Solubility Product Calculations

12 Key excerpts on "Solubility Product Calculations"

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  eBook - ePub

  Rapid 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

  sp

  Consider, for example, the slightly soluble salt CaF2 (s) in equilibrium with water. Its limited dissociation and ionization in water can be expressed as follows:

  CaF 2

  ( s )

  ⇌

  Ca

  2 +

  ( aq )

  + 2

  F −

  ( aq )

  (4.9)

  And its solubility product constant,K

  sp

  , as

  K

  s p

  =

  [

  Ca

  2 +

  ]

  [

  F −

  ]

  2

  (4.10)

  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:

  [

  Ca

  2 +

  ( aq )

  ]

  = 2

  [

  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

  sp

  The 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

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  General Chemistry for Engineers

  • Jeffrey Gaffney, Nancy Marley(Authors)
  • 2017(Publication Date)
  • Elsevier

    (Publisher)

  y is;

  A a

  B b

  s ⇄ a

  A

  x +

  aq + b

  B

  y –

    (10)

  The general expression for the solubility product constant for this dissociation is;

  K sp

  =

  A

  x +

  a

  B

  y −

  b

    (11)

  Example 12.4: Determining the K sp From Solubility Data

  Calculate the solubility product constant for lead(II) chloride if 50.0 mL of a saturated solution contains 0.2207 g of lead(II) chloride.

  1.  

  Write the chemical equation and the K sp expression for the dissolution of lead(II) chloride.

  Chemical equation: PbCl2 (s) → Pb2 + (aq) + 2Cl− (aq)

  Solubility product constant: K sp  = [Pb2 + ][Cl− ]2

  2.  

  Determine the concentration of PbCl2 .

  0.2207 g

  278.1 g / mol

  = 7.936 ×

  10

  – 4

  mol PbCl

  7.936 ×

  10

  – 4

  mol

  0.050 L

  = 0.0159 M

  PbCl 2

  3.  Determine the concentrations of the ions in solution.

  Since, after dissociation, there are two chloride ions for every lead ion:

  [Pb2 + ] = 0.159 M

  [Cl– ] = 2 × 0.159 M = 0.0318 M

  4.  

  Calculate the Ksp .

  K sp  = [Pb2 + ][Cl− ]2  = (0.159)(0.038)2   = 1.61 × 10− 5

  Since the solubility product constant is an equilibrium constant of a dissolution reaction, a solubility product quotient (Q sp ), similar to the reaction quotient described in Section 7.5 , would be the value of the expression for the K sp in Eq. (11) with concentrations at any point in the reaction. If the product of the ionic concentrations in Eq. (11)

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  Environmental 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

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  Concepts & Calculations in Analytical Chemistry, Featuring the Use of Excel

  • Henry Freiser, Monika Freiser(Authors)
  • 1992(Publication Date)
  • CRC Press

    (Publisher)

  This represents an application of the general criteria for predicting the direction of reactions and was developed earlier. For the purposes of this discus-sion, it will be convenient to call the product of the concentrations of the ions each raised to the appropriate power, i.e ., the right-hand side of the solubility product expression, Ks p = [M] a [X] h , which is the ion product. The ion product would be the actual value (not the equilibrium value), assuming that no precipitation occurred. The following relations will then be seen to apply. Ion product < Ks p : Solution is unsatur ated. No precipitate will form and precipitate present will dissolve. Chapter 6 Precipitation Equilibria 105 Ion product > Ks p : Ion product = Ks p : Solution is supersaturated. In time a precipitate will form, and precipitate present will not dissolve. Solution is saturated. In this equilibrium mixture, no precIpItate will form and precipitate present will not dissolve. Quantitative Relationship Between Solubility and Solubility Product It may readily be shown that the Ks p is not the solubility, but is, of course, related to it. In a saturated solution ofBaS0 4 of1.0 x 10-5 , molarity, the concentration of Ba ++ and of SO= 4 is each 1.0 x 10-5 , since BaS0 4 is completely dissociated. Hence, Ks p = 1.0 x 10-5 x 1.0 x 10-5 = 1.0 x 10-10 We can generalize this for any slightly soluble strong electrolyte MX whose molar solubility is S, that Ks p = [M][X] = S 2 . In the most general case where MaX b is the formula of a slightly soluble strong electrolyte, with a solubility S mollL, for every mole of M a X b dissolved a moles of M and b moles of X are formed. Hence the solubility product expression, Ks p = [M] a [X] b = (as) a ·(bS) b = a a .

