Solubility and Comple-ion Equilibria. Solubility and the Common-Ion Effect a. Eplain how the solubility of a salt is affected by another salt that has the same cation or anion. (common ion) b. Calculate the solubility of a slightly soluble salt in a solution of a common ion. Contents and Concepts Solubility Equilibria 1. The Solubility Product Constant. Solubility and the Common-Ion Effect. Precipitation Calculations. Effect of ph on Solubility Comple-Ion Equilibria 5. Comple-Ion Formation 6. Comple Ions and Solubility An Application of Solubility Equilibria 7. Qualitative Analysis of Metal Ions. Precipitation Calculations a. Define ion product. b. State the criterion for precipitation. c. Predict whether precipitation will occur, given ion concentrations. d. Predict whether precipitation will occur, given solution volumes and concentrations. e. Define fractional precipitation. f. Eplain how two ions can be separated using fractional precipitation. Learning Objectives Solubility Equilibria 1. The Solubility Product Constant a. Define the solubility product constant, sp. b. Write solubility product epressions. c. Define molar solubility. d. Calculate sp from the solubility (simple eample). e. Calculate sp from the solubility (more complicated eample). f. Calculate the solubility from sp.. Effect of ph on Solubility a. Eplain the qualitative effect of ph on solubility of a slightly soluble salt. b. Determine the qualitative effect of ph on solubility. c. Eplain the basis for the sulfide scheme to separate a miture of metal ions. 1
Solubility Equilibria Comple-Ion Equilibria 5. Comple-Ion Formation a. Define comple ion and ligand. b. Define formation constant or stability constant, f, and dissociation constant, d. c. Calculate the concentration of a metal ion in equilibrium with a comple ion. d. Define amphoteric hydroide. Many natural processes depend on the precipitation or dissolving of a slightly soluble salt. In the net section, we look at the equilibria of slightly soluble, or nearly insoluble, ionic compounds. Their equilibrium constants can be used to answer questions regarding solubility and precipitation. 6. Comple Ions and Solubility a. Predict whether a precipitate will form in the presence of the comple ion. b. Calculate the solubility of a slightly soluble ionic compound in a solution of the comple ion. An Application of Solubility Equilibria 7. Qualitative Analysis of Metal Ions a. Define qualitative analysis. b. Describe the main outline of the sulfide scheme for qualitative analysis. When an ionic compound is insoluble or slightly soluble, an equilibrium is established: MX(s) M X - The equilibrium constant for this type of reaction is called the solubility-product constant, sp. For the above reaction, sp [M ][X - ] To deal quantitatively with an equilibrium, you must know the equilibrium constant. We will look at the equilibria of slightly soluble (or nearly insoluble) ionic compounds and show how you can determine their equilibrium constants. Once you find these values for various ionic compounds, you can use them to answer questions about solubility or precipitation. The Solubility Product Constant In general, the solubility product constant is the equilibrium constant for the solubility equilibrium of a slightly soluble (or nearly insoluble) ionic compound. It equals the product of the equilibrium concentrations of the ions in the compound. Each concentration is raised to a power equal to the number of such ions in the formula of the compound. For eample, lead iodide, PbI, is another slightly soluble salt. Its equilibrium is: PbI (s) H O Pb I
