Chemistry
Solubility Product
Solubility product is a measure of the extent to which a solid substance dissolves in water to form a saturated solution. It is represented by the equilibrium expression for the dissolution of the solid, and its value depends on the specific compound and temperature. The solubility product constant (Ksp) is used to quantify the solubility of a substance.
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10 Key excerpts on "Solubility Product"
eBook - PDFSolution Chemistry
Minerals and Reagents
- Valeria Severino(Author)
- 2019(Publication Date)
- Arcler Press(Publisher)
The ability to perform appropriate calculations, based on all available physicochemical knowledge, in accordance with the basic laws of matter conservation, deepens our knowledge of the relevant systems. At the same time, it produces the ability to acquire relevant knowledge in an organized manner—not just imitative, but focused on heuristics. This viewpoint is in accordance with constructivist teaching, based on the belief that learning occurs, as learners are actively involved in a process of meaning and knowledge construction, as opposed to passively receiving information [19]. Solution Chemistry: Minerals and Reagents 4 DEFINITIONS AND FORMULATION OF Solubility ProductS The K sp value refers to a two-phase system where the equilibrium solid phase is a sparingly soluble precipitate, whose K sp value is measured/ calculated according to defined expression for the Solubility Product. This assumption means that the solution with defined species is saturated against this precipitate, at given temperature and composition of the solution. However, often a precipitate, when introduced into aqueous media, is not the equilibrium solid phase, and then this fundamental requirement is not complied, as indicated in examples of the physicochemical analyses of the systems with struvite MgNH 4 PO 4 [20, 21], dolomite CaMg(CO 3 ) 2 [22, 23], and Ag 2 Cr 2 O 7 . The values of Solubility Products K sp (usually represented by solubility constant pK sp = − log K sp value) are known for stoichiometric precipitates of A a B b or A a B b C c type, related to dissociation reactions: (1) (2) where A and B or A, B, and C are the species forming the related precipitate; charges are omitted here, for simplicity of notation. The Solubility Products for more complex precipitates are unknown in the literature. The precipitates A a B b C c are known as ternary salts [24], e.g., struvite, dolomite, and hydroxyapatite Ca 5 (PO 4 ) 3 OH .
eBook - ePubEnvironmental Engineering
Principles and Practice
- Richard O. Mines, Jr.(Authors)
- 2014(Publication Date)
- Wiley-Blackwell(Publisher)
2.7 Solubility (Solubility Product) So far, we have dealt with aqueous solutions in which the chemical species are highly soluble. In this section, our focus will be on liquid-solid species that are partially soluble or insoluble. All solids, no matter how seemingly insoluble, are soluble to some degree. When a solid is placed in water, the ions at the surface of the solid will migrate into the water. This is called dissolution. Simultaneously, ions in the solution will be redeposited on the surface of the solid; this is known as precipitation. Equilibrium will be reached between the crystals of the compound in the solid state and its ions in solution. In general, the solubility of most compounds increases with increasing temperature. Snoeyink & Jenkins (1980, page 251) indicate that the solubilities of and do not increase as temperature increases. Equation (2.112) shows the general equation of a solid compound dissolving in pure water to form its constituent ions. 2.112 The equilibrium expression is written as follows: 2.113 As described by Sawyer & McCarty (1994, page 37), at equilibrium or saturation, the surface area of the solid is the only portion that is in equilibrium with the ions in solution. Therefore, the concentration of solid as represented by in the denominator of Equation (2.113) can be considered a constant in equilibrium solubility problems. Equation (2.114) is rewritten to show the development of the solubility-product constant, 2.114 2.115 When the solution is saturated or at equilibrium. When the solution is under-saturated and no solids species are present. When the solution is super-saturated and solid species are being formed. The solubility-product constants for several solids of significance in environmental engineering are presented in Table 2.16. Partially soluble salts have small values, while soluble salts have relatively large values
