. [Ca ppt ] = [Ca 2+ before ] - [Ca 2+ after ] To use the above equation [Ca 2+ before ] must first be known. To determine [Ca 2+ after ], the charge balance equation derived previously will be used. [Ca 2+ after ] is the [Ca 2+ ] in the charge balance equation, thus

64 By substituting the carbonate system equilibria equations in the above equation and solving for, [Ca 2+ after ], produces The [H + ] in the previous equation is the saturation pH. Finally, the precipitation potential [CaCO3ppt] is: [CaCO 3ppt ] = [Ca ppt ] = [Ca 2+ before ] - [Ca 2+ after ]

65 2.5. Determination of Percent Blocking Potential of Pipes: Let Vol pipe 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 t is: = the mass density of the carbonate precipitate. = the detention time of the pipe segment.

66 t d = Vol pipe /Q pipe Q pipe = the rate of flow through the pipe. by substituting this expression for t d, the percent blocking potential P block after time t is: = 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.

67 Example 9

68

69 Example 10

70 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 3 /s. Determine the length of time it takes clog a section of the distribution main 1 km in length, if the diameter is 0.42 m. Solution:

71

72 Hand-off Any questions?

73 3. Recarbonation of Softened Water After the softening process, the pH is so high that reduction is necessary to prevent deposition of scales in distribution pipes. This can be accomplished inexpensively using carbon dioxide. We will therefore develop the method for determining the carbonic acid necessary to set the water to the equilibrium pH. In recarbonation, the available calcium ion in solution is prevented from precipitation.

74 Therefore, it remains to determine at what pH will the equilibrium condition be, given this calcium concentration. This determination is, in fact, the basis of the Langelier saturation pH. Adding carbonic acid will increase the acidity of the solution after it has neutralized any existing alkalinity. Let the current pH be pH cur and the pH to which it is to be adjusted (the destination pH) be pH to.

75 The hydrogen ion concentration corresponding to pH cur is 10 -pH cur gram moles per liter and that corresponding to pH to is 10 -pH to gram moles per liter. Assuming no alkalinity present, the total acid to be added is 10 -pH to pH cur gram moles per liter. Alkalinity is always present, however, so more acid must be added to counteract the natural alkalinity, [A cur ] geq. Thus, the total acidity to be added, [A cadd ] geq, is:

76 Where Ф a is the fractional dissociation of the hydrogen ion from the acid supplied. For strong acids, Ф a is unity; for weak acids, it may be calculated from equilibrium constants (Table 11.5).

77

78 Th anks

79 Any questions? Q & A

80 Thanks


Download ppt "Stabilization Water Stabilization. Water Stabilization As in water softening, when the concentrations of CaCO 3 and Mg(OH) 2 exceed their solubilities,"

Similar presentations


Ads by Google