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  Chemistry

  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.

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  Descriptive Inorganic Chemistry Researches of Metal Compounds

  • 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.

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  Chemistry

  The Molecular Nature of Matter

  • Neil D. Jespersen, Alison Hyslop(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  b Amorphous form. c Calcite form. d At 10 °C 17.1 Equilibria in Solutions of Slightly Soluble Salts 851 Ion Product, the Reaction Quotient for Slightly Soluble Salts In preceding chapters we described the value of the mass action equation as the reaction quotient, Q. For simple solubility equilibria like those discussed in this section, the mass action equation is a product of ion concentrations raised to appropriate powers, so Q is often called the ion product of the salt. Thus, for AgCl, ion product = [Ag + ][Cl − ] = Q At any dilution of a salt throughout the range of possibilities for an unsaturated solution, there will be varying values for the ion concentrations and, therefore, for Q. However, Q acquires a constant value, K sp , in a saturated solution. When a solution is less than saturated, the value of Q is less than K sp . Thus, we can use the numerical value of Q for a given solution as a test for saturation by comparing it to the value of K sp . Another aspect of solubility products is that many salts produce more than one of a given ion per formula unit when they dissociate, and this introduces exponents into the ion product equation. For example, when mercury(II) iodide, HgI 2 , precipitates (Figure 17.3), it enters into the following solubility equilibrium: HgI 2 (s) Hg 2+ (aq) + 2I − (aq) The equilibrium law is obtained following the procedure we developed in Chapter 14, using coefficients as exponents in the mass action equation: [ Hg 2+ ][ I − ] 2 = K sp Thus, the ion product contains the ion concentrations raised to powers equal to the number of ions released per formula unit. This means that to obtain the correct ion product equation, you have to know the formulas of the ions that make up the salt. For example, you have to realize that the ionic compound bismuth(III) iodide, BiI 3 , is composed of one Bi 3+ ion and three iodide, I − , anions.

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  Analytical Chemistry

  • Gary D. Christian, Purnendu K. Dasgupta, Kevin A. Schug(Authors)
  • 2020(Publication Date)
  • Wiley

    (Publisher)

  In fact this is simpler than the metal-ligand titration because unlike the formation constant for a metal-ligand complex ML, the expression for the corre- sponding equilibrium constant, the solubility product, K sp , has no denominator term. The two situations are identical (except in that the metal-ligand case the equilibrium constant by convention is written as an association constant, whereas the solubility product constant is written as a dissociation constant). For an insoluble salt XY, let us assume we took V X mL of a solution of concentration C x molar and it is being titrated with C Y molar solution of Y, V Y mL having been added at any point, with the total vol- ume being V T (which equals V X + V Y ). The implicit concentration units in the solubility product expression are in molar units. To keep all quantities in moles and molar units, we express volumes in liters. K sp = [X][Y] = (C X V X - p) V T (C Y V Y - p) V T (10.12) where p indicates the moles XY precipitated. This results in a quadratic equation in p: p 2 - (C X V X + C Y V Y )p + (C X V X C Y V Y - V T 2 K sp ) = 0 (10.13) As in the case of metal-ligand complexation, generally the negative root of this equation provides the meaningful solution for p. [X] and [Y] (and hence pX and pY) are then Pdf _Folio:3 69 370 CHAPTER 10 PRECIPITATION REACTIONS AND TITRATIONS readily calculated as (C X V X - p)∕V T and (C Y V Y-p )∕V T . We demonstrate the generation of Figure 10.1 by this approach in Figure 10.1.xlsx in the website supplement. Note that before any Y is added, there is no precipitation of XY and the solubility product equilibrium is inapplicable. For V Y = 0, p = 0 must be manually put in. Stepwise Precipitation Titrations If K sp values differ sufficiently, two analytes that can be precipitated by the same reagent can be titrated stepwise. Consider a mixture of iodide and chloride that is titrated with silver ion.