When an ecess of a slightly soluble ionic compound is mied with water, an equilibrium is established between the solid and the ions in the saturated solution. For the salt calcium oalate, CaC O, you have the following equilibrium. CaC O (s) H O Ca C O The equilibrium constant for this process is called the solubility product constant. sp [Ca ][C O ] Eactly 0.1 mg of AgBr will dissolve in 1.00 L of water. What is the value of sp for AgBr? Solubility equilibrium: AgBr(s) Ag Br - Solubility-product constant epression: sp [Ag ][Br - ] Write the solubility-product epression for the following salts: a. Hg Cl b. HgCl The solubility is given as 0.1 mg/1.00 L, but sp uses molarity: 0.1 g 1mol 7 7.08 M 1.00 L 187.77 g a.hg Cl Hg Cl (s) Hg Cl - sp [Hg ][Cl - ] Initial AgBr(s) Ag 0 Br - 0 b.hgcl HgCl (s) Hg Cl - sp [Hg ][Cl - ] Equilibriu m The epression for the solubility product constant is: sp [Pb ][I ] AgCl (s) Ag Cl - sp [Ag ][Cl - ] [Ag ] [Cl - ] 7.08-7 M sp (7.08-7 ) sp 5.0-1
Calculating sp from the Solubility A 1.0-L sample of a saturated calcium oalate solution, CaC O, contains 0.0061-g of the salt at 5 C. Calculate the sp for this salt at 5 C. We must first convert the solubility of calcium oalate from 0.0061 g/liter to moles per liter. 1 mol CaCO M CaCO (0.0061 g CaCO / L) 18g CaC O 5.8 mol CaCO / L Calculating sp from the Solubility By eperiment, it is found that 1. - mol of lead(ii) iodide, PbI, dissolves in 1.0 L of water at 5 C. What is the sp at this temperature? Starting Equilibrium Note that in this eample, you find that 1. - mol of the solid dissolves to give 1. - mol Pb and (1. - ) mol of I -. PbI (s) H O Pb I 0 0 1. - (1. -) 1. - (1. -) Calculating sp from the Solubility A 1.0-L sample of a saturated calcium oalate solution, CaC O, contains 0.0061-g of the salt at 5 C. Calculate the sp for this salt at 5 C. When.8-5 mol of solid dissolve it forms.8-5 mol of each ion. Substituting into the equilibrium-constant epression: sp [Pb ][I ] sp (1. )( (1. )) 9 sp 6.9 Starting Equilibrium CaC O (s) H O Ca CO 0 0.8-5.8-5.8-5.8-5 Table 17.1 lists the solubility product constants for various ionic compounds. If the solubility product constant is known, the solubility of the compound can be calculated. You can now substitute into the equilibriumconstant epression. sp sp [Ca (.8 ][C O 5 9 sp. ] )(.8 5 )
An eperimenter finds that the solubility of barium fluoride is 1.1 g in 1.00 L of water at 5 C. What is the value of sp for barium fluoride, BaF, at this temperature? Calomel, whose chemical name is mercury(i) chloride, Hg Cl, was once used in medicine (as a laative and diuretic). It has a sp equal to 1. 18. What is the solubility of Hg Cl in grams per liter? Solubility equilibrium: BaF (s) Ba F - Solubility equilibrium: Hg Cl (s) Hg Cl - Solubility-product constant epression: sp [Ba ][F - ] Solubility-product constant epression: sp [Hg ][Cl - ] The solubility is given as 1.1 g/1.00 L, but sp uses molarity: Initial Equilibriu m 1.1 g 1.00 L 1mol 175. g BaF (s) 6.7 Ba 0 M F - 0 Initial Equilibrium Hg Cl (s) sp () sp ( ) sp Hg 0 1. -18.5-19 6.88-7 M Cl - 0 [Ba ] 6.7 - M [F - ] (6.7 - ) 1.5 - M sp (6.7 - )(1.5 - ) sp 9.8-7 When sp is known, we can find the molar solubility. The molar solubility is 6.9-7 M, but we also need the solubility in g/l: -7 6.88 mol 7.086 g. g/l L 1mol 0. mg/l 5
