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Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 1 More Equilibria in Aqueous Solutions:

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1 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 1 More Equilibria in Aqueous Solutions: Slightly Soluble Salts and Complex Ions Chapter Sixteen

2 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 2 Solubility product constant, K sp : the equilibrium constant expression for the dissolving of a slightly soluble solid. The Solubility Product Constant, K sp K sp = [ Ba 2+ ][ SO 4 2– ] BaSO 4 (s) Ba 2+ (aq) + SO 4 2– (aq) Many important ionic compounds are only slightly soluble in water (we used to call them “insoluble” – Chapter 4). An equation can represent the equilibrium between the compound and the ions present in a saturated aqueous solution:

3 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 3

4 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 4 Example 16.1 Write a solubility product constant expression for equilibrium in a saturated aqueous solution of the slightly soluble salts (a) iron(III) phosphate, FePO 4, and (b) chromium(III) hydroxide, Cr(OH) 3.

5 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 5 Example 16.1 Write a solubility product constant expression for equilibrium in a saturated aqueous solution of the slightly soluble salts (a) iron(III) phosphate, FePO 4 and (b) chromium(III) hydroxide, Cr(OH) 3. Strategy We first write chemical equations for the solubility equilibrium, basing each equation on one mole of solid on the left side. The coefficients on the right side are then the numbers of moles of cations and anions per mole of the compound. As with other equilibrium constant expressions, coefficients in the balanced equation appear as exponents in the K sp expression. Write a K sp expression for equilibrium in a saturated aqueous solution of (a) MgF 2, (b) Li 2 CO 3, and (c) Cu 3 (AsO 4 ) 2. Exercise 16.1A Write a K sp expression for equilibrium in a saturated solution of (a) magnesium hydroxide (commonly known as Milk of magnesia), (b) scandium fluoride, ScF 3 (used in the preparation of scandium metal), and (c) zinc phosphate (used in dental cements). Exercise 16.1B Solution

6 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 6 K sp is an equilibrium constant Molar solubility is the number of moles of compound that will dissolve per liter of solution. Molar solubility is related to the value of K sp, but molar solubility and K sp are not the same thing. In fact, “smaller K sp ” doesn’t always mean “lower molar solubility.” Solubility depends on both K sp and the form of the equilibrium constant expression. K sp and Molar Solubility

7 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 7 Example 16.2 At 20 °C, a saturated aqueous solution of silver carbonate contains 32 mg of Ag 2 CO 3 per liter of solution. Calculate K sp for Ag 2 CO 3 at 20 °C. The balanced equation is Ag 2 CO 3 (s) 2 Ag + (aq) + CO 3 2– (aq) K sp = ? Example 16.3 From the K sp value for silver sulfate, calculate its molar solubility at 25 °C. Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 2– (aq) K sp = 1.4 x 10 –5 at 25 °C

8 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 8 Strategy To evaluate K sp, we need the molarities of the individual ions in the saturated solution, but we are given only the solubility of the solute. Thus, the heart of the calculation is to obtain ion molarities from a concentration in milligrams of solute per liter. Once we have these ion molarities, we can get a value for K sp by substituting them into the solubility constant expression. Example 16.2 At 20 °C, a saturated aqueous solution of silver carbonate contains 32 mg of Ag 2 CO 3 per liter of solution. Calculate K sp for Ag 2 CO 3 at 20 °C. The balanced equation is Solution From the equation for the solubility equilibrium, we see that 2 mol Ag + and 1 mol CO 3 2– appear in solution for every 1 mol Ag 2 CO 3 that dissolves. We can represent this fact by conversion factors (shown in blue).

9 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 9 Example 16.2 continued Now we can write the K sp expression for Ag 2 CO 3 and substitute in the equilibrium concentrations of the ions. K sp = [Ag + ] 2 [CO 3 2– ] = (2.3 x 10 –4 ) 2 (1.2 x 10 –4 ) = 6.3 x 10 –12 In Example 15.4, we determined the molar solubility of magnesium hydroxide from the measured pH of its saturated solution. Use data from that example to determine K sp for Mg(OH) 2. Exercise 16.2A A saturated aqueous solution of silver(I) chromate contains 14 ppm of Ag + by mass. Determine K sp for silver(I) chromate. Assume the solution has a density of 1.00 g/mL. Exercise 16.2B

10 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 10 Example 16.3 From the K sp value for silver sulfate, calculate its molar solubility at 25 °C. Strategy The equation shows that when 1 mol Ag 2 SO 4 (s) dissolves, 1 mol SO 4 2– and 2 mol Ag + appear in solution. If we let s represent the number of moles of Ag 2 SO 4 that dissolve per liter of solution, the two ion concentrations are [SO 4 2– ] = s and [Ag + ] = 2s We then substitute these concentrations into the K sp expression and solve for s, which, because 1 mol of Ag 2 SO 4 produces 1 mol of SO 4 2–, is the molar solubility we are seeking. Solution First, we write the K sp expression for Ag 2 SO 4 based on the solubility equilibrium equation. K sp = [Ag + ] 2 [SO 4 2– ] = 1.4 x 10 –5 Then, we substitute the values [Ag + ] = 2s and [SO 4 2– ] = s to obtain the equation that must be solved.