eBook - PDFChemistry
An Atoms First Approach
- Steven Zumdahl, Susan Zumdahl, Donald J. DeCoste, , Steven Zumdahl, Steven Zumdahl, Susan Zumdahl, Donald J. DeCoste(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
It is very important to distinguish between the solubility of a given solid and its Solubility Product. The Solubility Product is an equilibrium constant and has only one value for a given solid at a given temperature. Solubility, on the other hand, is an equilibrium position. In pure water at a specific temperature a given salt has a particu- lar solubility. On the other hand, if a common ion is present in the solution, the solubility varies according to the concentration of the common ion. However, in all cases the product of the ion concentrations must satisfy the K sp expression. The K sp values at 258C for many common ionic solids are listed in Table 15.1. The units are customarily omitted. Solving solubility equilibria problems requires many of the same procedures we have used to deal with acid–base equilibria, as illustrated in Examples 15.1 and 15.2. Calculating K sp from Solubility I Copper(I) bromide has a measured solubility of 2.0 3 10 24 mol/L at 258C. Calculate its K sp value. SOLUTION In this experiment the solid was placed in contact with water. Thus, before any reaction occurred, the system contained solid CuBr and H 2 O. The process that occurs is the dissolving of CuBr to form the separated Cu 1 and Br 2 ions: CuBrssd Δ Cu 1 saqd 1 Br 2 saqd where K sp 5 fCu 1 gfBr 2 g Initially, the solution contains no Cu 1 or Br 2 , so the initial concentrations are fCu 1 g 0 5 fBr 2 g 0 5 0 The equilibrium concentrations can be obtained from the measured solubility of CuBr, which is 2.0 3 10 24 mol/L. This means that 2.0 3 10 24 mole of solid CuBr dissolves per 1.0 L of solution to come to equilibrium with the excess solid. The reac- tion is CuBrssd ¡ Cu 1 saqd 1 Br 2 saqd Pure liquids and pure solids are never included in an equilibrium expression. K sp is an equilibrium constant; solubility is an equilibrium position.
eBook - PDFChemistry
The Molecular Nature of Matter
- Neil D. Jespersen, Alison Hyslop(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
After a period of time (shown in beaker B), due to the equilibrium that exists between the solid and the solution, the smaller piece of the solid slowly disappears and the larger piece increases in size. Through the dynamic equilibrium between the solid and the solution, the particles that dissolve are more likely to encounter and precipitate on the larger solid, rather than the smaller solid. the larger piece of silver chloride and precipitate there, in a process called Oswald ripening. The smaller piece will become smaller as more ions dissolve into solution, and the larger piece will increase in size until the small piece disappears, as shown in Figure 17.2, Flask B. Using the procedure developed in Section 14.4, we write the equilibrium law as shown in Equation 17.1, omitting the solid from the mass action expression. [ Ag + ][ Cl − ] = K sp (17.1) The equilibrium constant, K sp , is called the Solubility Product constant (because the system is a solubility equilibrium and the constant equals a product of ion concentrations). TOOLS Solubility Product constant, K sp AgCl Ag + Na + Cl – 36 Cl – A B C AgCl Ag 36 Cl AgCl Ag + Time Cl – A B 850 CHAPTER 17 Solubility and Simultaneous Equilibria It’s important that you understand the distinction between solubility and solubility prod- uct. The solubility of a salt is the amount of the salt that dissolves in a given amount of solvent to give a saturated solution. The Solubility Product is the mathematical expression that is the product of the molar concentrations of the ions in the saturated solution, raised to appro- priate powers (see below). The solubilities of salts change with temperature, so a value of K sp applies only at the temperature at which it was determined. Some typical K sp values are listed in Table 17.1 and in Appendix C. NOTE In this case, we use the mathematical meaning of “product” rather than the chemical meaning.