Calculating the Solubility from sp The mineral fluorite is calcium fluoride, CaF. Calculate the solubility (in grams per liter) of calcium fluoride in water from the sp (. -11 ) Let be the molar solubility of CaF. Starting Equilibrium CaF (s) H O Ca F 0 0 Quick Quiz 1. The solubility product constant for CaCO is.8-9. Calculate the mass of calcium carbonate that will dissolve in 1 liter of water.. The solubility product constant for Fe(OH) is.5-9. Calculate the moles of iron that will dissolve in 1 liter of water. You substitute into the equilibrium-constant equation [Ca ][F ] You now solve for. sp ()().. 11. -11 11.0 We can begin by identifying the value of sp for each compound: PbCrO 1.8-1 PbSO 1.7-8 PbS.5-7 Each salt produces two ions, so each has the same epression for the solubility-product constant: sp. The solubility will be largest for PbSO. Convert to g/l (CaF 78.1 g/mol). solubility.0 mol / L 1.6 g CaF / L 78.1g CaF 1 mol CaF Do Eamples 18.- See problems 18.,,,5,6,9 What effect does the presence of a common ion have on solubility? Given: MX(s) M X - Qualitatively, we can use Le Châtelier s principle to predict that the reaction will shift in the reverse direction when M or X - is added, reducing the solubility. In the net problem, we will eplore this situation quantitatively. 6
What is the molar solubility of silver chloride in 1.0 L of solution that contains.0 mol of HCl? First, using Table 17.1, we find that the sp for AgCl at 5 C is 1.8 -. Net, we construct the ICE chart with the initial [Cl - ] 0.00 M. We then solve for, the molar solubility. Solubility and the Common-Ion Effect In this section we will look at calculating solubilities in the presence of other ions. The importance of the sp becomes apparent when you consider the solubility of one salt in the solution of another having the same cation. For eample, suppose you wish to know the solubility of calcium oalate in a solution of calcium chloride. Each salt contributes the same cation (Ca ) The effect is to make calcium oalate less soluble than it would be in pure water. Initial Equilibrium AgCl(s) Ag 0 Cl - 0.00 0.00 A Problem To Consider What is the molar solubility of calcium oalate in 0.15 M calcium chloride? The sp for calcium oalate is. -9. Note that before the calcium oalate dissolves, there is already 0.15 M Ca in the solution. sp [Ag ][Cl - ] 1.8 - (0.00 ) CaC O (s) H O Ca C O We make the following simplifying assumption: 0.00 0.00. 1.8-0.00 Starting Equilibrium 0.15 0 0.15 The molar solubility is given by : 9.0-9 M Let s compare this result to the solubility of AgCl in water: sp 1.8 1. -5 M The solubility was reduced by a factor of about 100! You substitute into the equilibrium-constant equation [Ca ][CO ] sp (0.15 )(). Now rearrange this equation to give. 0.15 9. 0.15 9 We epect to be negligible compared to 0.15. 9 7
Now rearrange this equation to give. 0.15 1.5 9 8. 0.15 Therefore, the molar solubility of calcium oalate in 0.15 M CaCl is 1.5-8 M. In pure water, the molarity was.8-5 M, which is over 000 times greater. 9 We can use the reaction quotient, Q, to determine whether precipitation will occur. One form of kidney stones is calcium phosphate, Ca (PO ), which has a sp of 1.0 6. A sample of urine contains 1.0 M Ca and 1.0 8 M PO ion. Calculate Q c and predict whether Ca (PO ) will precipitate. Precipitation Calculations Precipitation is merely another way of looking at solubility equilibrium. Rather than considering how much of a substance will dissolve, we ask: Will precipitation occur for a given starting ion concentration? Ca (PO ) (s) Ca PO - sp [Ca ] [PO - ] sp 1.0-6 Q c (1.0 - ) (1.0-8 ) Q c 1.0-5 sp < Q c A precipitate will form. Criteria for Precipitation To determine whether an equilibrium system will go in the forward or reverse direction requires that we evaluate the reaction quotient, Q c. To predict the direction of reaction, you compare Q c with c (Chapter 15). The reaction quotient has the same form as the sp epression, but the concentrations of products are starting values. Consider the following equilibrium. PbCl (s) H O Pb Cl The Q c epression is Q c [Pb ] i[cl ] i where initial concentration is denoted by i. If Q c eceeds the sp, precipitation occurs. If Q c is less than sp, more solute can dissolve. If Q c equals the sp, the solution is saturated. 8