11 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 11 Example 16.3 continued Now, by recalling that s = [SO 4 2– ] and that [SO 4 2– ] is the same as the molar solubility of Ag 2 SO 4, we obtain our final result. Assessment In substituting ion concentrations into the K sp expression, it is important to note that (i) the factor 2 appears in 2s because the concentration of Ag + is twice that of SO 4 2–, and (ii) the exponent 2 appears in the term (2s) 2 because the concentration of Ag + (that is, 2s) must be raised to the second power in the K sp expression. Be alert for similar situations that might arise in other problems. Using the K sp value in Table 16.1, determine the quantity of I –, in parts per million, in a saturated aqueous solution of PbI 2. Assume the solution has a density of 1.00 g/mL. Exercise 16.3B Calculate the molar solubility of silver arsenate, given that Exercise 16.3A

12 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 12 Example 16.4 A Conceptual Example Without doing detailed calculations, but using data from Table 16.1, establish the order of increasing solubility of these silver halides in water: AgCl, AgBr, AgI.

13 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 13 Example 16.4 Without doing detailed calculations, but using data from Table 16.1, establish the order of increasing solubility of these silver halides in water: AgCl, AgBr, AgI. Analysis and Conclusions The key to solving this problem is to recognize that all three solutes are of the same type; that is, the ratio of number of cations to anions in each is 1:1. As suggested by the following equation for the solubility equilibrium of AgX, the ion concentrations are equal to each other and to the molar solubility, s. The order of increasing molar solubility is the same as the order of increasing K sp values: Increasing molar solubilities: AgI

14 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 14 The common ion effect affects solubility equilibria as it does other aqueous equilibria. The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution. The Common Ion Effect in Solubility Equilibria

15 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 15 Common Ion Effect Illustrated Na 2 SO 4 (aq) Saturated Ag 2 SO 4 (aq) Ag 2 SO 4 precipitates The added sulfate ion reduces the solubility of Ag 2 SO 4.

16 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 16 Common Ion Effect Illustrated When Na 2 SO 4 (aq) is added to the saturated solution of Ag 2 SO 4 … … [Ag + ] attains a new, lower equilibrium concentration as Ag + reacts with SO 4 2– to produce Ag 2 SO 4.

17 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 17 Example 16.5 Calculate the molar solubility of Ag 2 SO 4 in 1.00 M Na 2 SO 4 (aq).

18 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 18 Example 16.5 Calculate the molar solubility of Ag 2 SO 4 in 1.00 M Na 2 SO 4 (aq). Strategy The molar solubility of Ag 2 SO 4 will be influenced by the concentration of SO 4 2– in solution coming from the 1.00 M Na 2 SO 4 (aq). We will assume that the Na 2 SO 4 is completely dissociated and that the presence of Ag 2 SO 4 has no effect on this dissociation. As in Example 16.3, we will use s to represent the number of moles of Ag 2 SO 4 dissolved per liter of saturated solution. We will then represent [Ag + ] and [SO 4 2– ] in terms of s and solve the K sp expressions for s. Solution We can tabulate the relevant data, including the already present 1.00 mol SO 4 2– /L, in the ICE format introduced in Chapter 14.

19 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 19 Example 16.5 continued The equilibrium concentrations from the ICE table must satisfy the K sp expression. K sp = [Ag + ] 2 [SO 4 2– ] 1.4 x 10 –5 = (2s) 2 ( s) To simplify the equation, let us assume that s is much smaller than 1.00 M, so that ( s)  (2s) 2 (1.00) = 1.4 x 10 –5 4s 2 = 1.4 x 10 –5 We obtain our final result by solving for s. s 2 = 3.5 x 10 –6 s = molar solubility = (3.5 x 10 –6 ) 1/2 = 1.9 x 10 –3 mol Ag 2 SO 4 /L Assessment When we assumed that ( s)  1.00, we were saying that essentially all the common ion comes from the added source (1.00 M) and essentially none comes from the slightly soluble solute (s). The assumption is valid here: ( x 10 –3 ) = 1.00 (to two decimal places). This type of assumption will often be valid in calculations of this type. Notice that the solubility of Ag 2 SO 4 found here is only about one-eighth of that found for Ag 2 SO 4 in pure water in Example 16.3, illustrating, as we had intended, the common ion effect. Calculate the molar solubility of Ag 2 SO 4 in 1.00 M AgNO 3 (aq). Exercise 16.5A How many grams of silver nitrate must be added to mL of a saturated solution of Ag 2 SO 4 (aq) to reduce the solubility of the Ag 2 SO 4 to 1.0 x 10 –3 M (K sp of Ag 2 SO 4 = 1.4 x 10 –5 )? Exercise 16.5B

