Chemistry
Solubility Equilibria
Solubility equilibria refer to the dynamic balance between the dissolution and precipitation of a solute in a solvent. This equilibrium is governed by the solubility product constant, which is a measure of the solute's tendency to dissolve. Understanding solubility equilibria is crucial for predicting the formation of precipitates and determining the solubility of various compounds in solution.
Written by Perlego with AI-assistance
Related key terms
1 of 5
10 Key excerpts on "Solubility Equilibria"
eBook - PDF- Douglas Skoog, Donald West, F. Holler, Stanley Crouch, Douglas Skoog(Authors)
- 2021(Publication Date)
- Cengage Learning EMEA(Publisher)
What we mean is that the very small amount that does go into solution dissociates completely. ❯ Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 172 CHAPTER 7 Aqueous Solutions and Chemical Equilibria Using Equation 7-7, write K 5 3 Ba 21 43 IO 2 3 4 2 3 Ba 1 IO 3 2 2 1 s 24 The denominator represents the molar concentration of Ba 1 IO 3 2 2 in the solid, which is a phase that is separate from but in contact with the saturated solution. The con- centration of a compound in its solid state is, however, constant. In other words, the number of moles of Ba 1 IO 3 2 2 divided by the volume of the solid Ba 1 IO 3 2 2 is constant no matter how much excess solid is present. Therefore, the previous equation can be rewritten in the form K 3 Ba 1 IO 3 2 2 1 s 24 5 K sp 5 3 Ba 21 43 IO 2 3 4 2 (7-13) where the new constant is called the solubility-product constant or the solubility product. It is important to appreciate that Equation 7-13 shows that the position of this equilibrium is independent of the amount of Ba 1 IO 3 2 2 as long as some solid is present. In other words, it does not matter whether the amount is a few milligrams or several grams. A table of solubility-product constants for numerous inorganic salts is found in Appendix 2. The examples that follow demonstrate some typical uses of solubility- product expressions. Further applications are considered in later chapters. The Solubility of a Precipitate in Pure Water With the solubility-product expression, we can calculate the solubility of a sparingly soluble substance that ionizes completely in water.
eBook - PDF- Edward J. Neth, Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
- 2019(Publication Date)
- Openstax(Publisher)
Note that pure fluorite is colorless, and that the color in this sample is due to the presence of other metal ions in the crystal. CHAPTER OUTLINE rates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An understanding of the factors affecting compound solubility is, therefore, essential to the effective management of these processes. This section applies previously introduced equilibrium concepts and tools to systems involving dissolution and precipitation. The Solubility Product Recall from the chapter on solutions that the solubility of a substance can vary from essentially zero (insoluble or sparingly soluble) to infinity (miscible). A solute with finite solubility can yield a saturated solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For example, a saturated solution of silver chloride is one in which the equilibrium shown below has been established. In this solution, an excess of solid AgCl dissolves and dissociates to produce aqueous Ag + and Cl – ions at the same rate that these aqueous ions combine and precipitate to form solid AgCl ( Figure 15.2). Because silver chloride is a sparingly soluble salt, the equilibrium concentration of its dissolved ions in the solution is relatively low. FIGURE 15.2 Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag + and Cl – ions in equilibrium with undissolved silver chloride. The equilibrium constant for Solubility Equilibria such as this one is called the solubility product constant, K sp , in this case Recall that only gases and solutes are represented in equilibrium constant expressions, so the K sp does not include a term for the undissolved AgCl.