Predicting Whether Precipitation Will Occur The concentration of calcium ion in blood plasma is 0.005 M. If the concentration of oalate ion is 1.0-7 M, do you epect calcium oalate to precipitate? sp for calcium oalate is. -9. The ion product quotient, Q c, is: c [Ca ] i [CO ] i c (0.005) (1.0 - Q Q Q c.5-7 ) Fractional Precipitation Fractional precipitation is the technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another ion, and so forth. The solubility of an insoluble salt can be manipulated by adding a species that reacts with either the cation or the anion. Effect of ph on Solubility When a salt contains the conjugate base of a weak acid, the ph will affect the solubility of the salt. Do eercise 18.6 See problems 18.1 and When a problem gives the amounts and concentrations of two samples that are then mied, the first step in solving the problem is to calculate the new initial concentrations. Eactly 0.00 L of 0.50 M Pb and 1.60 L of.50 M Cl are mied together to form.00 L of solution. Calculate Q c and predict whether PbCl will precipitate. sp for PbCl is 1.6 5. This value is smaller than the sp, so you do not epect precipitation to occur. - Q c.5 < sp Fractional Precipitation Fractional precipitation is the technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so forth. (page 7) For eample, when you slowly add potassium chromate, CrO, to a solution containing Ba and Sr, barium chromate precipitates first. (0.500 M) (0.00 L) [Pb ] 0.0 M (.00 L) - (.50 M) (1.60 L) [Cl ] 0.000 M (.00 L) PbCl (s) Pb Cl - sp [Pb ] [Cl - ] 1.6-8 Q c (0.0)(0.000).00-5 sp < Q c A precipitate will form. Fractional Precipitation 0.1 M Ba and 0.1 M Sr [Ba ][CrO - ] sp 1. - [CrO - ] 1. -9 [Sr ][CrO - ] sp.5-5 [CrO - ].5 - Calculate the percent Ba remaining when Sr begins to Ppt 9
Fractional Precipitation Fractional precipitation is the technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so forth. After most of the Ba ion has precipitated, strontium chromate begins to precipitate. It is therefore possible to separate Ba from Sr by fractional precipitation using CrO. sp for MgC O is 8.5-5. sp for BaC O is 1.5-8. MgC O is more soluble. If a dilute acidic solution is added, it will increase the solubility of both salts. MgC O is still more soluble. We will qualitatively eplore the situation involving a generic salt, MX, where X is the conjugate base of a weak acid. It is possible to use these differences to separate compounds. MX(s) M X - This is common practice when the goal is to separate sulfides from one another. As the acid concentration increases, X - reacts with the H O, forming HX and reducing the X - concentration. As a result, more MX dissolves, increasing the solubility. The qualitative analysis scheme for the separation of metal ions uses sulfide solubility to separate Co, Fe, Mn, Ni, and Zn (Analytical Group III). Consider the two slightly soluble salts barium fluoride and silver bromide. Which of these would have its solubility more affected by the addition of strong acid? Would the solubility of that salt increase or decrease? Effect of ph on Solubility Sometimes it is necessary to account for other reactions aqueous ions might undergo. For eample, if the anion is the conjugate base of a weak acid, it will react with H O. HF is a weak acid, while HBr is a strong acid. BaF is more soluble in an acidic solution. AgBr is unaffected by an acidic solution. You should epect the solubility to be affected by ph.