eBook - ePubRapid Review of Chemistry for the Life Sciences and Engineering
With Select Applications
- Armen S. Casparian, Gergely Sirokman, Ann Omollo(Authors)
- 2021(Publication Date)
- CRC Press(Publisher)
ppm).4.12 Definition of Solubility Product Constant K
spConsider, for example, the slightly soluble salt CaF2 (s) in equilibrium with water. Its limited dissociation and ionization in water can be expressed as follows:(4.9)CaF 2( s )⇌Ca2 +( aq )+ 2F −( aq )And its Solubility Product constant,Ksp, as(4.10)K=s p[Ca]2 +[2F −]Note that [CaF2 (s)] does not appear in the Ksp expression since it represents the undissolved portion, and it is pure solid. It has no concentration.Also note that [Ca2+ ] represents the concentration of Ca2+ ion dissolved in water, while [F− ] represents the F− ion concentration dissolved in water.Note also that when a given amount of CaF2 (s) is dissolved in a known amount of water:[= 2]Ca2 +( aq )[=]F −( aq )[]CaF 2( aq )where CaF2 (aq) is the portion that is dissolved in water. In other words, whatever the proportion of CaF2 (s) that dissolves in water, the same molar concentration is present as Ca2+ (aq) ion, but twice that molar concentration is present as F− (aq) ion.The smaller the Ksp value, the lower the solubility of the salt. It follows that the Ksp value of a salt can be calculated by determining the solubility of each ion and then using Equation 4.10.4.13 Calculating the Molar Solubility from K
spThe molar solubility of any slightly soluble salt, along with the concentration of any of its ions, can be calculated from its Ksp value. The formula of the salt must be known, however, so that its dissociation and ionization can be written.It is important to note that simply comparing the Ksp
- Takashiro Akitsu(Author)
- 2017(Publication Date)
- IntechOpen(Publisher)
General Chemistry. 2nd ed. Minneapolis, MN: West; 1996. pp. 621-622 [83] Hawkes SJ. What should we teach beginners about solubility and Solubility Products? Journal of Chemical Education. 1998; 75 (9):1179-1181. DOI: 10.1021/ed075p1179 [84] Meites L, Pode JSF, Thomas HC. Are solubilities and Solubility Products related? Journal of Chemical Education. 1966; 43 :667-672 Solubility Products and Solubility Concepts http://dx.doi.org/10.5772/67840 133 [85] Šů cha L, Kotrlý S. Solution Equilibrium in Analytical Chemistry. London: Van Nostrand Reinhold Comp.; 1975 [86] Chen G, Tao D. Effect of solution chemistry on flotability of magnesite and dolomite. International Journal of Mineral Processing. 2004; 74 (1 – 4):343-357 [87] Ringbom A. Complexation in Analytical Chemistry. New York: Interscience Publishers; 1963 [88] Rossotti H. The study of ionic equilibria. An introduction. London: Longman; 1978 [89] Sillén LG, Martell AE, Stability Constants of Metal-Ion Complexes, The Chemical Society, London, 1964; Supplement No. 1, 1971. [90] Beck MT. Chemistry of Complex Equilibria. New York: Van Nostrand Reinhold Co.; 1970 [91] Janecki D, Doktór K, Micha ł owski T. Determination of stability constants of complexes of M i K j H k L type in concentrated solutions of mixed salts. Talanta. 1999; 48 :1191-1197 http:// www.sciencedirect.com/science/article/pii/S0039914098003452 [92] Janecki D, Doktór K, Micha ł owski T. Erratum to “ Determination of stability constants of complexes of M j K j H k L type in concentrated solutions of mixed salts ” [48 (1999) 1191]. Talanta. 1999; 49 :943. http://www.sciencedirect.com/science/article/pii/S0039914099001745 [93] Janecki D, Styszko-Grochowiak K, Micha ł owski T. The catenation and isomerisation effects on stability constants of complexes formed by some diprotic acids. Talanta. 2000; 52 :555-562. http://www.ncbi.nlm.nih.gov/pubmed/18968016 [94] https://en.wikipedia.org/wiki/Gibbs_free_energy [95] Zeidler E.