- Henry Freiser, Monika Freiser(Authors)
- 1992(Publication Date)
- CRC Press(Publisher)
The solubility product equilibrium just described is much more useful when applied to systems of slightly soluble electrolytes. With BaS0 4 , for example, which has a solubility of 1 .0 x 10-5 mol lL in water, we may write, as we did for NaCl: Chapter I) Precipitation Equilibria 1 03 (BaSO 4 )s ' Ba 2 + + SO 4 2 -(6-3) In the remainder of this chapter, concentration constants, Ks p , calculated from * Ks p by methods described in Chapter 3, will be employed. These will vary with ionic strength (see Equation 3-1 5). Factors That Affect the Solubility Product Constant The true or thermodynamic solubility product constant Ksp will, for most substances, increase with the temperature by an amount that depends on the heat of solution. The addit ion of any organic solvent such as alcohol to an aqueous solution will generally result in a lower * Ks p ' This may be easily understood in terms of the increased work of separation of ions in a mediu m orxlower dielectric constant. The particle size of the undissolved solid in equilibrium with the dissolved solute can also affect the solubility and solubility product constant. If the particle size is sufficiently small, the surface area per mole becomes large enough to require taking the surface energy into account in describing the equilibri um. In effect, the smaller the particle size (i.e., smaller than 1 0-4 em radius), the higher the solubility. For example, the molar solubil ity of lead chromate having particles of 9 x 1 0-5 cm radius was found to be 2 . 1 x 1 0-4 in contrast to a value of 1 .24 x 1 0-4 for particles having a radius of 3 .0 x 1 0-3 em or larger. In this case, the Ks p has changed by 250%. There are proba bly cases in which the effect of particle size on Ks p is much larger, but these are exceedingly difficult to verify experimentally. Calculations involving precipitation equil ibria involve yet another difficulty, namely the slow equilibrat ion of the solid phase with the solution.
eBook - PDFChemistry
Principles and Reactions
- William Masterton, Cecile Hurley(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 388 CHAPTER 15 Complex Ion and Precipitation Equilibria ▼ From Example 15.1 we see that the reaction for the conversion of Ag(NH 3 ) 2 1 to Ag(S 2 O 3 ) 2 3 2 has a large equilibrium constant. Looking at it in a slightly dif-ferent way, we can say that of these two complexes of Ag 1 , the Ag(S 2 O 3 ) 2 3 2 ion is the more stable. A similar situation applies to two complex ions of Co 3 1 : Co(NH 3 ) 6 3 1 ( aq ) 1 Cl 2 ( aq ) EF Co(NH 3 ) 5 Cl 2 1 ( aq ) 1 NH 3 ( aq ) K 5 K f Co s NH 3 d 5 Cl 2 1 K f Co s NH 3 d 6 3 1 5 2 3 10 28 1 3 10 23 5 2 3 10 5 Addition of Cl 2 ions to the complex Co(NH 3 ) 6 3 1 converts it to the more stable complex Co(NH 3 ) 5 Cl 2 1 , which has a pink color (Figure 15.1). 15-2 Solubility; Solubility Product Constant ( K sp ) In Chapter 4, we considered a compound soluble if it dissolved in water and insoluble if it did not. In this section, we look at the dissolution process in the context of an equilibrium existing between the component ions of an ionic solid and the solid itself. Consider SrCrO 4 in water (Figure 15.2). Much like the weak acid HF in water, HF( aq ) EF H 1 ( aq ) 1 F 2 ( aq ) an equilibrium is established in solution between the compound (SrCrO 4 ) and its corresponding ions (Sr 2 1 and CrO 4 2 2 ). The net ionic equation representing this equilibrium is SrCrO 4 ( s ) EF Sr 2 1 ( aq ) 1 CrO 4 2 2 ( aq ) K sp Expression The rules in Chapters 12 and 13 make it possible to write the equilibrium con-stant expression for the dissolving of SrCrO 4 . In particular, the solid does not appear in the expression; ▲ the concentration of each ion is raised to a power equal to its coefficient in the chemical equation. K sp 5 [Sr 2 1 ][CrO 4 2 2 ] The symbol K sp represents a particular type of equilibrium constant known as the solubility product constant .