20 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 20 Ions that are not common to the precipitate can also affect solubility. –CaF 2 is more soluble in M Na 2 SO 4 than it is in water. Increased solubility occurs because of interionic attractions. Each Ca 2+ and F – is surrounded by ions of opposite charge, which impede the reaction of Ca 2+ with F –. The effective concentrations, or activities, of Ca 2+ and F – are lower than their actual concentrations. Solubility and Activities

21 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 21 Q ip can then be compared to K sp. Precipitation should occur if Q ip > K sp. Precipitation cannot occur if Q ip < K sp. A solution is just saturated if Q ip = K sp. In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered. Q ip is the ion product reaction quotient and is based on initial conditions of the reaction. Q ip and Q c : new look, same great taste! Will Precipitation Occur? Is It Complete?

22 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 22 Example 16.6 If 1.00 mg of Na 2 CrO 4 is added to 225 mL of M AgNO 3, will a precipitate form? Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2– (aq) K sp = 1.1 x 10 –12

23 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 23 Strategy We can apply the three-step strategy just outlined. Example 16.6 If 1.00 mg of Na 2 CrO 4 is added to 225 mL of M AgNO 3 (aq), will a precipitate form? Solution Step 1: Initial concentrations of ions: The initial concentration of Ag + is simply 1.5 x 10 –4 M. To establish CrO 4 2–, we first need to determine the number of moles of CrO 4 2– placed in solution. We then determine [CrO 4 2– ] in the 225 mL (0.225 L) of solution that contains 6.17 x 10 –6 mol CrO 4 2–. Step 2: Evaluation of Q ip : Q ip = [Ag + ] initial 2 [CrO 4 2– ] initial = (1.5 x 10 –4 ) 2 (2.74 x 10 –5 ) = 6.2 x 10 –13

24 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 24 Example 16.6 continued Step 3: Comparison of Q ip and K sp : Q ip = 6.2 x 10 –13 is less than K sp = 1.1 x 10 –12 Because Q ip < K sp, we conclude that no precipitation occurs. If 1.00 g Pb(NO 3 ) 2 and 1.00 g MgI 2 are both added to 1.50 L of H 2 O, should a precipitate form (K sp for PbI 2 = 7.1 x 10 –9 )? Exercise 16.6A Will precipitation occur from a solution that has 2.5 ppm MgCl 2 and a pH of 10.35? Assume the solution density is 1.00 g/mL [K sp for Mg(OH) 2 = 1.8 x 10 –11 ]. Exercise 16.6B

25 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 25 Example 16.7 A Conceptual Example Pictured here is the result of adding a few drops of concentrated KI(aq) to a dilute solution of Pb(NO 3 ) 2. What is the solid that first appears? Explain why it then disappears. Example 16.8 If L of M MgCl 2 and L of M NaF are mixed, should a precipitate of MgF 2 form? MgF 2 (s) Mg 2+ (aq) + 2 F – (aq) K sp = 3.7 x 10 –8

26 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 26 Example 16.7 Figure 16.4 shows the result of adding a few drops of concentrated KI(aq) to a dilute solution of Pb(NO 3 ) 2 (aq). What is the yellow solid that appears at first? Explain why it then disappears. Describe how you might use observations of the type suggested by the photographs of Figure 16.4 to estimate the value of K sp for PbI 2. Exercise 16.7A Analysis and Conclusions Where the drops of concentrated KI(aq) first enter the Pb(NO 3 ) solution, there is a localized excess of iodide ion. In this region of the solution, [I – ] is large enough so that Q ip = [Pb 2+ ][I – ] 2 > K sp Because Q ip > K sp, yellow PbI 2 (s) precipitates. When the PbI 2 (s) that initially formed settles below the point of entry of the KI(aq), however, [I – ] is no longer large enough to maintain a saturated solution for the concentration of Pb 2+ present, and therefore the solid redissolves. Based on a uniform distribution of I – ion throughout the solution, the following holds true: Q ip = [Pb 2+ ][I – ] 2 < K sp Consider the following situation similar to Example 16.7: Initially, the mL of dilute solution in the beaker is 1 x 10 –4 M and the more concentrated solution added dropwise is 0.10 M. (a) If the dilute solution is Pb(NO 3 ) 2 (aq) and the more concentrated one is KI(aq), will the observations made resemble Figure 16.4? Explain. (b) What should be observed if the dilute solution is KI(aq) and the more concentrated one is Pb(NO 3 ) 2 ? Explain. Exercise 16.7B

27 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 27 Example 16.8 If L of M MgCl 2 and L of M NaF are mixed, should a precipitate of MgF 2 form? Strategy We can apply the same three-step strategy as in Example 16.6, but now we will incorporate into Step 1 the assumption that the volume of the mixture of solutions is L L = L. Solution Step 1: Initial concentrations of ions: First, we need to determine [Mg 2+ ] and [F – ] as they initially exist in this mixed solution.