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  Chemistry

  Principles and Reactions

  • William Masterton, Cecile Hurley(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  STRATEGY 1. Recall the definition of a saturated solution. It has the maximum number of grams of a solute that can be dissolved (solubility). 2. Use the solubility in g/L as a conversion factor to find the mass needed to prepare 298 mL of a saturated solution. 15.4 b c a ▼ continued 15-2 Solubility; Solubility Product Constant ( K sp ) 391 Copyright 2016 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. 392 CHAPTER 15 Complex Ion and Precipitation Equilibria SOLUTION Mass BaF 2 0.298 L 3 0.63 g 1 L 5 0.19 g END POINT You have to know either the molar solubility or the solubility in g/L to prepare a saturated solution. Frequently we find that the experimentally determined solubility of an ionic solid is larger than that predicted from K sp . Consider, for example, PbCl 2 , where the solubility calculated from the relation 4 s 3 5 K sp 5 1.7 3 10 2 5 is 0.016 mol/L. The measured solubility is considerably larger, 0.036 mol/L. The explanation for this is that some of the lead in PbCl 2 goes into solution in the form of species other than Pb 2 1 . In particular, we can detect the presence of the ions Pb(OH) 1 and PbCl 1 . The reverse of Example 15.4 involves finding K sp of a compound given its solubility. The solubilities of many ionic compounds are determined experimen-tally and tabulated in chemical handbooks. Most solubility values are given in grams of solute dissolved in 100 grams of water. To obtain the molar solubility in moles/L, we have to assume that the density of the solution is equal to that of water.

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  Solution Chemistry

  Minerals and Reagents

  • Valeria Severino(Author)
  • 2019(Publication Date)
  • Arcler Press

    (Publisher)

  About 20 years ago, Hawkes put in the title of his article [83] a dramatic question, corresponding to his statement presented therein that “the simple algorithms in introductory texts usually produce dramatic and often catastrophic errors”; it is hard not to agree with this opinion. In the meantime, Meites et al. [84] stated that “It would be better to confine illustrations of the solubility product principle to 1:1 salts, like silver bromide (…), in which the (…) calculations will yield results close enough to the truth.” The unwarranted simplifications cause confusion in teaching of chemistry. 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 stoichiometry 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 dissolution 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 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 Solution Chemistry: Minerals and Reagents 38 water, as were proved in Refs.

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  Chemistry

  The Molecular Nature of Matter

  • Neil D. Jespersen, Alison Hyslop(Authors)
  • 2014(Publication Date)
  • Wiley

    (Publisher)

