1. Copyright Sautter 2003

2. SOLUBILITY EQUILIBRIUM Solubility refers to the ability of a substance to dissolve. In the study of solubility equilibrium we generally deal with low solubility compounds (those which dissolve poorly) Low solubility compounds are classified as precipitates. A set of solubility rules indicate which combination of ions generally form precipitates. These rules allow us to identify low solubility substances and potential precipitates. The concentrations of the ions present however are the final determiners of precipitate formation.

3. SOLUBILITY RULES (1) Inorganic acids (those not consisting of carbon, hydrogen and oxygen) are soluble. Low molecular weight organic acids are soluble. (2) Compound containing alkali metals and ammonium (NH 4 + ) cations are soluble. (3) Compounds containing nitrates (NO 3 -), acetates (C 2 H 3 O 2 - ) and perchlorates (ClO 4 - ) are soluble. (4) Compounds containing chlorides (Cl - ) are soluble except for AgCl, PbCl 2 and Hg 2 Cl 2. Compounds containing bromides (Br - ) and iodides (I - ) have solubility similar to chlorides. (5) Sulfate containing compounds are soluble except for PbSO 4 and BaSO 4. CaSO 4 and Ag 2 SO 4 are slightly soluble. (6) Hydroxides are insoluble (precipitates) except Alkali Metal (I) and heavier Alkaline Earth Metal (II) hydroxides after Mg. (7) Carbonates (CO 3 -2 ), phosphates (PO 4 -3 ) and arsenates (AsO 4 -3 ) are insoluble except for Alkali Metal (I) and Alkaline Earth Metal (II) compounds (8) Sulfides (S -2 ) are insoluble except for Alkali Metal (I) and Alkaline Earth Metal (II) compounds.

4. ALL LOW SOLUBILITY SYSTEMS ARE EQUILIBRIUM SYSTEMS RATE OF DISSOLVING = RATE OF CRYSTALIZATION RATE OF FORWARD PROCESS = RATE OF REVERSE PROCESS The amount of substance dissolved remains constant The system appears macroscopically static.

5. Ksp AND SOLUBILITY Since solubility is an equilibrium phenomena, equilibrium concepts can readily be applied. If the system is at equilibrium, the solution must be saturated (no more solid can be dissolved). Equilibrium concepts cannot be applied to unsaturated solutions (more solid can still be dissolved) since equilibrium has not yet been reached. For a typical saturated solution of silver chloride, AgCl (s)  Ag + (aq) + Cl - (aq) the equilibrium expression is: K = [Ag + ] x [Cl - ] / 1 or simply [Ag + ] x [Cl - ] Since AgCl is and solid it is not included in the equilibrium expression! This K value is called the K sp. “sp” stands for solubility product. The product of the concentrations of the soluble ions.

6. CALCULATING Ksp Problem: A saturated solution of BaSO 4 contains 0.0025 grams in one liter. Calculate the Ksp for barium sulfate. Solution: Since Ksp requires concentrations is moles per liter we must first calculate the number of moles present. Moles = 0.0025 gram / 233 grams per mole = 1.1 x 10 -5 M BaSO 4(s)  Ba +2 (aq) + SO 4 -2 (aq) Each dissolved BaSO 4 forms one Ba +2 ion and one SO 4 -2 ion, therefore: [BaSO 4 ] = 1.1 x 10 -5 M, [Ba +2 ] = [SO 4 -2 ] = 1.1 x10 -5 M Ksp = [Ba +2 ] x [SO 4 -2 ] = (1.1 x 10 -5 ) 2 = 1.2 x10 -10

7. SOLUBILITY PRODUCTS OF SOME COMMON SALTS AT 25 0 C

8. Calculating Concentrations From Ksp Problem: Find the concentration of dissolved Al(OH) 3 in a saturated solution. Solution: Al(OH) 3 (s)  Al +3 (aq) + 3 OH - (aq) Ksp = [Al +3 ] x [OH - ] 3, Ksp = 1.9 x 10 -33 Each dissolved Al(OH) 3 gives one Al +3 and three OH - ions therefore if [Al +3 ] = X, [OH - ] = 3X and Ksp = X (3X) 3 = 27 X 4 = 3.0 x 10 -34 X = (3.0 x 10 -34 / 27) 1/4 = 1.8 x 10 -9 M

9. THE COMMON ION EFFECT The common ion effect is an application of LeChatelier’s Principle. Recall that equilibrium shifts are caused be stresses applied to the system. If additional products are added to an equilibrium system that system will shift towards the reactants. In a solubility equilibrium adding one of the dissolved ion components will cause a shift left and the solubility of the compound will decrease. Reducing the concentration of one of the dissolved ions will cause a shift to the right thereby allowing the compound to dissolve to a greater extend.