28 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 28 Example 16.8 continued Step 2: Evaluation of Q ip : Q ip = [Mg 2+ ][F – ] 2 = (5.0 x 10 –4 )(1.7 x 10 –2 ) 2 = 1.4 x 10 –7 Step 3: Comparison of Q ip and K sp : Because Q ip > K sp, we conclude that precipitation should occur. Q ip = 1.4 x 10 –7 is greater than K sp = 3.7 x 10 –8. Should a precipitate of MgF 2 (s) form when equal volumes of M MgCl 2 (aq) and M NaF(aq) are mixed? Exercise 16.8A How many grams of KI(s) should be dissolved in 10.5 L of M Pb(NO 3 ) 2 (aq) so that precipitation of PbI 2 (s) will just begin? Exercise 16.8B

29 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 29 A slightly soluble solid does not precipitate totally from solution … … but we generally consider precipitation to be “complete” if about 99.9% of the target ion is precipitated (0.1% or less left in solution). Three conditions generally favor completeness of precipitation: 1.A very small value of K sp. 2.A high initial concentration of the target ion. 3.A concentration of common ion that greatly exceeds that of the target ion. To Determine Whether Precipitation Is Complete

30 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 30 Example 16.9 To a solution with [Ca 2+ ] = M, we add sufficient solid ammonium oxalate, (NH 4 ) 2 C 2 O 4 (s), to make the initial [C 2 O 4 2– ] = M. Will precipitation of Ca 2+ as CaC 2 O 4 (s) be complete? CaC 2 O 4 (s) Ca 2+ (aq) + C 2 O 4 2– (aq) K sp = 2.7 x 10 –9

31 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 31 Example 16.9 To a solution with [Ca 2+ ] = M, we add sufficient solid ammonium oxalate, (NH 4 ) 2 C 2 O 4 (s), to make the initial [C 2 O 4 2– ] = M. Will the precipitation of Ca 2+ as CaC 2 O 4 (s) be complete? Strategy We can begin with the three-step approach to determine whether precipitation occurs. If no precipitation occurs, we need not proceed further—precipitation obviously will not be complete. If precipitation does occur, we will need to make further calculations, which we will do in two parts: First, we will treat the situation as if complete precipitation occurred. This will enable us to determine the excess [C 2 O 4 2– ] present. Then, we will solve for the solubility of a slightly soluble solute in the presence of a common ion in a calculation similar to that of Example Solution First, we use our usual three-step approach to determine if precipitation occurs. Step 1: Initial concentrations of ions: These are given. [Ca 2+ ] = M, and [C 2 O 4 2– ] = M. Step 2: Evaluation of Q ip : Q ip = [Ca 2+ ][C 2 O 4 2– ] = (5.0 x 10 –3 )(5.1 x 10 –3 ) = 2.6 x 10 –5

32 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 32 Example 16.9 continued Step 3: Comparison of Q ip and K sp : Q ip = 2.6 x 10 –5 exceeds K sp = 2.7 x 10 –9 Because Q ip > K sp, precipitation should occur. Next, we follow the two-part approach outlined in our basic strategy. Assume complete precipitation. We describe the stoichiometry of the precipitation in a format resembling the ICE format for equilibrium calculations (see Example 15.17b). The reaction:Ca 2+ (aq) + C 2 O 4 2– (aq)  CaC 2 O 4 (s) Initial concentrations, M: Changes, M:–0.0050– After precipitation, M: The solubility equilibrium. Here we enter the appropriate data into the ICE format for solubility equilibrium in the presence of a common ion.

33 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 33 Example 16.9 continued As usual, we assume that s << and that ( s)  (We test the validity of this assumption in the Assessment.) K sp = [Ca 2+ ][C 2 O 4 2– ] = s( s)  s x = 2.7 x 10 –9 s = [Ca 2+ ] = 3 x 10 –5 M We calculate the percentage of Ca 2+ remaining in solution as the ratio of the remaining Ca 2+ concentration to the initial Ca 2+ concentration (5 x 10 –3 M). Our rule for complete precipitation requires that less than 0.1% of an ion should remain in solution. We therefore conclude that precipitation is incomplete. Assessment In this problem, we made the assumption that the contribution of s to the term s was negligible. Such assumptions are usually valid in common ion situations because the concentration of the common ion is generally relatively large, but M is not a high concentration. If we had not made the simplifying assumption and solved a quadratic equation, we would have obtained s = 2 x 10 –5 M. From this result, the percent Ca 2+ remaining in solution is 0.4%, but our conclusion that precipitation is incomplete is still valid. If our task had been to calculate [Ca 2+ ] remaining in solution, we would not have made the simplifying assumption.

