Table

Solubility Product Constants of Respective Solids at 25°C

Ksp reaction of solid

Ksp

Significance

AgCl(s) — Ag+aq) + Cl(aq)

1.8(10(10)

Chloride analysis

Al (OH)3(s) — Alj^l 3OH(aq)

1.9(10-33)

Coagulation

BaSO4(s) — Ba2aq) + SO2(aq)

10(10

Sulfate analysis

CaCO3(s) — Ca2aq) + CO2(aq)

4.8(10(9)

Hardness removal, scales

Ca (OH )2( s) — C^ + 2OH(aq)

7.9(10(6)

Hardness removal

CaSO4( s) — Ca2aq) + SO2(aq)

2.4(10(5)

Flue gas desulfurization

Ca3 ( PO4 )2( s) — 3Ca2a(q) + 2PO2a(q)

Phosphate removal

CaHPO4( s) — Ca(aq) + HPO4(aq)

5.0(10(6)

Phosphate removal

CaF2(s) — Ca2aq) + 2F(aq)

3.9(10-11)

Fluoridation

Cr( OH)3(s) — G^q) + 3OH(aq)

6.7(10(31)

Heavy metal removal

Cu (OH )2( s) — Cu2a+q) + 2OH(aq)

5.6(10(20)

Heavy metal removal

Fe ( OH)3(s) — Fe3a+q) + 3OH(aq)

1.1(10 - 36)

Coagulation, iron removal, corrosion

Fe ( OH)2 — Fe2a+q) + 2OH(aq)

7.9(10-15)

Coagulation, iron removal, corrosion

MgCO3( s) — Mg2;, + CO^q)

10(5

Hardness removal, scales

Mg (OH)2 — Mg2a+q) + 2OH(aq)

1.5(10-11)

Hardness removal, scales

Mn ( OH)2 — Mn2; + 2OH(aq)

4.5(10-14)

Manganese removal

Ni(OH)2 — N^l 2OH(aq)

1.6(10(14)

Heavy metal removal

Zn (OH )2 — Zn2a+q) + 2OH(aq)

4.5(10-17)

Heavy metal removal

From A. P. Sincero and G. A. Sincero (1996). Enviromental Engineering: A Design Approach. Prentice Hall, Upper Saddle River, NJ, 42.

Another important application of the equilibrium constant K in general, [Equation (11)], is in the coagulation treatment of water using alum, Al2(SO4)3-14 H2O. (The "14" actually varies from 13 to 18.) In coagulating a raw water using alum, a number of complex reactions are formed by the Al3+ ion. These reactions are as follows:

13 34

The equilibrium constants KAl(OHc KAl7(OH-17c, KAl13(OH}34c, Ksp,Al(OH)3, KAl(OH-4c , and Kai (OH- c apply at 25°C. Note that the subscript c is used for the equilibrium constants of the complexes; it stands for "complex." Al(OH-3(s- is not a complex; thus, Ksp,Al(OH- does not contain the subscript c.

The previous equations can be used to determine the conditions that will allow maximum precipitation of the solid represented by Al(OH-3. The maximum precipitation of Al(OH-3 will produce the utmost clarity of the treated water. To allow for this maximum precipitation, the concentrations of the complex ions Al(OH-2+, Al7(OH-4+, Al13(OH-34, Al(OH--, and Al2(OH-2+ and Al3+ must be held to a minimum. This will involve finding the optimum pH of coagulation. This optimum pH may be determined as follows:

It is obvious that the complex ions contain the Al atom. Thus, the technique is to sum up their molar concentrations in terms of their Al atom content. Once they have been summed up, they are then eliminated using the previous K equations, Eqs. (13- to (18-, with the objective of expressing the resulting equation in terms of the constants and the hydrogen ion concentration. Because the K constants are constants, the equations would simply be expressed in terms of one variable, the hydrogen ion concentration, and the equation can then be easily differentiated to obtain the optimum pH of coagulation.

Gleaning from Eqs. (13) to (18), a molar mass balance on the aluminum atom may be performed. Let sp Al represent all the species that contain the aluminum atom standing in solution. Thus, the concentration of all the species containing the aluminum atom, is

[ sp A1 ] = [ Al3+] + [ Al ( OH )2+] + 7 [ AI7 ( OH )£] + 13 [ AlB( OH ) £]

Note that, because we are summing the species containing the aluminum atom, the coefficients of the terms of the previous equation contain the number of aluminum atoms in the respective species. For example, Al7(OH)|+ contains 7 aluminum atoms; thus, its coefficient is 7. Similarly, Al13(OH)3+ contains 13 aluminum atoms; therefore, its coefficient is 13. This explanation holds for the other species as well.

Using Eqs. (13) to (18) and the relations of molar concentration and activity, Equation (9), the following equations are obtained [for the purpose of eliminating the complex ions in Equation (19)]:

[A,3+] = {Al3 + } = Ksp,Al(OH)3 = Ksp,Al(OH)3{H } = Ksp,Al(OH)3?H[H ]

/Al /ai {OH-} /Al K w /ai Kw uunm2+] { Al(OH)2+ } KAl(OH)c{ Al3+} KAl(OH)cKsp,Al(OH)3 /h [H+]2 [ Al( OH ) ] = —-- = --7-7— = -—-

4+] { Al7(OH)17 } KAl7(OH)17c { Al3+} 7 kai7(oh)17c7a7i2a-3-27

17rTT+n17

^OH17c Y

Yai7( OH)i7c YAi7(OH)17c{ H+}17 Yai7(oh)17c yH7[H+]

KAl7(OH)17cKsp,Al(OH)3 7h[H ] YAl7(OH)17cK w

KAl13(OH)34cKsf),Al(OH)37H[H ]

7a113( OH)34cK w

4+1 { Al2( OH )2+} K Al2(0H)2c { Al3 + }2 K ai2( oh)2c Y Ai [Al3+]2

KAl2(0H)2cKsp,Al(0H)37H[H ]

7a12( 0H)2cKw

YAl, 7h, 7ai(oh)c, 7ai7(oh)17c, 7ai13(oh)34c, 7ai(oh)4c, 7ai2(oh)2c are, respectively, the activity coefficients of the aluminum ion and the hydrogen ion and the complexes Al(OH)2+, Al7(OH)47+, Ali3(OH)^4+, Al(OH)-, and Al2(OH)4+. Ksp,moH)3 is the solubility product constant of the solid Al(0H)3(s) and Kw is the ion product of water. KAi(OH)c, K ai7(oh)17c , K Ai13(OH)34c, Kai(oh)4c , and K ai2(oh)2c are, respectively, the equilibrium constants of the complexes Al(OH)2+, Al7(OH)47+, Al13(0H)5+, Al(OH)-, and Al2(OH)4+.

Let us explain the derivation of one of the above equations. Pick Equation (20). The first part of this equation is

This equation is simply the application of Equation (9), where sp is Al + and y is /Al and where the molar concentration is solved in terms of the activity.

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