Solubility Equilibria.notebook April 22, 2015
Solubility Equilibria and the Solubility Product One type of problem you will need to solve is to interconvert between solubility and Ksp. The keys to solving solubility problems are to Solubility refers to the amount of a salt that can be dissolved in water. Your text points out that the solubility of a salt is variable but the solubility product ("Ksp") is constant (at a given temperature). properly define solubility, and properly use the reaction stoichiometry The solubility equilibrium expression is set up as any other. For example, the equilibrium expression for the dissolution of 5 o Ag2S in water is Problem: Barium sulfate has a solubility of 3.9 X 10 M at 25 C. Calculate Ksp for BaSO4.
+ 2 49 Ag2S(s) = 2Ag (aq) + S (aq) Ksp = 1.6 X 10
+ 2 2 Ksp = [Ag ] [S ]
Remember the pure solid, Ag2S, is not included in the equilibrium expression.
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Note that relative solubilities among different compounds cannot be measured simply by comparing Ksp values. You must take the composition of the salt into account. Problem: Calculate the solubility of NiCO (K = 1.4 X 107) in moles per liter and grams per liter. 3 sp Problem: Which of the following compounds is the most soluble?
10 AgCl; Ksp = 1.5 X 10 12 Ag2CrO4 Ksp = 9.0 X 10 18 Ag3PO4 Ksp = 1.8 X 10
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Another type of problem you will encounter involves the common ion effect. Recall that LeChatelier's Principle predicts that adding a common ion to the solution shifts the equilibrium to the left in solubility equations. In other words adding a common ion (that does not react with other species in the solution) reduces the solubility.
10 Problem: Calculate the solubility of SrF2 (Ksp = 7.9 X 10 ) in Keep in mind that substances (such as acids or bases) that can combine with one of the ions of the salt will increase the solubility of (a) pure water the salt by removing product, thus forcing the reaction to the right.
2+ For example, Mg(OH)2(s) = Mg (aq) + 2OH (aq)
+ If we add acid (H3O ) to the above saturated solution, the equilibrium will shift right as [OH ] decreases because it reacts + (b) 0.100 M Sr(NO3)2 with H3O to form water.
(c) 0.200 M NaF
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3 5 Problem: A 200.0 mL solution of 1.3 X 10 M AgNO3 is mixed with 100.0 mL of a 4.5 X 10 M Na2S. Will 49 precipitation occur? (The Ksp for silver sulfide is 1.6 X 10 ).
Precipitation
In precipitation reactions we answer two questions:
1. If two solutions are mixed, will a precipitate form? 2. If it forms, what will be the concentrations of each ion at equilibrium?
Recall the idea of "Q" the calculated equilibrium constant for initial conditions. In this case if Q > Ksp then precipitation will occur.
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Problem: Calculate the equilibrium concentration of each ion in a solution obtained by mixing 50.0 mL of 6.0 X 103 11 M CaCl2 with 30.0 mL of 0.040 M NaF. (Ksp for CaF2 = 4.0 X 10 ).
Once we determine that precipitation will occur, we are faced with the problem of determining the equilibrium concentrations of each of our ions of interest. The general strategy involves assuming that because Ksp is so low, we can assume that if a precipitate forms, it will do so quantitatively. We can then use the equilibrium expression involving the solubility of the salt (with a common ion) to solve for the equilibrium concentration of each ion.