  For example, for CuS the equilib- rium is best written as CuS(s) + H 2 O m Cu 2+ (aq) + HS - (aq) + OH - (aq) (17.5) This equation yields the ion product [Cu 2+ ][HS - ][OH - ], and the solubility product con- stant for CuS is expressed by the equation, K sp = 3 Cu 2+ 4 3 HS - 4 3 OH - 4 Values of K sp of this form for a number of metal sulfides are listed in the last column in Table 17.2. Notice particularly how much the K sp values vary—from 2 Ž 10 -53 to 3 Ž 10 -11 , a spread of a factor of 10 42 . ■ In the laboratory, the qualitative analysis of metal ions uses the precipitation of metal sulfides to separate some metal ions from others. Figure 17.4 | The colors of some metal sulfides. Metal Ions Separable by Selective Precipitation of Sulfides a Metal Ion Sulfide K spa K sp Acid-Insoluble Sulfides Hg 2+ HgS (black form) 2 Ž 10 -32 2 Ž 10 -53 Ag + Ag 2 S 6 Ž 10 -30 6 Ž 10 -51 Cu 2+ CuS 6 Ž 10 -16 6 Ž 10 -37 Cd 2+ CdS 8 Ž 10 -7 8 Ž 10 -28 Pb 2+ PbS 3 Ž 10 -7 3 Ž 10 -28 Sn 2+ SnS 1 Ž 10 -5 1 Ž 10 -26 Base-Insoluble Sulfides (Acid-Soluble Sulfides) Zn 2+ a-ZnS 2 Ž 10 -4 2 Ž 10 -25 b-ZnS 3 Ž 10 -2 3 Ž 10 -23 Co 2+ CoS 5 Ž 10 -1 5 Ž 10 -22 Ni 2+ NiS 4 Ž 10 1 4 Ž 10 -20 Fe 2+ FeS 6 Ž 10 2 6 Ž 10 -19 Mn 2+ MnS (pink form) 3 Ž 10 10 3 Ž 10 -11 MnS (green form) 3 Ž 10 7 3 Ž 10 -14 a Data are for 25 °C. See R. J. Meyers, J. Chem. Ed., vol. 63, 1986, p. 687. TABLE 17.2 CuS CdS As 2 S 3 SnS 2 Sb 2 S 3 MnS ZnS FeS OPC, Inc. 17.3 | Equilibria in Solutions of Metal Oxides and Sulfides 833 Example 17.7 Molar Solubility of Iron(II) Sulfide What pH is needed to achieve a molar solubility of FeS equal to 0.075 M? Analysis: We will use the desired concentration of FeS to determine the concentrations of the Fe 2+ and H 2 S and then solve for the hydrogen ion concentration.

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  Encyclopedia of Pharmaceutical Technology

  Volume 6

  • James Swarbrick(Author)
  • 2013(Publication Date)
  • CRC Press

    (Publisher)

  The equiva-lence point is reached when sufficient reagent to complete precipitation has been added. In practice, the insoluble reaction product formed will be very slightly soluble and to an extent that depends on the amount of solvent present as well as on the nature and amounts of other ions and compounds present. In the simplest general case of a slightly soluble salt (AB) formed by the reaction of the oppositely charged univalent anion (A ) and cation (B þ ) of two soluble salts: A þ B þ ¼ AB and Kppn ¼ ½ A Š½ B þ Š = ½ AB Š where Kppn is the precipitation constant. Assuming that interferences are small enough to be neglected, the concentration of the insoluble salt AB may be regarded as being constant: Kppn ½ AB Š ¼ ½ A Š½ B þ Š ¼ constant ¼ S AB where the constant S is the solubility product. If the solution is pure, equivalent concentrations of A and B þ will be present, and therefore: ½ A Š ¼ ½ B þ Š or ½ A Š 2 ¼ ½ B þ Š 2 ¼ S AB : In this case, if the product [A ][B þ ] exceeds ( S AB ) 0.5 , the solution is saturated with respect to AB, and this substance separates as a precipitate. [1] Titration Curves Titration curves are based on the negative logarithm (to the base 10) of the molar concentration of the species; p -values or p -functions are useful for deducing the properties required of an indicator as well as the titration error that its use is likely to cause. The general shape of a titration curve for the pre-cipitation titration of a solution containing one anion, Cl , is shown in Fig. 1. Curves plotted from calculated p X values are always symmetrical. A curve plotted by using experimentally obtained values for [Cl ] is not symmetrical. The lack of symmetry is due to those ions that are being adsorbed in excess by the precipi-tate in different amounts. This fact is not taken into account in theoretical calculations.

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