34 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 34 Example 16.9 continued Consider a solution in which [Ca 2+ ] = M. If we add sufficient solid ammonium oxalate, (NH 4 ) 2 C 2 O 4 (s), so that the solution is also [C 2 O 4 2– ] = M, will the precipitation of Ca 2+ as CaC 2 O 4 (s) be complete? Exercise 16.9A The two members of each pair of ions are brought together in solution. Without doing detailed calculations, arrange the pairs according to how close to completeness the cation precipitation is—farthest from completeness first and closest to completeness last. (a) [Ca 2+ ] = M and [SO 4 2– ] = M (b) [Pb 2+ ] = 0.12 M and [Cl – ] = 0.25 M (c) [Mg 2+ ] = 0.12 M and [SO 4 2– ] = 0.15 M (d) [Ag + ] = M and [Cl – ] = M Exercise 16.9B

35 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 35 Selective Precipitation AgNO 3 added to a mixture containing Cl – and I –

36 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 36 Example An aqueous solution that is 2.00 M in AgNO 3 is slowly added from a buret to an aqueous solution that is M in Cl – and also M in I –. a.Which ion, Cl – or I –, is the first to precipitate from solution? b.When the second ion begins to precipitate, what is the remaining concentration of the first ion? c.Is separation of the two ions by selective precipitation feasible? AgCl(s) Ag + (aq) + Cl – (aq) K sp = 1.8 x 10 –10 AgI(s) Ag + (aq) + I – (aq) K sp = 8.5 x 10 –17

37 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 37 Example An aqueous solution that is 2.00 M in AgNO 3 is slowly added from a buret to an aqueous solution that is M in Cl – and M in I – (Figure 16.5). (a)Which ion, Cl – or I –, is the first to precipitate from solution? (b)When the second ion begins to precipitate, what is the remaining concentration of the first ion? (c)Is separation of the two ions by selective precipitation feasible?

38 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 38 Example continued Strategy The molarity of the silver nitrate solution is 200 times greater than [Cl – ] or [I – ], which means that only a small volume of AgNO 3 (aq) is required in the precipitation. Because of this, we can neglect throughout the calculation the slight dilution that occurs as the AgNO 3 (aq) is added to the beaker. Solution (a)We must find the [Ag + ] necessary to precipitate each anion. Precipitation will begin when the solution has just been saturated, a point at which Q ip = K sp. To precipitate Cl –. When Q ip is equal to K sp for silver chloride, [Cl – ] = M. We then solve the K sp expression for [Ag + ]. To precipitate I –. When Q ip is equal to K sp for silver chloride, [I – ] = M. We then solve the K sp expression for [Ag + ].

39 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 39 Example continued Precipitation of AgI(s) begins as soon as [Ag + ] = 1.8 x 10 –15 M, a concentration much too low for AgCl(s) to precipitate. Therefore I – is the first ion to precipitate. (b)Next, we determine [I – ] remaining in solution when [Ag + ] = 1.8 x 10 –8 M, the [Ag + ] at which AgCl(s) begins to precipitate. To determine this quantity, we rearrange the K sp expression for silver iodide and solve for [I – ]. (c)Now, let us see what percent of the I – is still in solution when AgCl(s) begins to precipitate. The amount of I – remaining is far below the 0.1% we require for completeness of precipitation. Thus, complete precipitation of I – occurs before Cl – precipitation begins. Separation of the two anions is feasible. Assessment This example illustrates a general rule: The greater the difference in magnitude of the relevant K sp values, the more likely it is that ion separations by selective precipitation can be achieved. The K sp values of AgI and AgCl differ by a factor of about 2 x 10 6.

40 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 40 If the anion of a precipitate is that of a weak acid, the precipitate will dissolve somewhat when the pH is lowered: If, however, the anion of the precipitate is that of a strong acid, lowering the pH will have no effect on the precipitate. Added H + reacts with, and removes, F – ; LeChâtelier’s principle says more F – forms. H + does not consume Cl – ; acid does not affect the equilibrium. Effect of pH on Solubility CaF 2 (s) Ca 2+ (aq) + 2 F – (aq) AgCl(s) Ag + (aq) + Cl – (aq)

41 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 41 Example What is the molar solubility of Mg(OH) 2 (s) in a buffer solution having [OH – ] = 1.0 x 10 –5 M, that is, pH = 9.00? Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH – (aq) K sp = 1.8 x 10 –11 Example A Conceptual Example Without doing detailed calculations, determine in which of the following solutions Mg(OH) 2 (s) is most soluble: (a) 1.00 M NH 3 (b) 1.00 M NH 3 /1.00 M NH 4 + (c) 1.00 M NH 4 Cl.

