Solubility Equilibria

Transcription

1 Every electrolyte (in aqueous solution), consists partly of active (in electrical and chemical relation), and partly of inactive molecules... Svante Arrhenius Solubility Equilibria Pure water is actually a solution that contains tiny amounts of hydrogen and hydroxide ions in addition to water molecules. Although mole per liter of these ions may seem insignificant, slight variations in their concentrations can result in major changes in many important properties of the solution. For example, the hydrogen ion concentration in your blood is normally mole per liter, but if it changes to mole per liter, death results in only a few seconds. Clearly, an understanding of these tiny concentrations is an important aspect of chemistry. A similar situation exists in solutions of so-called insoluble salts. Even though these salts were introduced to you as being insoluble, there are actually tiny amounts of the ions that make up the salts in their solutions. For example, you may recall that the solubility rules said that all chlorides are soluble with the exception of certain heavy metal chlorides such as silver chloride. Thus, you classified silver chloride as insoluble. In fact, a concentration of mole per liter of silver and chloride ions exist in such a solution. These small concentrations of dissolved ions can be important to industrial and analytical chemists, as well as in medicine and geology. The Solubility Product Constant If undissolved silver chloride is present in a water solution, some will dissolve. When the process reaches equilibrium, the macroscopic concentrations of silver and chloride ions will remain constant, and the process can be represented by the equation AgCl(s) Ag + (aq) + Cl (aq) Recalling that solids are not included in an equilibrium constant expression, the K expression for this reaction is K sp = [Ag + ] [Cl ] where the subscript sp on the K refers to the term solubility product. K sp is called the solubility product constant. K sp expressions are written in the same manner as any other equilibrium expression. Just keep in mind that solids are not included in those expressions. If we have a solution with lead(ii) chloride, the dissolving equation and solubility product constant expression are

2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) K sp = [Pb 2+ ] [Cl ] 2 Recall that ion concentrations are raised to the power of their stoichiometric coefficient in equilibrium constant expressions. As with all equilibrium constant expressions, the solubility product constant is temperature dependent. We will restrict our interest to reactions occurring at 25 C, the standard thermodynamic temperature, in this lesson. Determination of Solubility Product Constant Values If we know the solubility of a salt in water, we can calculate its K sp value. We follow a four-step procedure: 1. Write the balanced equation representing the dissolving process. 2. Write the K sp expression, based on the equation from Step Calculate the mole per liter (M) concentration of each of the ions represented in the K sp expression. 4. Substitute the ion concentrations into the K sp expression and calculate its value. EXAMPLE 1 It is experimentally determined that g of silver bromide will dissolve in ml of water at 25 C. What is the solubility product constant for silver bromide? SOLUTION The first step is to write the balanced equation for dissolving silver bromide: AgBr(s) Ag + (aq) + Br (aq) Now we write the solubility product constant expression: K sp = [Ag + ] [Br ] The third step requires changing concentration in g/100 ml to mol/l: g AgBr 100 ml 1 mol AgBr g AgBr 1 mol Ag + 1 mol AgBr 1000 ml L = mol Ag + L = M Ag + Since one bromide ion is formed for each silver ion formed, [Br ] = [Ag + ] = M.

3 Solubility Equilibria Now we can substitute the mole per liter concentrations into the K sp expression and solve: K sp = [Ag + ] [Br ] = ( ) ( ) = Determination of Solubility Solubility product constant values are known for a large number of compounds, so in the majority of cases, they can be obtained from reference sources. Therefore, a more common calculation is to find solubility from known K sp values. The procedure for determining solubility from K sp is: 1. Write the balanced equation representing the dissolving process. 2. Write the K sp expression, based on the equation from Step Assign the algebraic variable x to one of the ions on the right side of the equilibrium equation that has the same stoichiometric coefficient as the solid on the left side of the equation. 4. Assign algebraic variables for the other ions on the right side of the equilibrium equation in terms of their relationship to the ion from Step Substitute the algebraic variables into the K sp expression, solve for x, and determine the solubility in the units requested in the problem statement. EXAMPLE 2 Determine the g/100 cm 3 solubility of lead(ii) bromide, given that one reference source gives its K sp as SOLUTION The first two steps are to write the equation and K sp expression: PbBr 2 (s) Pb 2+ (aq) + 2 Br (aq) K sp = [Pb 2+ ] [Br ] 2 We want to know the solubility of PbBr 2 (s). There is a 1:1 stoichiometric relationship between PbBr 2 (s) and Pb 2+ (aq), so if we find the [Pb 2+ ], we have the solubility of PbBr 2 (s). Thus we let x = [Pb 2+ ]. Since the coefficient on Br (aq) is twice that on Pb 2+ (aq), [Br ] = 2x. This completes Steps 3 and 4. Now we substitute and solve: K sp = [Pb 2+ ] [Br ] 2 = = (x) (2x) = 4x 3 x = mol/l = [Pb 2+ ] = solubility of PbBr 2 (s) To complete the problem, we need to convert moles per liter to grams per 100 cm 3 :