10. THE COMMON ION EFFECT Add NaCl to the system System shifts left!

11. THE COMMON ION EFFECT Problem: Calculate the solubility of silver chloride in pure water and in a 0.10 M NaCl solution. Compare the results. Solution: AgCl (s)  Ag + (aq) + Cl - (aq) Ksp = [Ag + ] x [Cl - ] = 1.8 x 10 -10 Each dissolved AgCl gives one Ag + and one Cl - ion therefore if [Ag + ] = X, [Cl - ] = X Ksp = X 2 = 1.8 x 10 -10, X = (1.8 x 10 -10 ) 1/2 = 1.3 x 10 -5 M In 0.10 M NaCl the concentration of Cl - = 0.10 M and Ksp = [Ag + ] (0.10)* = 1.8 x 10 -10, [Ag + ] = 1.8 x 10 -10 / 0.10 ]Ag + ] = 1.8 x 10 -9 M (the presence of the common Cl - ion in solution has shifted the equilibrium hard to the left and made the silver chloride much less soluble than in pure water *Note: the Cl - from the AgCl in the common ion system is so small as compared to the 0.10 M Cl - from the NaCl that is was neglected.

12. WILL A PRECIPITATE FORM? In order to decide whether a precipitate will form when two solutions are mixed we must first calculate the value of Q (reaction quotient). Q is calculated with the same format as Ksp (an ion concentration product) If the value of Q exceeds the Ksp value, a precipitate will form. If the value of Q is less than Ksp, a precipitate will not be formed. If Q = Ksp then the solution is a saturated solution.

13. WILL A PRECIPITATE FORM?

14. 0.0001 M AgNO 3 0.00002 M NaCl 100 ml200 ml WILL A PRECIPITATE FORM?

15. Problem: If 100 ml of 0.0001 M AgNO 3 is mixed with 200 ml of 0.0002 M NaCl, will a precipitate form? Solution: The potential precipitate is AgCl. (See the solubility rules). Q = [Ag + ] x [Cl - ], when the two solutions are mixed, the concentrations of both the Ag + ion and the Cl - ion will change due to dilution of one solution by the other. M 1 V 1 = M 2 V 2 (the dilution formula) For Ag +, 0.0001 (100) = [Ag + ] (100 + 200) [Ag + ] = 3.33 x 10 -5 M For Cl -, 0.0002 (100) = [Cl - ] (100 + 200) [Cl - ] = 6.67 x 10 -5 M Q = (3.33 x 10 -5 ) (6.67 x 10 -5 ) = 2.22 x 10 -9, Ksp = 1.8 x 10 -10 Q > Ksp therefore a precipitate will form

16. DISSOLVING PRECIPITATES Several methods of dissolving low solubility materials can be used. One involves the use of weak acid conjugate bases. Systems that can be affected by these changes are said to be pH sensitive. Another method involves the formation of complex ions from one of the dissolved ions. Recall that a weak acid is an equilibrium system in which the acid dissociates poorly and its conjugate base readily forms the molecular acid. For example in hydrosulfuric acid H 2 S (aq)  2 H + (aq) + S -2 (aq) sulfide ion (S -2 ) will readily be consumed by H + ion to form molecular hydrosulfuric acid (H 2 S). Precipitates containing sulfide ion are pH sensitive since added H + ion will cause a shift towards the molecular acid thereby consuming S -2 ion and allow further dissolving of the precipitate.

17. DISSOLVING PRECIPITATES ADD AN ACID (pH IS LOWERED) SOLUBILITY OF PbS INCREASES AS A RESULT OF DECREASED pH

18. DISSOLVING PRECIPITATES ADDING A BASE (OH - ) RAISES PH AND EQUILIBRIUM SHIFTS LEFT MAKING Al(OH) 3 LESS SOLUBLE

19. COMPLEX ION FORMATION & SOLUBILITY Complex ions are formed when simple ions are combined with other ions to form new, more complex ions. Also simple ions may combine with molecules in solution which when attached to the ion are called ligands. Examples of complex ion formation include: Ag + (aq) + 2 NH 3 (aq)  Ag(NH 3 ) 2 + (aq) Fe +3 (aq) + SCN - (aq)  FeSCN +2 (aq) Cu +2 (aq) + 4 NH 3(aq)  Cu(NH 3 ) 4 +2 (aq) Complex ion formation reactions generally have large K values and occur readily in solution.