42 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 42 Strategy First, note that because the solution is a buffer, any OH – coming from the dissolution of Mg(OH) 2 (s) is neutralized by the acidic component of the buffer. Therefore, the pH will remain at 9.00 and the [OH – ] at 1.0 x 10 –5 M. We will use the K sp expression and [OH – ] = 1.0 x 10 –5 M to calculate [Mg 2+ ] in the solution. Example What is the molar solubility of Mg(OH) 2 (s) in a buffer solution having [OH – ] = 1.0 x 10 –5 M, that is, pH = 9.00? Strategy For every mole of Mg 2+ (aq) appearing in solution, 1 mol of Mg(OH) 2 (s) must have dissolved. Thus the molar solubility of Mg(OH) 2 and the equilibrium concentration of Mg 2+ are the same, meaning that the molar solubility of Mg(OH) 2 at pH 9.00 is 0.18 mol Mg(OH) 2 /L.

43 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 43 Equilibria Involving Complex Ions Silver chloride becomes more soluble, not less soluble, in high concentrations of chloride ion.

44 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 44 A complex ion consists of a central metal atom or ion, with other groups called ligands bonded to it. The metal ion acts as a Lewis acid (accepts electron pairs). Ligands act as Lewis bases (donate electron pairs). The equilibrium involving a complex ion, the metal ion, and the ligands may be described through a formation constant, K f : Complex Ion Formation Ag + (aq) + 2 Cl – (aq) [AgCl 2 ] – (aq) [AgCl 2 ] – K f = –––––––––– = 1.2 x 10 8 [Ag + ][Cl – ] 2

45 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 45 Complex Ion Formation Concentrated NH 3 added to a solution of pale-blue Cu 2+ … … forms deep-blue Cu(NH 3 ) 4 2+.

46 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 46

47 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 47 Complex Ion Formation and Solubilities AgCl is insoluble in water. But if the concentration of NH 3 is made high enough … … the AgCl forms the soluble [Ag(NH 3 ) 2 ] + ion.

48 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 48 Example Calculate the concentration of free silver ion, [Ag + ], in an aqueous solution prepared as 0.10 M AgNO 3 and 3.0 M NH 3. Ag + (aq) + 2 NH 3 (aq) [Ag(NH 3 ) 2 ] + (aq) K f = 1.6 x 10 7 Example If 1.00 g KBr is added to 1.00 L of the solution described in Example 16.13, should any AgBr(s) precipitate from the solution? AgBr(s) Ag + (aq) + Br – (aq) K sp = 5.0 x 10 –13

49 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 49 Example Calculate the concentration of free silver ion, [Ag + ], in an aqueous solution prepared as 0.10 M AgNO 3 and 3.0 M NH 3. Strategy First, because K f is such a large number, we can assume that the formation of [Ag(NH 3 ) 2 ] + goes to completion. We can then establish the concentrations of [Ag(NH 3 ) 2 ] + and free NH 3 through a straightforward stoichiometric calculation. Next, we can return to the formation reaction for the complex ion and consider the extent to which it proceeds in the reverse direction. For this calculation, we will use the K f expression and solve for x = [Ag + ]. Solution We complete the bottom row of our ICE tabulation to establish the concentrations of complex ion and free ammonia, assuming that the formation reaction goes to completion.

50 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 50 Example continued Now, we restate the question in this way: What is [Ag + ] in a solution that is 0.10 M [Ag(NH 3 ) 2 ] + and 2.8 M NH 3 ? To answer this question, we need to focus on the reverse direction in the reaction for [Ag(NH 3 ) 2 ] + formation. When we substitute the equilibrium concentrations into the K f expression, we obtain an algebraic equation that we must solve for x = [Ag + ]. Because the formation constant is so large, practically all the silver will be present as [Ag(NH 3 ) 2 ] + and very little as Ag +. We can therefore assume that x is very much smaller than 0.10 and that 2x is very much smaller than 2.8.

51 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 51 Example If 1.00 g KBr is added to 1.00 L of the solution described in Example 16.13, should any AgBr(s) precipitate from the solution? Strategy We need to compare Q ip for the described solution with K sp for silver bromide. In the Q ip expression, [Ag + ] is that calculated in Example We can establish [Br – ] from data given in this example. Solution From Example 16.13, we know that [Ag + ] = 8.0 x 10 –10 M. We determine [Br – ] when 1.00 g KBr is dissolved in 1.00 L of the solution with the equation Because Q ip > K sp, we conclude that some AgBr(s) should precipitate from the solution.