eBook - ePubDeveloping Solid Oral Dosage Forms
Pharmaceutical Theory and Practice
- Yihong Qiu, Yisheng Chen, Geoff G.Z. Zhang, Lawrence Yu, Rao V. Mantri(Authors)
- 2016(Publication Date)
- Academic Press(Publisher)
Eq. (1.44) . The ability of in situ salt formation thus would depend on the pK a of the acid and pK a of the conjugate acid of the base.These equations are expressed in terms of concentration, with the assumption that the activity coefficient for all the species is 1.0. This assumption may be reasonable for neutral species, but for charged species in solutions at high ionic strength, the activity coefficients are less than 1.0. Therefore, the assumption that activity coefficients are unity should be verified.49 The expressions generated here are valid regardless of whether the excess solid that is added at the beginning of the experiment is in free acid or salt form. The solubility at any given pH, therefore, is governed by pH, the intrinsic solubility of the free acid, the Solubility Product of the salt, and the concentration of the common ions. The various ionization and solid/liquid equilibria during solubilization of weak acid are shown in Scheme 1.1 . It is imperative that the solid phase be analyzed after equilibration during solubility experiments, as the solid phase exists either as the free acid (AH ) or its salt (A − M + ), except at pHmax .Scheme 1.1 Ionization and solid–liquid equilibria.Source: Drawn based on Serajuddin ATM and Pudipeddi M. Handbook of pharmaceutical salts . Wiley-VCH; 2002. p. 135–160 and Chowhan ZT. J Pharm Sci
eBook - ePub- Simon Gaisford, Mark Saunders(Authors)
- 2012(Publication Date)
- Wiley-Blackwell(Publisher)
o ). This is important, because if the drug is ionisable it will ionise to a greater or lesser extent with solution pH and this will affect the observed solubility, as discussed in Chapter 2. If the structure of the compound is known, then it should be clear whether solubility will exhibit a dependence upon pH. If the structure is not known then measuring solubility over a range of pH will show whether an ionisable moiety is present (although care must be taken when selecting the buffer to ensure salts are not accidentally formed). If the compound has an ionisable group, then modification of solubility by preparation of a salt is a possible formulation strategy (see Chapter 6).From a thermodynamic perspective, the energy input required in order to break any solute–solute interactions must be equal to the enthalpy of fusion required to melt the solid (since the same interactions must be overcome). Unlike melting, however, in the case of dissolution there is an additional change in enthalpy where solvent–solvent intermolecular interactions are broken and solute–solvent interactions are formed, termed the enthalpy of mixing (Δmix H ). The net enthalpy of solution (Δsol H ) is then the sum of the enthalpies of fusion and mixing:(4.2)The overall process of dissolution can be represented more simply as(4.3)In this case the equilibrium constant (K ) can be written as(4.4)where a denotes the activity of the drug in each phase. Since the activity of a solid is defined as unity and activity approximates to concentration (saturation in this case) in dilute solution then(4.5)where x 2 denotes the saturated concentration of drug in mole fraction units (x 1 being the mole fraction of the solvent). It should now be possible to see from Equation (4.5 ) the reason for the term equilibrium solubility noted earlier.It appears from Equation (4.2 ) that the crystal lattice energy might affect solubility. It also seems from Equation (4.1 ) that there should be an effect of temperature on solubility, since the position of equilibrium will change. Both of these effects can be explored further through the concept of ideal solubility
eBook - PDF- Allan Blackman, Steven E. Bottle, Siegbert Schmid, Mauro Mocerino, Uta Wille(Authors)
- 2022(Publication Date)
- Wiley(Publisher)