eBook - PDF- Young, William Vining, Roberta Day, Beatrice Botch(Authors)
- 2017(Publication Date)
- Cengage Learning EMEA(Publisher)
AlbertSmirnov/iStockphoto.com 19 Precipitation and Lewis Acid–Base Equilibria Unit Outline 19.1 Solubility Equilibria and K sp 19.2 Using K sp in Calculations 19.3 Lewis Acid–Base Complexes and Complex Ion Equilibria 19.4 Simultaneous Equilibria In This Unit… This is the third unit in our study of chemical equilibria. After studying acid–base equilibria in some depth, we now turn to equilibria involv-ing sparingly soluble compounds and the equilibria of Lewis acid–base complexes. You were first introduced to sparingly soluble compounds in Chemical Reactions and Solution Stoichiometry (Unit 9) , when we cov-ered precipitation reactions and the solubility of ionic compounds. We will discover in this unit that even insoluble ionic compounds dissolve in water to a small extent and that this solubility can be affected by a variety of chemical species, including Lewis bases. Lewis acids and bases were briefly introduced in Acids and Bases (Unit 17). In this unit we take a closer look at equilibria involving Lewis acid–base complexes and how they can be used to influence the solubility of ionic compounds. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 Unit 19 Precipitation and Lewis Acid–Base Equilibria 608 19.1 Solubility Equilibria and K sp 19.1a Solubility Units This unit is about the chemistry of binding ions and molecules together. The tendency of different species to bind together and our ability to control these binding processes is impor-tant. For example, biological systems are rife with examples of precipitation chemistry. The formation of seashells and coral involves precipitation of calcium carbonate, CaCO 3 . The for-mation of caves and the interesting structures within them is the result of a combination of precipitation reactions coupled with Lewis acid–base reactions and Brønsted–Lowry acid– base reactions.
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 - PDF- L. Pataki, E. Zapp, R. Belcher, D Betteridge, L Meites(Authors)
- 2013(Publication Date)
- Pergamon(Publisher)
Thus the metal ions are hydrated in their solutions. In conse-quence, complex formation of a metal ion with another ligand can be described by reactions such as: M 2 +(H 2 0) 4 + Y 4 - ;± MY 2 + 4H 2 0. Accordingly, chelate formation tends to increase the disorder of the system (two entities converted to five entities), that is, the entropy is increased. As this causes a decrease in enthalpy, it also increases the stability of chelates. CHEMICAL EQUILIBRIA IN SOLUTION 53 1.4 Precipitate formation reactions 1.4.1 Solubility of precipitates, solubility product Depending on their nature, substances have different solubilities, and at a given temperature only a limited amount of them can be dissolved in a given solvent. The solid phase and the solution saturated with the solute are in dynamic equilibrium, that is the amount dissolved per unit time is equal to the amount precipitated in this time. The equilibrium can be characterized by the concentration of the saturated solution. This value is called the solubility and denoted by S. In general, analytical precipitates are salts, or other ionic compounds with low solubilities, which dissociate in their solutions. Thus, for a saturated solution in contact with the solid phase BA, the following simultaneous equilibria apply: BA(s) ;± BA(soln.) ;± B+ + A. For the second, homogeneous equilibrium, the law of mass action holds: J2ìli*3-*, wo) [BA](soln.) As the solution is in equilibrium with the solid phase, [BA](soln.) = K'. Substituting this constant into equation (130) gives: [ B + JI A ] = K d , that is [B + ][A-] = K s (131) K. where K s is the solubility product. In general, for compound B^AQ, K s = [B+HA-]«. (132) According to this equation the product of the ionic concentrations raised to the appropriate powers, in the saturated solution of a poorly soluble precipitate is constant at a given temperature.
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)
CHAPTER 21 CHAPTER 15 15.1 Solubility Equilibria and the Solubility Product Relative Solubilities Common Ion Effect pH and Solubility 15.2 Precipitation and Qualitative Analysis Selective Precipitation Qualitative Analysis 15.3 Equilibria Involving Complex Ions Complex Ions and Solubility Solubility and Complex Ion Equilibria Stalactites and stalagmites. These formations are created when carbonate minerals dissolve in groundwater acidified by carbon dioxide and then solidify when the water evaporates. (Jason Patrick Ross/Shutterstock.com) 619 Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. M ost of the chemistry of the natural world occurs in aqueous solution. We have already introduced one very significant class of aqueous equilibria, acid–base reactions. In this chapter we will consider more applications of aqueous equilibria, those involving the solubility of salts and those involving the formation of complex ions. The interplay of acid–base, solubility, and complex ion equilibria is often important in natural processes, such as the weathering of minerals, the uptake of nutrients by plants, and tooth decay. For example, limestone (CaCO 3 ) will dissolve in water made acidic by dissolved carbon dioxide: CO 2 saqd 1 H 2 Os l d Δ H 1 saqd 1 HCO 3 2 saqd H 1 saqd 1 CaCO 3 ssd Δ Ca 21 saqd 1 HCO 3 2 saqd This two-step process and its reverse account for the formation of limestone caves and the stalactites and stalagmites found therein.