52 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 52 Example A Conceptual Example Figure shows that a precipitate forms when HNO 3 (aq) is added to the solution in the beaker on the right in Figure Write the equation(s) to show what happens. Example What is the molar solubility of AgBr(s) in 3.0 M NH 3 ? AgBr(s) + 2 NH 3 (aq) [Ag(NH 3 ) 2 ] + (aq) + Br – (aq) K c = 8.0 x 10 –6

53 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 53 Example What is the molar solubility of AgBr(s) in 3.0 M NH 3 ? Strategy The key relationship in the dissolution equilibrium is that for every formula unit of AgBr(s) that is dissolved, one ion each of [Ag(NH 3 ) 2 ] + and Br – is formed. The molar solubility is the same as the molarity of either of these ions at equilibrium. We calculate these ion molarities in an equilibrium calculation based on the K c expression. Solution We can use the ICE format to establish equilibrium concentrations in terms of the initial concentration of NH 3 (aq) and the unknown equilibrium concentration of [Ag(NH 3 ) 2 ] + (aq) = s. Now we write the equilibrium constant expression for the reaction and substitute the above data.

54 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 54 Example continued At this point we could assume that s << 3.0 and proceed in the usual fashion, but we can simplify our calculation just by taking the square root of each side of the equation. Then, we can complete the calculation. s = (3.0 – 2s)(2.8 x 10 –3 ) = (8.4 x 10 –3 ) – 5.6 x 10 –3 s s = 8.4 x 10 –3 s = [Ag(NH 3 ) 2 ] + = [Br – ] = 8.4 x 10 –3 M The molar solubility is therefore 8.4 x 10 –3 mol AgBr/L. What is the molar solubility of AgBr(s) in M Na 2 S 2 O 3 (aq)? Exercise 16.15A Without doing detailed calculations, determine in which of the following M solutions AgI(s) should be most soluble: NH 3 (aq), Na 2 S 2 O 3 (aq), or NaCN(aq). Explain. Exercise 16.15B

55 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 55 Water molecules are commonly found as ligands in complex ions (H 2 O is a Lewis base). The electron-withdrawing power of a small, highly charged metal ion can weaken an O—H bond in one of the ligand water molecules. The weakened O—H bond can then give up its proton to another water molecule in the solution. The complex ion acts as an acid. Complex Ions in Acid–Base Reactions [Na(H 2 O) 4 ] + [Al(H 2 O) 6 ] 3+ [Fe(H 2 O) 6 ] 3+

56 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 56 Ionization of a Complex Ion [Fe(H 2 O) 6 ] 3+ + H 2 O [Fe(H 2 O) 5 OH] 2+ + H 3 O + The highly-charged iron(III) ion withdraws electron density from the O—H bonds. K a = 1 x 10 –7

57 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 57 Certain metal hydroxides, insoluble in water, are amphoteric; they will react with both strong acids and strong bases. Al(OH) 3, Zn(OH) 2, and Cr(OH) 3 are amphoteric. Amphoteric Species

58 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 58 Acid–base chemistry, precipitation reactions, oxidation– reduction, and complex ion formation all apply to an area of analytical chemistry called classical qualitative inorganic analysis. “Qualitative” signifies that the interest is in determining what is present. –Quantitative analyses are those that determine how much of a particular substance or species is present. Although classical qualitative analysis is not used as widely today as are instrumental methods, it is still a good vehicle for applying all the basic concepts of equilibria in aqueous solutions. Qualitative Inorganic Analysis

59 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 59 Qualitative Analysis Outline In acid, H 2 S produces very little S 2–, so only the most-insoluble sulfides precipitate. In base, there is more S 2–, and the less-insoluble sulfides also precipitate. Some hydroxides also precipitate here.

60 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 60 If aqueous HCl is added to an unknown solution of cations, and a precipitate forms, then the unknown contains one or more of these cations: Pb 2+, Hg 2 2+, or Ag +. These are the only ions to form insoluble chlorides. Any precipitate is separated from the mixture and further tests are performed to determine which of the three Group 1 cations are present. The supernatant liquid is also saved for further analysis (it contains the rest of the cations). If there is no precipitate, then Group 1 ions must be absent from the mixture. Cation Group 1

61 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 61 Precipitated PbCl 2 is slightly soluble in hot water. The precipitate is washed with hot water, then aqueous K 2 CrO 4 is added to the washings. If Pb 2+ is present, a precipitate of yellow lead chromate forms, which is less soluble than PbCl 2. (If all of the precipitate dissolves in the hot water, what does that mean?) Cation Group 1 (cont’d) Analyzing for Pb 2+