4 0.013 mol PbBr 2 L g PbBr 2 mol PbBr L [100 cm 3 ] = 0.5 g PbBr 2 /100 cm 3 The Ion Product The ion product, IP, for a slightly soluble salt has the same form as the expression for the solubility product constant, but it is free from the restriction of only applying to an equilibrium situation. The ion product is valid whether or not an equilibrium situation exists, whereas K sp, an equilibrium constant, only refers to an equilibrium situation. Returning to our silver chloride example, we have AgCl(s) Ag + (aq) + Cl (aq) K sp = [Ag + ] [Cl ] IP = [Ag + ] [Cl ] The ion product is particularly useful in predicting whether or not precipitation will occur when solutions containing ions that, when combined, will form a slightly soluble salt. If we allow a saturated solution of silver chloride to come to equilibrium, IP will be equal to K sp : IP = [Ag + ] [Cl ] = K sp In other words, the ion product and the equilibrium constant are the same for an equilibrium situation. On the other hand, if we were to begin with a non-equilibrium condition, and then give the system time to come to equilibrium, that equilibrium will occur at concentrations such that IP = K sp. What about when the system is not at equilibrium? First, let s consider the case when IP is less than K sp, IP < K sp. In this case, the solution is unsaturated because there is not a great enough concentration of ions for crystallization to exceed dissolving. No precipitate will form. The other potential nonequilibrium situation occurs when the ion product is greater than the equilibrium constant, IP > K sp. In this case, the concentration of the aqueous ions in solution is greater than the equilibrium concentration. Therefore, for the system to come to equilibrium, the solution ion concentrations must decrease. Solid will precipitate until IP = K sp. Thus if solutions are combined such that IP > K sp, precipitation will occur. In summary, if solutions are combined such that the IP is greater than K sp, the slightly soluble salt will precipitate. If the ion product does not exceed K sp, no precipitation will occur. EXAMPLE 3 In Example 1, we determined that the K sp of silver bromide is If 50.0 ml M silver nitrate and ml M potassium bromide are combined, will a precipitate form?

5 Solubility Equilibria SOLUTION As with all problems involving slightly soluble salts, we start with the equilibrium equation and the K sp expression. This time, however, the reaction for the potential precipitation of the slightly soluble salt is not immediately obvious. Let s consider what ions we have in solution: Ag + (aq) and NO 3 (aq) from the silver nitrate solution and K + (aq) and Br (aq) from the potassium bromide solution. A double replacement reaction is possible, forming AgBr and KNO 3. From the solubility rules, we know that potassium and nitrate salts are soluble, and that bromide salts are generally soluble, but AgBr is an exception. Thus the precipitation is Ag + (aq) + Br (aq) AgBr(s) K sp = [Ag + ] [Br ] = The ion product expression follows from the K sp expression: IP = [Ag + ] [Br ] Next, we need the concentrations of the silver and bromide ions. When the two solutions are combined, each original solution is diluted, and thus we use the formula for calculating a dilution, M c V c = M d V d, where the subscript c stands for concentrated and the subscript d stands for dilute, M is molarity, and V is volume. For silver ion, M d = M c V c V d = M Ag ml ml = M Ag + For bromide ion, M d = M c V c V d = M Br ml ml = M Br Note that the total volume of solution after the two solutions are combined, V d, is the sum of the solution volumes: 50.0 Ml ml = ml. Now we can calculate the ion product and compare it to the K sp value: IP = [Ag + ] [Br ] = (0.033) (0.0013) = > = K sp IP > K sp, therefore, we predict that a precipitate of AgBr will form when the solutions are combined.

6 Worksheet: Solubility Equilibria I Name(s) 1. An experiment is conducted to determine the solubility of magnesium fluoride, and it is found that g of the compound will dissolve in ml water. What is K sp for MgF 2? 2. The solubility product constant for silver sulfate is at 25 C. What is the solubility (g/100 ml water) of the compound? 3. A solution that is M in barium chloride is added to a solution that is M in sodium sulfate. Equal volumes of both solutions were combined. Given that K sp = for barium sulfate, will precipitation occur or will the solution remain unsaturated?