In this chapter, we will focus on the energetics of formation of solutions and the quantification of physical properties of solutions. FIGURE 10.2 A saturated solution. In a saturated solution, a dynamic equilibrium exists between the undissolved solute and the dissolved solute in the solution. A solution can be defined as a homogeneous mixture of two or more pure substances (‘homogeneous’ means that all regions of the solution have exactly the same composition). Despite the fact that we often think of a solution as involving some type of liquid, we can have gaseous solutions com- prising two or more gases (e.g. air) and even solid solutions comprising two or more solids (e.g. alloys such as brass and solder) in addition to the ‘usual’ solutions containing a gas, liquid or solid dissolved in a liquid. In the case of liquid solutions, we call the liquid the solvent, while the dissolved substance is called the solute; the solute is present in smaller amounts than the solvent. A solute is said to be soluble in a particular solvent if it dissolves completely in that solvent at the specified temperature. We define the solubility of the solute in a particular solvent as the maximum amount of the solute that dissolves completely in a given mass or volume of the solvent at a particular temperature, T, and a particular pressure, p. A saturated solution is one in which no more solute will dissolve. The most common type of saturated solution we will encounter is that illustrated in figure 10.2, in which excess solid solute is in equilibrium with its dissolved form. The process of dissolving a solute in a solvent to give a homogeneous solution is called dissolution. We therefore talk of the dissolution of a solute in a particular solvent. 10.2 Gaseous solutions LEARNING OBJECTIVE 10.2 Describe why gases mix spontaneously. We will begin our investigation of solutions by looking at mixing two gases to form a gaseous solution.
- Frank R. Foulkes(Author)
- 2012(Publication Date)
- CRC Press(Publisher)
23-6 SOLUBILITY EQUILIBRIA = ( ) ( ) a a i i + + < < + < m m 5 ( ) ( ) a a i i + + < < + < c c = [ ( )] [ ( )] a i a i i i + + < < + < s s = a a i i i i i i i + < + < + < + < u u u u s . . . [6] where i = i i + < + . The mean activity coefficient a ± for the salt is defined as a a a i i i ± + < > u ( ) + < 1 / . . . [7] Therefore, it can be seen that a a i i + < + < u = a i ± . . . [8] Substituting Eqn [8] into Eqn [6] gives K SP = a i i i i i i ± + < u u u + < s Rearranging: s i = K SP a i i i i i ± + < u u + < = 1 K SP a i i i i i ± + < £ ¤ ² ¥ ¦ ´ u + < s = 1 K SP 1 a i i i i i ± + < + < u £ ¤ ² ¥ ¦ ´ / . . . [9] Eqn [9] gives the general relationship between the solubility s of a salt and its Solubility Product K SP , corrected for nonideal behavior in the solution. The mean activity coefficient can be obtained using the Debye-Hückel equation for a complete salt: log a ± = < + < 0.5108 z z I | | . . . [10] Note that Eqn [10], the Debye-Hückel equation for a complete salt, essentially is the same as the Debye-Hückel equation for a single ion, except that the factor z i 2 has been replaced with the abso-lute value | z z + < |, which incorporates the valences of both ions. As in the case of the equation for a single ion, Eqn [10] is valid only in dilute aqueous solutions at 25°C. 5 Example 23-2 The solubility of Ag 2 SO 4 in water at 25°C is 0.030 mol L –1 . Calculate its Solubility Product, taking account of activity effects. 5 Strictly speaking, when the Debye-Hückel equation contains the factor 0.5108 , the ionic strength should be calcu-lated using molalities rather than molarities. In dilute solutions, however, the two are essentially the same. SOLUBILITY EQUILIBRIA 23-7 Solution (a) Ag 2 SO 4 A 100% 2Ag + + SO 4 2 < 0.03 M A 0.06 M 0.03 M K SP = a a Ag 2 SO 4 2 + < u = ( m ) ( m ) 2 a a + + < < = ( . ) ( . ) 0 06 0 03 2 2 a a + < = 1.08 × 10 –4 a ± 3 I = 1 2 c z c z 2 2 + + < < + [ ] = 1 2 [(0.06)(+1) 2 + (0.03)(–2) 2 ] = 0.09 log a ± = – 0 5108 .
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