Consider the following equilibrium. CaC O (s) H O Ca C O Because the oalate ion is conjugate to a weak acid (HC O - ), it will react with H O. H O C O HO HCO HO(l) According to Le Chatelier s principle, as C O - ion is removed by the reaction with H O, more calcium oalate dissolves. Therefore, you epect calcium oalate to be more soluble in acidic solution (low ph) than in pure water. Comple-Ion Formation Some cations form soluble comple ions. Their formation increases the solubility of a salt containing those cations. Separation of Metal Ions by Sulfide Precipitation Many metal sulfides are insoluble in water but dissolve in acidic solution. Qualitative analysis uses this change in solubility of the metal sulfides with ph to separate a miture of metal ions. By adjusting the ph in an aqueous solution of H S, you adjust the sulfide concentration to precipitate the least soluble metal sulfide first. Qualitative analysis is covered in Section 18.7. Metal ions that form comple ions include Ag, Cd, Cu, Fe, Fe, Ni, and Zn. Compleing agents, called ligands, are Lewis bases. They include CN -, NH, S O -, and OH -. In each case, an equilibrium is established, called the comple-ion formation equilibrium. Ag NH Ag(NH ) ) ] ] [Ag(NH f [Ag ] [NH Zn OH - Zn(OH) - - [Zn(OH) ] f - [Zn ] [OH ] 11
Comple-Ion Equilibria Many metal ions, especially transition metals, form coordinate covalent bonds with molecules or anions having a lone pair of electrons. This type of bond formation is essentially a Lewis acid-base reaction (Chapter 16). For eample, the silver ion, Ag, can react with ammonia to form the Ag(NH ) ion. Ag (: NH ) (HN : Ag : NH) Comple-Ion Formation The formation constant, f, is the equilibrium constant for the formation of a comple ion from the aqueous metal ion and the ligands. The formation constant for Ag(NH ) is: [Ag(NH [Ag ][NH f ) ] ] The value of f for Ag(NH ) is 1.7 7. Comple-Ion Equilibria A comple ion is an ion formed from a metal ion with a Lewis base attached to it by a coordinate covalent bond. A comple is defined as a compound containing comple ions. A ligand is a Lewis base (an electron pair donor) that bonds to a metal ion to form a comple ion. The large value means that the comple ion is quite stable. When a large amount of NH is added to a solution of Ag, you epect most of the Ag ion to react to form the comple ion. Table 17. lists formation constants of some comple ions. Comple-Ion Formation The aqueous silver ion forms a comple ion with ammonia in steps. Ag NH Ag(NH) Ag(NH ) NH Ag(NH) When you add these equations, you get the overall equation for the formation of Ag(NH ). Ag NH Ag(NH) Formation constants are shown to the right. Note that all the values are quite large, which means that the equilibrium favors the comple ion. 1
Comple-Ion Formation The dissociation constant, d, is the reciprocal, or inverse, value of f. The equation for the dissociation of Ag(NH ) is Ag(NH ) Ag NH The equilibrium constant equation is 1 [Ag ][NH ] [Ag(NH ) ] d f Substituting into this equation gives: ()(0.98 ) (0.0 ) 1 1.7 7 If we assume is small compared with 0.0 and 0.98, then ( )(0.98) 8 and (0.0) 8 5.9 5.9 (0.0) (0.98) 6.1 The silver ion concentration is 6.1 - M. Equilibrium Calculations with f What is the concentration of Ag ion in 0.0 M AgNO that is also 1.00 M NH? The f for Ag(NH ) is 1.7 7. In 1.0 L of solution, you initially have 0.0 mol Ag from AgNO. This reacts to give 0.0 mol Ag(NH ), leaving (1.00- ( 0.0)) 0.98 mol NH. You now look at the dissociation of Ag(NH ). Silver chloride usually does not precipitate in solutions of 1.00 M NH. However, silver bromide has a smaller sp than silver chloride. Will silver bromide precipitate from