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Complex Ion Equilibria Note: formation of a complex ion increases the solubility of an "insoluble" salt. The addition of a Lewis base will cause it to form a comlex ion with the cation (metal ion) in equilibrium with the salt. This will decrease the concentration of the cation, and via The main idea here is that introducing a Lewis base into a solution enhances the solubility of an otherwise insoluble salt. You Le Chatelier’s principle, the equilibrium will shift right to replace the cation. As a result, the solid will have to dissolve. should know the following terms: Complex Ion Formation complex ion: a charged species consisting of a metal ion surrounded by ligands. + + Ag + 2NH3 = Ag(NH3)2 ligand: typically an anion or neutral molecule that has an unshared electron pair (lone pair) that can be shared with an + empt metal ion orbital to form a metalligand bond. Ligands are Lewis bases and some common examples are H2O, Ag + 2CN = Ag(CN)2 NH3, l and CN . 2+ 2+ Cu + 4NH3 = Cu(NH3)4 coordination number: the number of ligands attached to a metal ion. The most common coordination numbers are six (as 2+ 2+ 2 2+ 2+ 2+ 2+ in Co(OH2)6 and Ni(NH3)6 ), four (as in CoCl4 and Cu(NH3)4 ), and two (as in Ag(NH3) , although others are Zn + 4NH3 = Zn(NH3)4 known. 2+ 2 Zn + 4OH = Zn(OH)4 formation constant: metal ions add ligands one at a time in steps characterized by equilibrium constants called 3+ 3 formation constants Fe + 6CN = Fe(CN)6
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Amphoterism Further proton removals are possible, but each successive reaction occurs less readily than the one before (as the charge on the ion becomes more negative it is increasingly more difficult to remove a positively charged proton. Many metal hydroxides and oxides that are relatively insoluble in neutral water dissolve in strongly acidic and strongly basic media. These subbstances because are soluble in strong acids and bases because they themselves are capable of beahving as either an acid Addition of acid reverses the reaction. The proton adds in a stepwise fashion to convert the OH groups to H2O, eventually 3+ 3+ 2+ 2 3+ or a base; they are amphoteric. Examples of amphoteric substances include the hydroxides and oxides of Al , Cr , Zn and Sn reforming Al(H2O)6 (aq). +. The common practice is to simplify the equations for these reactions by excluding the bound H2O molecules. Thus, we usually write 3+ 3+ This behavior results from the formation of complex anions containing several (typically four) hydroxides bound to the metal ion: Al instead of Al(H2O)6 , Al(OH)3 instead of Al(H2O)3(OH)3, Al(OH)4 instead of Al(H2O)2(OH)4 , and so forth.
Al(OH)3(s) + OH (aq) = Al(OH)4 (aq) The extent to which an insoluble metal hydroxide reacts with either acid or base varies with the particular metal ion involved. Many metal hydroxides for example, Ca(OH)2, Fe(OH)2, and Fe(OH)3 are capable of dissolving in acidic solution but do not react Amphoterism is often interpreted in terms of the behaqvior of the water molecules that surround the metal ion and that are bonded to with excess base. These hydroxides are not amphoteric. 3+ 3+ it by Lewis acidbase interactions. For example, Al (aq) is more accurately represented as Al(H2O)6 (aq); six water molecules 3+ 3+ are bonded to the Al in aqueous solution. This species acts as a weak acid. As a strong base is added, Al(H2O)6 loses protons in a tepwise fashion, eventually forming neutral and waterinsoluble Al(H2O)3(OH)3. This substance then dissolves upon removal of an additional proton to form the anion Al(H2O)2(OH)4 . The reactions that occur are as follows:
3+ + Al(H2O)6 (aq) + OH (aq) = Al(H2O)5(OH)2 (aq) + H2O(l)
2+ + Al(H2O)5(OH) (aq) + OH (aq) = Al(H2O)4(OH)2 (aq) + H2O(l)
+ Al(H2O)4(OH)2 (aq) + OH (aq) = Al(H2O)3(OH)3(s) + H2O(l)
Al(H2O)3(OH)3(s) + OH (aq) = Al(H2O)2(OH)4 (aq) + H2O(l)
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Qualitative Analysis Flowchart
Group I cations = insoluble chlorides
+ 2+ 2+ Ag (aq) Hg2 (aq) Pb (aq)
add HCl(aq)
AgCl(s) Hg2Cl2(s) PbCl2(s)
heat (hot water)
2+ Pb (aq) AgCl(s) Hg2Cl2(s)
2 Add CrO4 (aq) add NH3(aq)
+ PbCrO4(s) Ag(NH3)2 (aq) Hg(l) and (yellow ppt) and Cl (aq) HgNH2Cl2(s) (white ppt)
HNO3(aq)
AgCl(s)
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Group II and Group III Separation
Hg2+, Cd2+, Bi3+, Cu2+, Sn4+, Co2+, Zn2+, Mn2+, Ni2+, Fe2+, Cr3+, Al3+
(in acid)
H2S
2+ 2+ 2+ 2+ 2+ 3+ HgS, CdS, Bi2S3, CuS, SnS2 Co , Zn , Mn , Ni , Fe , Cr
(Group II insoluble sulfides in acidic solution) NaOH
CoS, ZnS, MnS, NiS, FeS, Cr(OH)3, Al(OH)3
(Group III insoluble sulfides in basic solution)
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