Chapter 15: Equilibria PDF

Chapter 15: Equilibria  Solubility Equilibria All ionic compounds dissolve in water to some degree. However, many compounds have such low solubility in water that they are classified as insoluble. The concepts of equilibrium can be applied to salts dissolving, and...

Chapter 15: Equilibria  Solubility Equilibria All ionic compounds dissolve in water to some degree. However, many compounds have such low solubility in water that they are classified as insoluble. The concepts of equilibrium can be applied to salts dissolving, and using the equilibrium constant Ksp, their relative solubilities in water can be determined. 2  Solubility Product Constant (Ksp) The equilibrium constant for the solubility equilibrium of a slightly soluble (or nearly insoluble) ionic compound. The equilibrium constant for the dissociation of a solid salt into its aqueous ions. A dissociation reaction is written for the slightly soluble ionic compound: M𝑝 X𝑞 𝑠 ⇌ 𝑝M 𝑚+ 𝑎𝑞 + 𝑞X 𝑛− 𝑎𝑞 Solubility product: 𝐾𝑠𝑝 = [𝑀𝑚+ ]𝑝 [𝑋 𝑛− ]𝑞 3 Table of Solubility Products: Ksp 4  Example Write the solubility product expressions for the following salts: AgCl Pb3(AsO4)2 5  Molar Solubility (s) Molar solubility: The moles of solute that will dissolve in a liter of solution. The molarity of the dissolved solute in a saturated solution. Molar solubility is related to Ksp. Ksp values of compounds cannot always be compared. For Ksp values to be compared, the compounds must have the same dissociation stoichiometry. 6  Example A liter of a solution saturated at 298 K with calcium oxalate, CaC2O4, is evaporated to dryness, giving a 0.0061-g residue of CaC2O4. Calculate the molar solubility of CaC2O4. 7  Example By experiment, it is found that 1.2 x 10-3 mol of lead(II) iodide, PbI2, dissolves in 1 L of aqueous solution at 298 K. What is the solubility product constant at this temperature? 8  Solubility and the Common-Ion Effect Addition of a soluble salt that contains one of the ions of the slightly soluble salt decreases the solubility of the slightly soluble salt. For example: The addition of Pb(NO3)2 to the solubility equilibrium of solid PbCrO4 decreases the solubility of PbCrO4. PbCrO4(s) ⇌ Pb2+(aq) + CrO42−(aq) Addition of Pb2+ shifts the equilibrium to the left. 9  Example What is the molar solubility of calcium oxalate in 0.15 M calcium chloride? The solubility product for calcium oxalate is 2.3 x 10-9. 10  Precipitation Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound. By comparing the ion product, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur. If Q = Ksp, the solution is saturated and no precipitation will occur. If Q < Ksp, the solution is unsaturated and no precipitation will occur. If Q > Ksp, the solution would be above saturation. The salt above saturation will precipitate. Note: Q for a solubility reaction is called the ion product instead of reaction quotient since it is the product of ion concentrations in a solution, each concentration raised to a power equal to the number of ions in the formula of the ionic compound. 11  Fractional Precipitation The technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so forth. A successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different. A solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions but not the others. (Selective precipitation) 12  Effect of pH on Solubility For insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide. And the lower the pH, the higher the solubility. Higher pH = increased [OH−] M(OH)n(s) ⇌ Mn+(aq) + nOH−(aq) For insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility. M2(CO3)n(s) ⇌ 2 Mn+(aq) + nCO32−(aq) H3O+(aq) + CO32−(aq) ⇌ HCO3−(aq) + H2O(l) 13