20. COMPLEX ION FORMATION & SOLUBILITY AgCl dissolves well in ammonia (NH 3 ) Solubility due to equilibrium shift caused by the complex formation NH 3 is added

21. COMPLEX ION FORMATION & SOLUBILITY Problem: What is the solubility of AgCl in 0.10 M NH 3 solution? Solution: Equ (1) AgCl (s)  Ag + (aq) + Cl - (aq) and Equ (2) Ag + (aq) + 2 NH 3 (aq)  Ag(NH 3 ) 2 + (aq) Adding equ (1) & (2): AgCl (s) + 2 NH 3 (aq)  Ag(NH 3 ) 2 + (aq) + Cl - (aq) Ksp AgCl = 1.8 x 10 -10, K f for the complex formation is 1.6 x 10 7 K (equ 1 + equ 2) = Ksp x K f = (1.8 x 10 -10 )(1.6 x 10 7 ) = 2.72 x 10 -3 K = ([Ag(NH 3 ) 2 + ] [Cl-]) / [NH 3 ] 2 = 2.72 x 10 -3 Let X = [Ag(NH 3 ) 2 + ] e, and since [Ag(NH 3 ) 2 + ] e = [Cl - ] e and two NH 3 are used to form one Ag(NH 3 ) 2 + [Ag(NH 3 ) 2 + ] e = X, [Cl - ] e = X and [NH 3 ] e = (0.10 – 2X) X 2 / (0.10 – 2X) 2 = 2.72 x 10 -3 and taking the square root of both sides of the equation we get: X / (0.10 – 2X) = (2.72 x 10 -3 ) 1/2 X = [Ag(NH 3 ) 2 + ] e = 4.7 x 10 -3 M (the solubility of AgCl in 0.10 M NH 3 )

23. SOLUBILITY AND LABORATORY ANALYSIS A common laboratory analysis involves the identification of unknown components of a solution. The procedure is called Qualitative Analysis meaning that just the identity of the components is sought not the amounts presence. An analysis of a solution may often require the determination of the presence of silver, lead or mercury. These three ions are referred to a Group I cations. In the case of Qualitative Analysis, Group I does not pertain to column I on the periodic table (the alkali metals). It refers only to Ag +, Pb +2 and Hg 2 +2 ions. The procedures involved in Group I analysis rely on the solubility characteristic of the three ions of the group.

24. SOLUBILITY AND LABORATORY ANALYSIS A common property of all three Group I ions is that they form precipitates with chloride ion. Ag + (aq) + Cl - (aq)  AgCl (s) Pb +2 (aq) + Cl - (aq)  PbCl 2(s) Hg 2 +2 (aq) + Cl - (aq)  Hg 2 Cl 2(s) In the second step of the analysis the temperature dependent properties of the precipitates allows separation of the three solids. Lead chloride is soluble in hot water while the solubility of the other solids are independent of temperature.(Solubility is temperature dependent for many precipitates. Some are unaffected by temperature, some dissolve better at higher temperatures and some dissolve more poorly at elevated temperatures)

25. GROUP I ANALYSIS Step I – Chloride Precipitation Ag + Pb +2 Hg 2 +2 Cl - (chloride ion) PRECIPITATES AgCl PbCl 2 Hg 2 Cl 2

26. SOLUBILITY CURVES FOR VARIOUS SOLIDS

27. GROUP I ANALYSIS Step II – Precipitate Separation Hot Water PRECIPITATES AgCl PbCl 2 Hg 2 Cl 2 Lead Chloride Dissolves PbCl 2(s)  Pb +2 (aq) + 2 Cl - (aq)

28. SOLUBILITY AND LABORATORY ANALYSIS The formation of complex ions is used to separate AgCl from the remaining Hg 2 Cl 2 precipitate. Recall that Ag + ion forms the complex Ag(NH 3 ) 2 + and therefore the addition of aqueous ammonia solution will dissolve AgCl solid leaving the mercury precipitate behind. A confirmatory test for the Ag + involves change the pH of the solution in which the silver complex is dissolved. Adding nitric acid neutralizes the ammonia ligands on the complex and the AgCl solid again precipitates as a white solid.

29. GROUP I ANALYSIS Step III – Complex Ion Formation Ammonia Solution (NH 3 ) PRECIPITATES AgCl Hg 2 Cl 2 Silver Chloride Dissolves AgCl (s) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) + Cl - (aq)

30. GROUP I ANALYSIS Step IIII – Precipitation by pH Change Nitric Acid (HNO 3 ) COMPLEX ION SOLUTION Ag(NH 3 ) 2 + Silver Chloride Precipitates Ag(NH 3 ) 2 + (aq) + Cl - (aq) + H + (aq)  AgCl (s) + NH 4 + (aq)