Chapter 16: Solubility and Complex Ion Equilibria PDF

Chapter 16 Solubility and Complex Ion Equilibria Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.1 Solubility Equilibria and the Solubility Product Equilibria ▪ When a typical ionic solid diss...

Chapter 16 Solubility and Complex Ion Equilibria Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.1 Solubility Equilibria and the Solubility Product Equilibria ▪ When a typical ionic solid dissolves in water, it separates into cations and anions ▪ Example CaF2 ( s ) ⎯⎯⎯ H O → Ca 2+ ( aq ) + 2F− ( aq ) 2 ▪ Ions formed - Ca2+ and F– ▪ In this reaction, when solid salt is first added, no ions are present ▪ As dissolution proceeds, the ionic concentration increases Copyright ©2017 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 2 Section 16.1 Solubility Equilibria and the Solubility Product Equilibria (Continued) ▪ The dissolution reaction and its reverse occurs, simultaneously Ca 2+ ( aq ) + 2F− ( aq ) → CaF2 ( s ) ▪ Ultimately, the solution attains a state of saturation (dynamic equilibrium) CaF2(s) ⇌ Ca2+(aq) + 2F–(aq) ▪ Equilibrium expression - Ksp = [Ca2+][F–]2 Copyright ©2017 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 3 Section 16.1 Solubility Equilibria and the Solubility Product Solubility Equilibria ▪ Solubility product constant (Ksp) ▪ Equilibrium expression constant that represents the dissolution of an ionic solid in water ▪ Known as solubility product ▪ Solubility equilibrium is unaffected by: ▪ Excess solid formed ▪ Size of particles present Copyright ©2017 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 4 Section 16.1 Solubility Equilibria and the Solubility Product Solubility Equilibria (Continued) ▪ Differences between the solubility of a given solid and its solubility product ▪ Solubility is an equilibrium position ▪ Solubility product is an equilibrium constant ▪ Has only one value for a given solid at a given temperature Copyright ©2017 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 5 Section 16.1 Solubility Equilibria and the Solubility Product Interactive Example 16.2 - Calculating Ksp from Solubility II ▪ Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0×10–15 mol/L at 25°C Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.1 Solubility Equilibria and the Solubility Product Exercise ▪ Approximately 0.14 g nickel(II) hydroxide, Ni(OH)2 (s), dissolves per liter of water at 20°C ▪ Calculate Ksp for Ni(OH)2 (s) at this temperature Copyright ©2017 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 7 Section 16.1 Solubility Equilibria and the Solubility Product Interactive Example 16.3 - Calculating Solubility from Ksp ▪ The Ksp value for copper(II) iodate, Cu(IO3)2, is 1.4×10–7 at 25°C ▪ Calculate its solubility at 25°C Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.1 Solubility Equilibria and the Solubility Product Exercise ▪ Calculate the solubility of each of the following compounds in moles per liter a. Ag3PO4 Ksp = 1.8×10–18 b. CaCO3 Ksp = 8.7×10–9 Copyright ©2017 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 9 Section 16.1 Solubility Equilibria and the Solubility Product Relative Solubilities ▪ One must be careful in using Ksp values to predict the relative solubilities of a group of salts ▪ Possible cases ▪ Salts being compared produce the same number of ions ▪ Ksp values can be compared to determine relative solubilities ▪ Salts being compared produce varying number of ions ▪ Cannot predict relative solubilities using Ksp values Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.1 Solubility Equilibria and the Solubility Product Relative Solubilities ▪ Salts being compared produce the same number of ions ▪ AgI (s) Ksp = 1.5 x 10-16 ▪ CuI (s) Ksp = 5.0 x 10-12 ▪ CaSO4 (s) Ksp = 6.1 x 10-5 Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.1 Solubility Equilibria and the Solubility Product Relative Solubilities ▪ Salts being compared produce varying number of ions ▪ CuS (s) Ksp = 8.5 x 10-45 ▪ Ag2S (s) Ksp = 1.6 x 10-49 ▪ Bi2S3 (s) Ksp = 1.1 x 10-73 Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.1 Solubility Equilibria and the Solubility Product Common Ion Effect ▪ The solubility of a solid becomes low when the solution already contains ions common to the solid A potassium chromate solution being added to aqueous silver nitrate, forming silver chromate Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.1 Solubility Equilibria and the Solubility Product Interactive Example 16.4 - Solubility and Common Ions ▪ Calculate the solubility of solid CaF2 (Ksp = 4.0×10–11) in a 0.025-M NaF solution Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.1 Solubility Equilibria and the Solubility Product pH and Solubility ▪ Increase in pH Mg(OH)2 (s)→ Mg+2 (aq) + 2(OH-) (aq) ▪ Decreases solubility ▪ Forces the equilibrium to the left ▪ Decrease in pH ▪ Increases solubility ▪ Equilibrium shifts to the right ▪ If the anion X– is an effective base, the salt MX will show increased solubility in an acidic solution Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.1 Solubility Equilibria and the Solubility Product pH and Solubility (Continued) ▪ Exception - AgCl has the same solubility in