Chemistry 132 NTSolubility and Complex-ion EquilibriaSolubility EquilibriaThe Solubility Product ConstantSlide 5Slide 6Slide 7Slide 8Calculating Ksp from the SolubilitySlide 10Slide 11Slide 12Slide 13Slide 14Slide 15Calculating the Solubility from KspSlide 17Slide 18Slide 19Solubility and the Common-Ion EffectSlide 21A Problem To ConsiderSlide 23Slide 24Slide 25Slide 26Precipitation CalculationsCriteria for PrecipitationSlide 29Slide 30Slide 31Predicting Whether Precipitation Will OccurSlide 33Fractional PrecipitationSlide 35Effect of pH on SolubilitySlide 37Slide 38Separation of Metal Ions by Sulfide PrecipitationOperational SkillsHomeworkPowerPoint Presentation11111Chemistry 132 NTThe most difficult thing to understand is the income tax.Albert Einstein22222Solubility and Complex-ion EquilibriaChapter 17Module 1Sections 17.1, 17.2, 17.3 and 17.4Reaction of zinc metal with hydrochloric acid.33333Solubility EquilibriaMany natural processes depend on the precipitation or dissolving of a slightly soluble salt.In the next section, we look at the equilibria of slightly soluble, or nearly insoluble, ionic compounds.Their equilibrium constants can be used to answer questions regarding solubility and precipitation.44444The Solubility Product ConstantWhen an excess of a slightly soluble ionic compound is mixed with water, an equilibrium is established between the solid and the ions in the saturated solution.For the salt calcium oxalate, CaC2O4, you have the following equilibrium.(aq)OC(aq)Ca )s(OCaC242242H2O55555The Solubility Product ConstantWhen an excess of a slightly soluble ionic compound is mixed with water, an equilibrium is established between the solid and the ions in the saturated solution.The equilibrium constant for this process is called the solubility product constant.]OC][[CaK2422sp66666The Solubility Product ConstantIn general, the solubility product constant is the equilibrium constant for the solubility equilibrium of a slightly soluble (or nearly insoluble) ionic compound.It equals the product of the equilibrium concentrations of the ions in the compound.Each concentration is raised to a power equal to the number of such ions in the formula of the compound.77777The Solubility Product ConstantIn general, the solubility product constant is the equilibrium constant for the solubility equilibrium of a slightly soluble (or nearly insoluble) ionic compound.For example, lead iodide, PbI2, is another slightly soluble salt. Its equilibrium is:(aq)2I(aq)Pb )s(PbI22H2O88888The Solubility Product ConstantIn general, the solubility product constant is the equilibrium constant for the solubility equilibrium of a slightly soluble (or nearly insoluble) ionic compound.The expression for the solubility product constant is:22sp]I][[PbK(see Exercise 17.1 and Problem 17.19)99999Calculating Ksp from the SolubilityA 1.0-L sample of a saturated calcium oxalate solution, CaC2O4, contains 0.0061-g of the salt at 25 oC. Calculate the Ksp for this salt at 25 oC.We must first convert the solubility of calcium oxalate from 0.0061 g/liter to moles per liter.42424242OCaC g128OCaC mol 1)L/OCaC g 0061.0(OCaC M L/OCaC olm 108.44251010101010Calculating Ksp from the SolubilityA 1.0-L sample of a saturated calcium oxalate solution, CaC2O4, contains 0.0061-g of the salt at 25 oC. Calculate the Ksp for this salt at 25 oC.When 4.8 x 10-5 mol of solid dissolve it forms 4.8 x 10-5 mol of each ion.(aq)OC(aq)Ca )s(OCaC242242H2O 4.8 x 10-5+4.8 x 10-50 0Starting4.8 x 10-5Equilibrium+4.8 x 10-5Change1111111111Calculating Ksp from the SolubilityA 1.0-L sample of a saturated calcium oxalate solution, CaC2O4, contains 0.0061-g of the salt at 25 oC. Calculate the Ksp for this salt at 25 oC.You can now substitute into the equilibrium-constant expression.]OC][[CaK2422sp)108.4)(108.4(K55sp9sp103.2K1212121212Calculating Ksp from the SolubilityBy experiment, it is found that 1.2 x 10-3 mol of lead(II) iodide, PbI2, dissolves in 1.0 L of water at 25 oC. What is the Ksp at this temperature?Note that in this example, you find that 1.2 x 10-3 mol of the solid dissolves to give 1.2 x 10-3 mol Pb2+ and (2 x (1.2 x 10-3)) mol of I-.1313131313Calculating Ksp from the SolubilityBy experiment, it is found that 1.2 x 10-3 mol of lead(II) iodide, PbI2, dissolves in 1.0 L of water at 25 oC. What is the Ksp at this temperature?Starting 0 0Change +1.2 x 10-3+2 x (1.2 x 10-3)Equilibrium 1.2 x 10-32 x (1.2 x 10-3)The following table summarizes.(aq)2I(aq)Pb )s(PbI22H2O1414141414Calculating Ksp from the SolubilityBy experiment, it is found that 1.2 x 10-3 mol of lead(II) iodide, PbI2, dissolves in 1.0 L of water at 25 oC. What is the Ksp at this temperature?Substituting into the equilibrium-constant expression:22sp]I][[PbK233sp))102.1(2)(102.1(K9sp109.6K1515151515Calculating Ksp from the SolubilityBy experiment, it is found that 1.2 x 10-3 mol of lead(II) iodide, PbI2, dissolves in 1.0 L of water at 25 oC. What is the Ksp at this temperature?Table 17.1 lists the solubility product constants for various ionic compounds.If the solubility product constant is known, the solubility of the compound can be calculated.1616161616Let x be the molar solubility of CaF2. x+x0 0Starting2xEquilibrium+2xChange(aq)2F(aq)Ca )s(CaF22H2OCalculating the Solubility from KspThe mineral fluorite is calcium fluoride, CaF2. Calculate the solubility (in grams per liter) of calcium fluoride in water from the Ksp (3.4 x 10-11)1717171717Calculating the Solubility from KspThe mineral fluorite is calcium fluoride, CaF2. Calculate the solubility (in grams per liter) of calcium fluoride in water from the Ksp (3.4 x 10-11)You substitute into the equilibrium-constant equationsp22K]F][[Ca 112104.3(x)(2x)113104.34x1818181818Calculating the Solubility from KspThe mineral fluorite is calcium fluoride, CaF2. Calculate the solubility (in grams per liter) of calcium fluoride in water from the Ksp (3.4 x 10-11)You now solve for x./LCaF mol 100.24103.4x24311-1919191919Calculating the Solubility from KspThe mineral fluorite is calcium fluoride, CaF2. Calculate the solubility (in grams per liter) of calcium fluoride in water from the Ksp (3.4 x 10-11)Convert to g/L (CaF2 78.1 g/mol).224CaF mol 1CaF g1.78L/mol100.2solubility L/CaF g106.1222020202020Solubility