[HgClM [Hg2I[cr4

Here the subscript, on the equilibrium constant has the same meaning as in Eqs. (2.38) to (241). It can easily be shown that /3, = Kh = Kfa, fa = KXK2KZ, and = KyKtK-iK^. Sometimes the reciprocal of the overall formation constant of the highest complex is given as the instability constant for that reaction. Thus,

[Hgcin

example 2.10

1 Thè chloride concentration in a typical freshwater stream is IO"3 M. If the HgGl2(a<?) con' cehtration is: lO;r8 M (about the accepted limit for Hg in drinking water), what will Be the-concentrations of Hg5+, HgCl+, HgClj", and HgCir9

filSllll:^

, [HgClM = ^(HgCl, HOT] = 9 33(lO"10)(lO~3) = 9.3 X 1<T13 M

; These calculations: indicate that most of the mercury in natural waters occurs as the neu.-;. tràl HgCl2(aq) species Little of the mercury occurs in ioni2ed forms The sorptive behav-■ibr Of inetals is significantly affected by complexation, pH, and charge, as well as the na-: ture of the adsorbent (solid). Additional details are given in Sees. 3.12 and 4.9. .

Other soluble molecules or ions can act as ligands to form complexes with metals such as mercury. Among the ligands are H+, OH", COf~, NH3, F~, CN~, SjOi-, and many other inorganic and organic species. NH3 complexes with metals are common. For example, with silver

All such ions are readily destroyed by creating conditions, physically or chemically, that will remove one of the dissociation products.

The silver-ammonia complex ion can be destroyed by adding a source of hydrogen ions. In this case destruction is caused by the formation of a more stable complex ion, NH|. The ammonium ion (NH4+) exists in equilibrium, written in the formation direction, as follows:

Addition of a strong base such as sodium hydroxide will decrease the [H+] concentration through the formation of poorly ionized water, and the equilibrium for the silver-ammonia complex will be shifted far to the left but not completely. The equilibrium may be completely destroyed by boiling the solution to expel ammonia. This is the basis of the separation and determination of ammonia nitrogen by the distillation technique.

Solubility Product

A fundamental concept is that all solids, no matter how insoluble, are soluble to some degree. For example, silver chloride and barium sulfate are considered to be very insoluble. However, in contact with water they do dissolve, slightly, and form the following equilibria:3

Crystals of compounds consist of ions arranged in an orderly manner. Thus, when crystals of a compound are placed in water, the ions at the surface migrate

'the (s) represents solid or precipitated material.

into the water and will continue to do so until the salt is completely dissolved or a condition of saturation is attained. With so-called insoluble substances, the saturation value is very small and is reached quickly. In silver chloride, barium sulfate, and other insoluble compounds, the ionic concentrations that can exist in equilibrium with the solid of crystalline material are very small.

The equilibrium that exists between crystals of a compound in the solid state and its ions in solution is amenable to consideration under the equilibrium relationship and can be treated mathematically as though the equilibrium were homogeneous in nature. For example, consider silver chloride at equilibrium, as shown in Eq. (2.49) and using molar concentrations instead of activities:

but [AgCl(s)] represents the silver chloride that exists in the solid state. Thus, in Eq. (2.52), [AgCl(s)] is taken to equal 1; so

where the constant is called the solubility-product constant.

For more complex substances, such as tricalcium phosphate, that ionize as follows:

the solubility-product expression in accordance with Eq. (2.53) is

Solubility products for nearly all insoluble substances may be obtained by reference to qualitative and quantitative textbooks or chemical handbooks. Typical solubility-product constants of interest in water chemistry are listed in Table 2.5.

A prediction of relative solubilities of compounds cannot be made by a simple comparison of solubility-product values because of the squares and cubes that enter into the calculation when more than two ions are derived from one molecule, as shown in Eqs. (2.54) and (2.55). Barium sulfate, which yields two ions, and calcium fluoride, which yields three ions, may be used to illustrate the point. If we let S represent the solubility of the metal, then for BaS04

[Ba2+] = 5 = [SOn and [Ba2+)[SOf] = $2 = = 1 X lCTi0 (2.56)

chapter 2 Basic Concepts from Générai Chemistry 39

fable 2.5 1 Typical solubility-product constants

■ 25:C ■ "

Significance ill environmental • ;

MgC03(i)^Mg2++C0|~

4 X 10~5

Hardness removal, scaling

Mg(OH)2(i)^Mgî+ + 20H"

9 X 10"12

Hardness removal, scaling

CaC03{i)^Ca2+ + C0r

5 X 10~9

Hardness removal, scaling

Ca(OH)2(i)^Caî+ + 20H"

5 X 10

Hardness removal

CaSO40)^Ca2+ + SOj"

2 X 10™5

Flue gas desulfurization

AgCl(i)^Ag++Cr

3 X Î0"10

Chloride analysis, heavy metal removal

Ag2C03(i)^2Ag+ + C0|-

8.5 X 10"12

Heavy metal removal and fate

AgOH(s) ^ Ag* + OH"

2 X 10~8

Heavy metal removal

Ag2S04(i) ^ 2Ag+ + SOf

1.6 X 10_s

Heavy metal removal

Ag2S0)^2Ag+ + S2~

1 x 10""49

Heavy metal removal and fate

CdC03(,ï) ^ Cdi+ + C03~ .