acid as in pure water ▪ Cl– ion is a weak base ▪ No HCl molecules are formed ▪ Adding H+ ions to a solution that contains Cl– ions does not affect: ▪ Concentration of the Cl– ion ▪ Solubility of the chloride salt Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.2 Precipitation and Qualitative Analysis Precipitation ▪ Ion product (Q) ▪ Defined similar to the expression for Ksp for a given solid ▪ Exception - Initial concentrations are used instead of equilibrium concentrations Copyright ©2017 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 17 Section 16.2 Precipitation and Qualitative Analysis Relationship between Q and Ksp ▪ One can predict the possibility of precipitation by considering the relationship between Q and Ksp ▪ Q > Ksp ▪ Precipitation occurs and will continue until the concentrations are reduced to the point that they satisfy Ksp ▪ Q < Ksp ▪ No precipitation occurs Copyright ©2017 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 18 Section 16.2 Precipitation and Qualitative Analysis Interactive Example 16.5 - Determining Precipitation Conditions ▪ A solution is prepared by adding 750.0 mL of 4.00×10–3 M Ce(NO3)3 to 300.0 mL of 2.00×10–2 M KIO3 ▪ Will Ce(IO3)3 (Ksp = 1.9×10–10) precipitate from this solution? Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.2 Precipitation and Qualitative Analysis Calculating Equilibrium Concentrations after Precipitation ▪ Step 1 - Determine if the product is formed when the solutions are mixed ▪ Calculate the concentration of the ions in the mixed solution to determine Q ▪ Step 2 - Run the reaction to completion ▪ Note - If a reaction virtually goes to completion when two solutions are mixed, it is necessary to conduct stoichiometric calculations prior to equilibrium calculations Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.2 Precipitation and Qualitative Analysis Calculating Equilibrium Concentrations after Precipitation Occurs (Continued) ▪ Step 3 - Allow the system to adjust to equilibrium and determine the concentrations of ions in the solution ▪ Step 4 - Substitute the expressions derived in step 3 to determine Ksp Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.2 Precipitation and Qualitative Analysis Interactive Example 16.6 - Precipitation ▪ A solution is prepared by mixing 150.0 mL of 1.00×10–2 M Mg(NO3)2 and 250.0 mL of 1.00×10–1 M NaF ▪ Calculate the concentrations of Mg2+ and F– at equilibrium with solid MgF2 (Ksp = 6.4×10–9) Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.2 Precipitation and Qualitative Analysis Selective Precipitation ▪ Method used to separate mixtures of metal ions in aqueous solution ▪ Involves using a reagent whose anion forms a precipitate with only one or a few of the metal ions in the mixture Copyright ©2017 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 23 Section 16.2 Precipitation and Qualitative Analysis Selective Precipitation - Example ▪ Consider a solution containing Ba2+ and Ag+ ions ▪ NaCl is added to this solution ▪ Result ▪ AgCl precipitates as a white solid ▪ BaCl2 is soluble, so Ba2+ ions remain in the solution Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.2 Precipitation and Qualitative Analysis Example 16.7 - Selective Precipitation ▪ A solution contains 1.0×10–4 M Cu+ and 2.0×10–3 M Pb2+ ▪ If a source of I– is added gradually to this solution, will PbI2 (Ksp = 1.4×10–8) or CuI (Ksp = 5.3×10–12) precipitate first? ▪ Specify the concentration of I– necessary to begin precipitation of each salt Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.2 Precipitation and Qualitative Analysis Exercise ▪ A solution is 1×10–4 M in NaF, Na2S, and Na3PO4 ▪ What would be the order of precipitation as a source of Pb2+ is added gradually to the solution? ▪ The relevant Ksp values are: Ksp(PbF2) = 4 × 10–8 Ksp(PbS) = 7 × 10–29 Ksp[Pb3(PO4)2] = 1 × 10–54 Copyright ©2017 Cengage Learning. All Rights Reserved. Section 16.3 Equilibria Involving Complex Ions Complex Ion Equilibria ▪ Complex ion: Charged species consisting of a metal ion surrounded by ligands ▪ Ligand - Lewis base ▪ Molecule or ion that has a lone pair that can be donated to an empty orbital on the metal ion to form a covalent bond ▪ Coordination number - Number of ligands attached to a metal ion Copyright ©2017 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 27 Section 16.3 Equilibria Involving Complex Ions Formation (Stability) Constant ▪ Equilibrium constant for each step of the formation of a complex ion by the addition of an individual ligand to a metal ion or complex ion in aqueous solution ▪ Usually, the total concentration of the ligand is larger than the total concentration of the metal ion Copyright ©2017 Cengage Learning. All Rights Reserved. 28 Section 16.3 Equilibria Involving Complex Ions Interactive Example 16.8 - Complex Ions ▪ Calculate the concentrations of Ag+, Ag(S2O3)–, and Ag(S2O3)23– in a solution prepared by mixing 150.0 mL of 1.00×10–3 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3 ▪ The stepwise formation equilibria are as follows: Ag + + S2 O2− 3 ⇌ Ag(S2 O3 ) − K1 = 7.4 × 108 − 2− 3− Ag(S2 O3 ) + S2 O3 ⇌ Ag(S2 O3 )2 K 2 = 3.9 × 104 Copyright ©2017 Cengage Learning. All Rights Reserved.