- Takashiro Akitsu(Author)
- 2017(Publication Date)
- IntechOpen(Publisher)
Students will trust us enough to believe that a calculation we have taught must be generally useful. The theory of electrolytic systems, perceived as the main problem in the physicochemical studies for many decades, is now put on the side. It can be argued that the gaining of quantitative chemical knowledge in the education process is essentially based on the stoichi-ometry and proportions. Overview of the literature indicates that the problems of dissolution and solubility calculation are not usually resolved in a proper manner; positive (and sole) exceptions are the studies and practice made by the authors of this chapter. Other authors, e.g., [13, 85], rely on the simplified schemes (ready-to-use formulas), which usually lead to erroneous results, expressed by disso-lution denoted as s * [mol/L]; the values for s * are based on stoichiometric reaction notations and expressions for the solubility product values, specified by Eqs. (1) and (2). The calculation (a) (b) -4 -3.8 -3.6 -3.4 -3.2 -3 -2.8 -2.6 -2.4 -2.2 0 5 10 15 20 log(s) V stage 3 -4 -3.8 -3.6 -3.4 -3.2 -3 -2.8 -2.6 -2.4 -2.2 0 5 10 15 20 log(s) V stage 4 Figure 11. Solubility s of CuI within stage 3 (a) and stage 4 (b). Solubility Products and Solubility Concepts http://dx.doi.org/10.5772/67840 125 of s * contradicts the common sense principle; this was clearly stated in the example with Fe ( OH ) 3 precipitate. Equation (27) was applied to struvite [50] and dolomite [86], although these precipitates are nonequilibrium solid phases when introduced into pure water, as were proved in Refs. [20 – 23]. The fact of the struvite instability was known at the end of nineteenth century [49]; nevertheless, the formula s * = ( K sp ) 1/3 for struvite may be still encountered in almost all textbooks and learning materials; this problem was raised in Ref. [15]. In this chapter, we identified typical errors involved with s * calculations, and indicated the proper manner of resolution of the problem in question.
eBook - PDFChemistry
The Molecular Nature of Matter
- Neil D. Jespersen, Alison Hyslop(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
Using the same argument, but for acid salts, adding a base will increase the amount that dissolves. Simultaneous Equilibria Simultaneous equilibrium describes a situation where two or more equilibrium reactions are occurring at the same time in a given mixture. In those cases, the mathematical rela- tionships of all equilibrium equations must be satisfied at the same time. The way to do this is to combine the chemical equations to obtain an expression of the overall reaction and then use our concepts for manipulating equilibrium constants (Section 14.2) to deter- mine the value of the equilibrium constant. Practice Exercise 17.11 Practice Exercise 17.12 17.2 | Solubility of Basic Salts Is Influenced by Acids 829 We can treat the simultaneous equilibria of a slightly soluble salt and a weak acid by adding equations 17.2 and 17.3 to get the overall reaction FePO 4 (s) + H + (aq) m Fe 3+ (aq) + HPO 4 2- (aq) (17.4) The equilibrium constant for Equation 17.2 is the K sp and for Equation 17.3 it is 1 K a for the weak acid HPO 4 2- , which is the third ionization reaction for H 3 PO 4 , or K a 3 . The overall equilibrium constant and mass action expression for Equation 17.4 is K overall = K sp 1 K a 3 = 3 Fe 3+ 4 3 HPO 4 2- 4 3 H + 4 = 9.9 Ž 10 -16 a 1 4.2 Ž 10 -13 b = 2.3 Ž 10 -3 Solving this equation for the solubility of FePO 4 is beyond the scope of this book. However, the multi-concept problem that follows shows that solutions for acidic solvents can be obtained if we know the concentration of the acid beforehand. ■ We are adding Equation 17.2 and 17.3 so we multiply the K sp and K a 3 . Analyzing and Solving Multi-Concept Problems W hat is the solubility of CaF 2 in a 1.50 M solution of nitric acid? Analysis: Even a quite simply stated question can be a multi- concept problem as we’ll see. In this situation (1) the salt has to be classified as neutral, acidic, or basic. (2) We need to identify all the equilibrium reactions that occur and combine them appropriately.
Index pages curate the most relevant extracts from our library of academic textbooks. They’ve been created using an in-house natural language model (NLM), each adding context and meaning to key research topics.