1.8 x 10""14

Heavy metal removal and fate

Cd(OH)2(i) ^ Cd2* + 20H"

2 X 10™14

Heavy metal removal

CdS(i) ^ Cd2+ + $2~

1.4 X 10"29

Heavy metal removal and fate

Cr(OH)3(î)^Cr3+ + 30H~

6 X 10~31

Heavy metal removal

CuC03(i) ^ Cu!+ + cof

2.3 X 10"'°

Heavy metal removal and fate

Cu(OH)j(j) Cu2+ + 20H"

2 X 10"

Heavy metal removal

CuS(i)^Cu2+ +S2~

1 X 10"3S

Heavy metal removal and fate

NiC03(i)^Ni2+ 4-COl"

1.5 X 10~7

Heavy metal removal and fate

Ni(OH)2(,ï) Ni" + 20H"

2 X 10~,s

Heavy metal removal

NiS(i) Ni21' + S2"

1.4 X 10~24

Heavy metal removal and fate

PbC03{ï)^Pbî+ + C0r

7.4 X KTi4

Heavy metal removal and fate

Pb(OH)2(i)==±Pb2+ + 20H"

2.5 X 10"16

Heavy metal removal

PbS(i)i±Pb2+ + S2~

1 X I0"28

Heavy metal removal and fate

PbS04{j) Pb2+ + SO|~

1.6 X 10~8

Heavy metal removal

ZnC03(H20) (s) ^ Zn,+ + CO|~ + B,0

5.5 X 10'"

Heavy metal removal and fate

Zn(OH)2(i)^Zn2+ + 20H"

8 x 10"1S

Heavy metal removal

ZnS(ï) Zn2+ + S2"

1 X 10 "22

Heavy metal removal and fate

Ai(OH)3(i)i±Al3+ + 30H"

1 X 10-32

Coagulation

Fe(OH)3(i)^Fe3+ + 30H"

6 X 10~38

Coagulation, iron removal, corrosion

CCont.)

Table 2.5 I Typical solubility-product constants (Cont.)

ii ii i< hi

^ : !® 25(':: ■ ' ■•;

. SteiiificimwMn environmental . imiymienng and science

Fe(OH)2(i)^Fe2+ + 20ET

5 x 10"15

Coagulation, iron removal, corrosion

FeCOjts) ^ Fe2+ + CO^

2.8 X 10"l!

Iron removal and fate

FeSts)^Feî+ + S2~

5 X 10":18

Jron removal and fate; abiotic reductant

Mn(OH)3(i) ^ Mn3+ + 30H"

1 x 10~36

Manganese removal

Mn(OH)2C0 ^ Mil" + 20H"

5 X 1CT15

Manganese removal

AIPO,,!» ^ Ai3++por

3.2 X 10~2Î

Phosphate removal

Ca3(P04)2(s) ^ 3Caî+ + 2PO^~

1 X 10""

Phosphate removal

CaHP04(i) ^±Ca2+ 4- HPOf

3 X 10"7

Phosphate removal

Ca5{P04)30H(i)i±5Ca2+ + 3POÎ" + OH"

8 X 1CT33

Phosphate removal

FeP04(i)i±Fei+ + POf

1.3 X 10~22

Phosphate removal

Fe3(P04)2(j) ^ 3Fe1+ + 2PO|~

1 X 10~36

Phosphate removal

CaF2(j) Ca2+ + 2F"

3 X 10"11

Fluoridation

BaS04Cs)^Ba2+ + SOf

1 X HT110

Sulfate analysis

Thus, S = 1 X 10™3 and the solubility of BaS04(s) is 1 X 10 5 M. Similarly, for CaF2(i)

and [Ca2+][F~]2 = (S)(2Sf = Ksp = 3 X 10~" (2.57)

Thus, 5 = 1.96 X 10~4 and the solubility of CaF,0) is 1.96 X 10~4 M.

It will be noted that calcium fluoride is about 20 times more soluble than barium sulfate. From this it is obvious that the most soluble material CaF2(i) has the smallest solubility product because of the squaring of the fluoride concentration. The case of compounds that yield more than three ions is even more exaggerated.

There are two corollary statements related to the solubility-product principle, an understanding of which is basic to explaining the phenomena of precipitation and solution of precipitates. They may be expressed as follows:

1. In an unsaturated solution, the product of the molar concentration of the ions is less than the solubility-product constant, or for a species AB, [A+][B""] < iTsp.

2. In a supersaturated solution, the product of the molar concentration of the ions is greater than the solubility-product constant, or [A+][B~] > Ksp, In the former case, if undissolved AB is present, it will dissolve to the extent that

[A+][B~] = Km, and a saturated solution results. In the second case, nothing will happen until such time as crystals of AB are introduced into the solution or internal forces allow formation of crystal nuclei; then precipitation will occur until the ionic concentrations are reduced equal to those of a saturated solution.

Common Ion Effect

The advantage of relating solubilities to the equilibrium relationship is that this allows mathematical treatment of the equilibrium and prediction of the effect of adding a common ion to a solution containing a slightly soluble salt. For example, consider a solution that has been saturated with barium sulfate. As indicated in Eq- (2.56), both [Ba2+] and [SOf] would equal 1 X 1<TS M. Now, if the barium ion concentration should be increased by addition from an outside source, such as BaCI2, the concentration of sulfate ion must decrease and the amount of precipitated BaS04(j) must increase in order for Ksp to remain the same. To illustrate, assume that 10 X 10~s mol/L of BaCl2 is added to the solution saturated with barium sulfate. This will result in the formation of an additional x moles of precipitated BaS04. The following changes in [Ba2+] and [SO2"] must then take place:

IOxW1-x}

According to the solubility-product principle,

By solving for x, it is found that an additional 0.90 X 10~3 mol/L of precipitated BaS04 is formed and that the new equilibrium concentrations of barium and sulfate ions are

[Ba1+] = (11.0 X I0-5) - (0.90 X I0"5) = 3.0.1 X 10~5

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