a solution containing 0.0 M AgNO, 0.0 M NaBr, and 1.00 M NH? Calculate the Q c value and compare it with silver bromide s sp of 5.0 1. Starting Equilibrium The following table summarizes. Ag(NH ) 0.0-0.0- The dissociation constant equation is: [Ag ][NH [Ag(NH ] ) ] d Ag NH 0 0.98 0.98 1 f We ll begin with the comple-ion formation, and then find the concentration of Ag in solution. Finally, we ll find the value of Q c for AgBr and compare it to sp. 1.00 L of solution contains 0.0 mol Ag and 1.00 mol NH. Ag NH Ag(NH ) (aq ) Initial 0.0 1.00 0-0.0 -(0.0) 0.0 Equilibrium 0 0.98 0.0 1
[Ag(NH ) ] 0.00 M; [NH ] 0.998 M Ag(NH ) (a NH Initial q) 0.0 Ag 0 0.980 - Equilibrium 0.0-0.980 [Ag ] [NH ] 1 1 d 5. 9 7 [Ag(NH 1.7 ) ] 5.9 8 f (0.980 ) ( 0.0 ) 8 Calculate the molar solubility of AgBr in 1.0 M NH at 5 C. We will first combine the two equilibria and find the combined equilibrium constant. AgBr(s) Ag Br - ; sp Ag NH Ag(NH ) ; f AgBr(s) NH Br - NH sp f (5.0-1 )(1.7 7 ) 8.5-6 We can assume that 0.980 0.980 and 0.0 0.0. 8 (0.980) 5.9 0.0 5.9 8 96.0 6.1 M The assumptions are valid. [Ag ] 6.1 - M Now, we ll use the combined equilibrium to find the solubility of AgBr in 1.0 M NH. Initial Equilibriu m AgBr(s) - ) ] [Br ][Ag(NH [NH 8.5 NH (a Br - q) 1.0 0-1.0 6 (1.0 ) ] Ag(NH0 ) )aq) Net we ll use the solubility equilibrium to find Q c. Q c [Ag ] [Br - ] Q c (6.1 - )(0.0) Q c 6.1-1 sp 5.0-1 sp < Q c The precipitate forms. The right side of the equation is a perfect square..9 (1.0 ) 0.009 0.0058 0.009 1.0058.9 The molar solubility of AgBr in 1.0 M NH is.9 - M. 1
The combination of solubility and comple-ion equilibria can be applied to separate metal ions. Cations can be separated into groups according to their precipitation properties. In each major step, one group of cations precipitates out. After separating the precipitate, the remaining solution is treated with the net reagent, precipitating the net group of cations. When a strong base is slowly added to a solution of ZnCl, a white precipitate of Zn(OH) first forms. Zn OH Zn(OH) (s) But as more base is added, the white preciptate dissolves, forming the comple ion Zn(OH) -. See Figure 17. Other common amphoteric hydroides are those of aluminum, chromium(iii), lead(ii), tin(ii), and tin(iv). Amphoteric Hydroides An amphoteric hydroide is a metal hydroide that reacts with both acids and bases. For eample, zinc hydroide, Zn(OH), reacts with a strong acid and the metal hydroide dissolves. Zn(OH) (s) HO Zn HO(l) With a base however, Zn(OH) reacts to form the comple ion Zn(OH) -. Zn(OH) (s) OH Zn(OH) Do Eercises 18.9,, and 11 Look at Problems 18.57-6 15
Figure 17.8: Demonstration of the amphoteric behavior of zinc hydroide CH CSNH H O CH CONH H S Photo by James Scherer. Houghton Mifflin Company. All rights reserved. Qualitative Analysis Qualitative analysis involves the determination of the identity of substances present in a miture. In the qualitative analysis scheme for metal ions, a cation is usually detected by the presence of a characteristic precipitate. Figure 18.8 summarizes how metal ions in an aqueous solution are separated into five analytical groups. Operational Skills Writing solubility product epressions Calculating sp from the solubility, or vice versa. Calculating the solubility of a slightly soluble salt in a solution of a common ion. Predicting whether precipitation will occur Determining the qualitative effect of ph on solubility Calculating the concentration of a metal ion in equilibrium with a comple ion Predicting whether a precipitate will form in the presence of the comple ion Calculating the solubility of a slightly soluble ionic compound in a solution of the comple ion Figure 17.8 16