62 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 62 Next, any undissolved precipitate is treated with aqueous ammonia. If AgCl is present, it will dissolve, forming Ag(NH 3 ) 2 + (the dissolution may not be visually apparent). If Hg 2 2+ is present, the precipitate will turn dark gray/ black, due to a disproportionation reaction that forms Hg metal and HgNH 2 Cl. The supernatant liquid (which contains the Ag +, if present) is then treated with aqueous nitric acid. If a precipitate reforms, then Ag + was present in the solution. Cation Group 1 (cont’d) Analyzing for Ag + and Hg 2 2+

63 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 63 PbCl 2 precipitates when HCl is added. Hg 2 Cl 2 reacts with NH 3 to form black Hg metal and HgNH 2 Cl. The presence of lead is confirmed by adding chromate ion; yellow PbCrO 4 precipitates. Group 1 Cation Precipitates

64 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 64 Once the Group 1 cations have been precipitated, hydrogen sulfide is used as the next reagent in the qualitative analysis scheme. H 2 S is a weak diprotic acid; there is very little ionization of the HS – ion and it is the precipitating agent. Hydrogen sulfide has the familiar rotten egg odor that is very noticeable around volcanic areas. Because of its toxicity, H 2 S is generally produced only in small quantities and directly in the solution where it is to be used. Hydrogen Sulfide in the Qualitative Analysis Scheme

65 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 65 The concentration of HS – is so low in a strongly acidic solution, that only the most insoluble sulfides precipitate. These include the eight metal sulfides of Group 2. Five of the Group 3 cations form sulfides that are soluble in acidic solution but insoluble in alkaline NH 3 /NH 4 +. The other three Group 3 cations form insoluble hydroxides in the alkaline solution. The cations of Groups 4 and 5 are soluble. Group 4 ions are precipitated as carbonates. Group 5 does not precipitate; these must be determined by flame test. Cation Groups 2, 3, 4, and 5

66 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 66 Cumulative Example A solid mixture containing 1.00 g of ammonium chloride and 2.00 g of barium hydroxide is heated to expel ammonia. The liberated NH 3 is then dissolved in L of water containing 225 ppm Ca 2+ as calcium chloride. Will a precipitate form in this water?

67 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 67 Cumulative Example A solid mixture containing 1.00 g of ammonium chloride and 2.00 g of barium hydroxide is heated to expel ammonia. The liberated NH 3 (g) is then dissolved in L of water containing 225 ppm Ca 2+ as calcium chloride. Will a precipitate form in this water? Strategy First, the precipitate, if there is one, will be Ca(OH) 2 (s). The Ca 2+ is present in the water, and after the liberated NH 3 (g) dissolves in the water, the OH – is produced by ionization of the weak base NH 3 (aq): If a precipitate is to form, the final ion product in the water, Q ip = [Ca 2+ ][OH – ] 2, must exceed K sp for Ca(OH) 2 : To determine the amount of NH 3 (g) liberated, we must perform a limiting-reactant calculation (recall Example 3.20) for the reaction The final steps in the calculation involve obtaining [OH – ] from equation (a), converting ppm Ca 2+ to [Ca 2+ ], and comparing Q ip and K sp for reaction (b).

68 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 68 Cumulative Example continued Solution To determine the amount of NH 3 (g) available, we complete two stoichiometric calculations based on reaction (c). Assuming NH 4 Cl is the limiting reactant: Assuming Ba(OH) 2 is the limiting reactant: Based on the smaller of the two values just calculated, we determine the molarity of NH 3. Now we use the method of Example 15.8 to determine [OH – ], where [OH – ] = x and [NH 3 ] = – x. We assume that x << and solve for x = [OH – ].

69 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Sixteen 69 Cumulative Example continued The quantity of liquid into which we inject the NH 3 (g) is L. Because this liquid is nearly pure water with d = 1.00 g/mL, its mass is 500 g. We can then calculate the mass of Ca 2+ in this 500 g water, based on the literal meaning of 225 ppm. By converting from mass to moles of Ca 2+ and dividing by the solution volume, we determine [Ca 2+ ]. Now we can evaluate the ion product, Q ip, and compare it with K sp for Ca(OH) 2. Q ip = [Ca 2+ ][OH – ] 2 = 5.64 x 10 –3 x (8.2 x 10 –4 ) 2 = 3.8 x 10 –9 When we compare Q ip with K sp, we conclude that no precipitate will form. Q ip = 3.8 x 10 –9 is considerably smaller than K sp = 5.5 x 10 –6. Assessment In determining [OH – ], we made the assumption that it would be much smaller than [NH 3 ], and this assumption is valid: x = 8.2 x 10 –4 is only about 2% of Note that even if a precipitate had formed, its quantity would have been very small. The maximum yield from g Ca 2+ would be g Ca(OH) 2 [that is, g Ca 2+ x (74.1 g Ca(OH) 2 / 40.1 g 2 Ca 2+ ]. With incomplete precipitation, the quantity would have been smaller still. Yet even a small amount of precipitate may be discernible as turbidity in a solution.


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