chemical principles revisited
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MURIELBISHOP Clemson University Clemson,SCZ9634
Chemical Equilibrium VIII. Precipitates Adon A. Gordus The University of Michigan, Ann Arbor, MI 48109 I n this last article in the series' on chemical equilibrium we will consider the calculation of solubilities in water and in buffered solutions. We will focus on certain ionic precipitates in water: selected salts of strong and weak acids and selected hydroxides. We consider buffered solutions because their constant pH allows easier calculation of solubiltities. Even for examples of precipitates in water, we keep the algebra relatively simple by making various assumptions. We assume that no secondary species form from the atoms in the precipitate, and we limit the discussion concerning secondam reactions to those concerninz the generation of hydroxidk ion. We calculate solubilities~only&hen the hydroxide ion due to the precipitate or to the water equilibrium or to both is negligible.We will use molarity as the unit of concentration in writing the solubility product constant K., If the solution is nonideal, then the solubility product constant that uses concentration is expressed as
where Kt is the thermodynamic solubility product constant (given in activity units) and Kf is the activity coefficient constant? If the solution is ideal K f = 1.00 andKsp =Kt
simple asCaSOI will yield not only Ca2-and SO:-' but also CaSOdaa~in solution?Others. such a s AKI. will vield not only the ;ndissociated ion pa& Agcl(aq)lbut also ?I- complexes: [AgC121-, [AgCl3I2; [AgClal", etc. The concentrations of these secondary species are oRen small compared with the concentrations of the principal ions that arise from the precipitate. Thus, introductory texts nsuallv omit them. We too will omit them because considering them is beyond the scope of most introductory texts and lecture oresentations. However. it cannot be emphasized too strbngly that neglecting these secondary species could introduce serious errors. If we assume that there are no secondary species of the type discussed above (complexes or undissociated dissolved molecules) or commbn ions. and if the ions from the orecioitate do not undergo hydklysis (or if i t is assume'd that hydrolysis reactions are unimportant), then the solubility product can be written in terms of the solubility in water, s in molediter (mol/L), of the precipitate. For hydroxide precipitates we will further assume that we can disregard the OH-due to the water equilibrium. Then the expressions relating K, and s will depend on the stoichiometry of the solubility process. If the formula for the precipitate is AB, and the positive charge on A is equal in value, but opposite in sign, to the negative charge on B, then when s m o m of AB dissolve, they produce s m o m of An+and s m o m of B"-. Thus Ksp = s2
Simplifying Assumptions
No Secondary Species Almost all saturated solutions conuiin species in addition to the ions of the dissolved precipitate. One of the most common is the undissociated ion pair. Even a substance a s I The previous articles are: Gordus, A. A. "I. Thermodynamic EquilibriurnConstant'J.Chem. Educ.1991, 68, 13&140;"11. Deriving an Exact Equilibrium Equation" J. Chem. Educ. 1991, 68, 215-217; "Ill. A Few Math Tricks' J. Chem. Educ. 1991,68,291-293:"IV.Weak Acids and Bases'' J. Chem. Educ. 1991, 68,566567; "V. Seeing an Endpoint in Acid-BaseTitrations" J. Chem. Educ. 1991, 68,556668; "VI. Buffer Solutions'' J. Chem. Educ. 1991, 68, 656458; "VII. pH Approximations in Acid-Base Titrations" J. Chem. Educ 1991, 68,
If the formula for the precipitate is AB2 or A2B, then the ion of the atom that appears once in the molecule will have concentration s, whereas the ion of the atom that appears twice will have a concentration of 2s. Since the ion that appears twice has its concentration squared in the solnbility product
If the formula for the precipitate is &B2, when s m o m of precipitate dissolve
759-761 . -.. .. .
Refer to article I of footnote 1. See, for example, the letter by J. H. Carpenter in J. Chem. Educ. 1989, 66, 184, and references therein. The algebra relating to the solubilitv of CaSOd when all secondary reactions as well as activity coefficients are taken into account is described by Meites. L.; ode, J. S. F.: Thomas. H. C. J. Chem. Educ. 1966.43.667-672.It is not a trivial problem.