7 Solubility Equilibria Worksheet: Solubility Equilibria II Name(s) 1. A number of factors can complicate calculations of equilibria involving slightly soluble salts, and these complications form the theme of this workshop. In this question, we want you to consider how the solubility of a salt is effected when it is dissolved in pure water versus when it is dissolved in a solution that contains one of the ions found in the slightly soluble salt. K sp = for copper(ii) carbonate. (a) What is the solubility of copper(ii) carbonate in pure water? Express your answer in g/100 ml. (b) Use Le Chatelier s principle to predict the shift that would occur in the copper(ii) carbonate equilibrium if carbonate ion were added to the system. (c) What is the solubility of copper(ii) carbonate in M sodium carbonate? (d) How does your prediction from Part (b) correspond with the numerical results from Parts (a) and (c)?

8 2. The solubility of slightly soluble salts can be effected by the formation of complex ions. A complex ion is composed of a central positively charged ion and attached electron-pair donor species. In a solution containing silver and chloride ions, for example, the dichloroargentate(i) ion forms: Ag + (aq) + 2 Cl (aq) AgCl 2 (aq) K = Of course, the equilibrium between the slightly soluble salt and its ions occurs simultaneously in the solution: AgCl(s) Ag + (aq) + Cl (aq) K sp = How will the solubility of a slightly soluble salt be effected if its ions form a complex ion? To answer this question, perform the following calculations: (a) Determine the g/100 ml solubility of silver chloride neglecting the effect of the complex ion formation. (b) Add the two equations to get the net overall reaction and calculate the resulting equilibrium constant. (c) Determine the g/100 ml solubility for the overall process. (d) Compare the solubilities. What effect does complex ion formation have on the solubility of the salt?

9 Solubility Equilibria 3. Consider these qualitative experimental results: (1) Very dilute solutions of Ba(NO 3 ) 2 (aq) and Na 2 CO 3 (aq) are poured into a large beaker half filled with pure water. A hazy cloud forms. (2) Very dilute solutions of Ba(NO 3 ) 2 (aq) and Na 2 CO 3 (aq) are poured into a large beaker half filled with KCl(aq) solution. The solution remains clear. a) What causes the hazy cloud in Experiment 1? b) Experiment 2 is an example of a phenomenon called the salt effect, where the solubility of a slightly soluble salt is increased by the presence of ions in the solution that are not common with the slightly soluble salt. In this case, potassium ions are attracted to the carbonate ions in the solution. Similarly, chloride ions are attracted to the barium ions. This keeps the barium and carbonate ions from combining to form a precipitate, as long as they are in low concentration relative to their solubility limit. Sketch a particulate-level diagram of this effect, and explain how your sketch illustrates the salt effect.

10 4. If we use the solubility product constant to predict the solubility of silver bromide in pure water, the result of the theoretically-based calculation matches the experimental result. On the other hand the experimentally determined solubility of silver carbonate does not match the solubility determined from a K sp calculation. a) K sp = for silver carbonate. Use this value to determine the theoretical solubility (g/100 ml) in pure water. b) Bromide ion is the conjugate base of a strong acid; carbonate ion is the conjugate base of a weak acid. How do these facts correlate with the theory/experiment results for silver bromide and silver carbonate? Do you expect silver carbonate to be more or less soluble than the K sp calculation would predict? Why? c) Assume that the carbonate ion reacts to form its conjugate acid in solution: CO 3 2 (aq) + H 2 O(l) HCO 3 (aq) + OH (aq) K = Calculate the g/100 ml solubility that should result from the combination of the two equilibrium systems.

11 Solubility Equilibria 5. Reconsider dissolving silver bromide and silver carbonate. If your objective was to get each of these slightly soluble compounds to dissolve, and you had available concentrated solutions of hydrochloric acid and hydrobromic acid, what would be the effect on each of the salts? Qualitatively consider all four possible combinations.

12 6. Another complicating factor in predicting solubilities from K sp values involves the formation of ion pairs in solution. An ion pair is an aqueous pair of ions that acts like a single particle in solution. The pair is formed by oppositely charged ions in the solution. When solid calcium sulfate is placed in pure water and allowed to come to equilibrium, the expected dissolving occurs to a small extent: CaSO 4 (s) Ca 2+ (aq) + SO 4 2 (aq) In addition, the solid dissolves to form ion pairs: CaSO 4 (s) Ca 2+ SO 4 2 (aq) where Ca 2+ SO 4 2 (aq) represents the ion pair formed from the calcium and sulfate ion. The ions in solution can also form ion pairs: Ca 2+ (aq) + SO 4 2 (aq) Ca 2+ SO 4 2 (aq) All three equilibria must be considered in order to understand the solubility of calcium sulfate. a) Does ion-pair formation increase or decrease the solubility of a slightly soluble salt? Explain. b) Ion-pair formation is greatest for highly charged ions and for small ions. List a few slightly soluble salts that you would expect to have a relatively large ion-pair effect and a few for which you would expect this effect to be relatively small.