so that K , = [ ~ ~ + l =~( 3[. ~~) ~ (=7 %108s5 ~ )~
Similar expressions relating the solubility product to s can be derived for other formulas, such as A3B or AB3 (K, = 27s4),ABC (K, = s3), etc. Volume 68 Number 11 November 1991
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Relatively InsolubleHydroxides. The amount of OHproduced when the precipitate of these compounds dissolves is less than 1.0 x 10.' M. The OH- from the precipitate can be neglected compared with the OH- from the water eq~ilibrium,~ and the [OH] term in the K, expresM. sion is set equal to 1.00 x Almost all positive trivalent metal ion hydroxides M(OH)3 fall into this group. The solubility s for a M(OH)3 precipitate is thus given by K, = (s)(l.OOx d l 3 = 1.00 x lo-% We can specify a general requirement for the value forK, of these M(OHI3 precipitates. Since we require6 that [OH-] from the precipitate be less than 1.0 x 104M and since this source of OH- equals 3s, [M3+1= s must be less than one-third of the OH- from the precipitate or 0.33 x 104 M. Thus K, < (0.33 x 104)(1.00x 1 0 Y = 3.3 x
Most M(OH13precipitates have a K, less than 3.3 x lo3' and thus fall into this group. The corresponding requirement for M(OHIZprecipitates is that the K, be less than 5.0 x
When the calculations become complicated, it is usually because the degree of hydrolysis cannot be specified until the entire set of equations for the simultaneous equilibria is solved. There are, however, circumstances in which one of the variables is specified. As a result, the calculations bewme much easier. This occurs, for example, when the solution is buffered and the pH-and thus the concentration of hydroxyl anion-is fixed. (We assume that the buffer mixture does not contain ions that react with the anions or cations from the precipitate.) Hydroxide precipitates are rather trivial cases because the buffer [OH]-is used directly in the K, equation to solve for the metal ion concentration, which is the solubility. If the precipitate is the salt of a weak acid so that the anion undergoes hydrolysis, we can then calculate, for any pH, the fraction F of the anion (from the precipitate) that will exist in the buffered solution as that specific anion8 For instance, when Zm(P04)~dissolves the PO:- it will undergo hydrolysis according to the reactions
Thus K, = 1.00x
lo-%
Of the common M(OH12 precipitates only two have K, values that small: P~(oH),
(K, = 1
Pt(OHh
(K, = 1 x
OtherHydroxides. These are hydroxides for which the water equilibrium must be considered, but that do not fall into either of the first two groups. Of the common hydroxide precipitates, only two fall into this category. BdOHh (KSp= 7 x lo-") ( K, = 6 x lo3') CdOH), Summary. Almost all common M(OH)z precipitates have K, values greater than 5.0 x lo-", and thus are sufficiently soluble that K, = 4s3 can be used to calculate the solubility. Only Pd(OHIZand Pt(OH)z among the common M(OH)? recipitates are sufficientlyinsoluble that Ksp = 1.00 x 10-P4 s can be used to calculate the solubility. Almost all common M(OHh precipitates have K, values and thus aresufficientlyinsoluble that less than 3 . 3 lo3', ~ K, = (1.00 x 10"')s can be used to calculate the solubility. Of the common precipitates, only Be(0H)z and Cr(OHI8 do not fall in any of the previous categories and require consideration of the water equilibrium when performing the solubility calculation.
The more acidic the solution (i.e., the lower the OHconcentration), the larger the fraction of the original POP that has been converted to the higher acidic forms. This principle is used to dissolve salts of weak acids: Their solubility can be increased by acidifying the solution. For a triprotic acid H3X, the fraction of the total X in solution that is present as X3- is8 F3=
K'K2K3 [H+13+ K ~ [ H ++IKIKz[Wl ~ + K1KS3
(1) whereKl, Kz, audK3are the acid dissociation constants for the H3X acid and [H+Iis obtained from the pH of the buffer solution. This equation can be used, for example, to solve for the solubility of Zna(PO& in a solution buffered to pH 5.00. When s m o m of Zn3(PO& dissolve, 3s mol1L of Zn2+and a total of 2s m o m of phosphate will be produced. Due to hydrolysis, the phosphate will be present in all four forms
The fraction of phosphate present as PO:- will equal F3 as given by eq 1, so that a t equilibrium
Buffered Solutions As seen above, hydrolysis can play an important role in the calculation of the solubilities of precipitates in water. Using for H3P04the values
.~
See. for examole. ~. Skoaa. ~ ~D. - -A.:. West. D. M.: Holler. J. A. '~naiyiical~nemstry", 5tn ea.: Sa~noersCo ege: ~ d YwO ; ~ : 1988: p 227-228,or the reference n footnote 5, pp 11 1-1 12. Tne notat on usLally Jsed Involves des gnat ng tne s~bscrpt to naicate thecnarge on the species. Thus, for an H X acid, Fois the fraction of X present and F3 as X3-. FO+ F, + F2 + F3 = as H3X; F1 as H2X-: F2 as HX~-: nn ,."".
and for the buffer [H+l= 1.0 x 10"
4
'There is areat unceltaintv in the literature about the exact value of KO about 10-13 to- 10-19. There - - - - ranbe . .are -. < for - H , S ; T ~-~values - - from a so ~ncefiinttes abou the K, va Jes of some nso Lole s~lfioes. Thus, the exam ca.c~lat~on of me F, factor for h,S ana me re% tan1 w~lldepena mardeo y on the val~esof K~sea. so ~oil~ty ~
~
Thus, we get F3 = 3.0 x lo-'' That is, only 3.0 x lWIO of the total phosphate from the precipitate will be present as POP in a solution buffered to Volume 68 Number 11 November 1991
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pH = 5.00. K, = 9.1 x loA3for Zn3(P04)2.SolvingKSpfor the solubility in this buffered solution gives s = 9.9 x lo4 mob%
Expressions analogousto eq 1can be obtained for diprotic and monoprotic acids. For diprotic acid H2X,the fraction of X present as X2-, denoted now as Fz, will be F2=
K1Kz IH+lZ+ KIIH+I+ K,K2
(2)
For monoprotic acid HX,the fraction of X present as X-, now denoted as F1, will be K F,=IH+I+K (3) The solubilities of HgS, CuS, and other very insoluble salts of very weak acids are calculated in a manner similar to that used for the Zn3(P04)2.The solution is not buffered
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Journal of Chemical Education
in the usual sense: Due to the very small amount of extra OH- from the hydrolysis of the sulfide, the pH will equal 7.00. Thus, F2 for HzS can be calculated fmm eq 2 using [Ht1 = 1.0 x 10-7M and K, = (s)(sFz)= s2Fzfor these MS sulfides. For very insoluble MzS sulfides such as Ag2S,the equation is
For sulfides such as BizS3
These calculations for the solubility of sulfides are usually done incorrectly in texts. When the Fz factor is included, the calculated solubility is usually much greatepthan the calculated solubility when F 2 is omitted.
