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## Solubility Product Constants

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**Solubility Product Constants**• Silver chloride, AgCl,is rather insoluble in water. • Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water. • The equilibrium constant expression for this dissolution is called a solubility product constant. • Ksp = solubility product constant**Solubility Product Constants**• In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as:**Solubility Product Constants**• The same rules apply for compounds that have more than two kinds of ions. • One example of a compound that has more than two kinds of ions is calcium ammonium phosphate.**Determination of Solubility Product Constants**• One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25oC. Calculate the molar solubility of, and Ksp for, AgCl. • The molar solubility can be easily calculated from the data: • The equation for the dissociation of silver chloride, the appropriate molar concentrations, and the solubility product expression are:**Determination of Solubility Product Constants**• One liter of saturated calcium fluoride solution contains 0.0167 gram of CaF2 at 25oC. Calculate the molar solubility of, and Ksp for, CaF2. • Calculate the molar solubility of CaF2. • From the molar solubility, we can find the ion concentrations in saturated CaF2. Then use those values to calculate the Ksp. • Note: You are most likely to leave out the factor of 2 for the concentration of the fluoride ion!**Uses of Solubility Product Constants**• The solubility product constant can be used to calculate the solubility of a compound at 25oC. • Calculate the molar solubility of barium sulfate, BaSO4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25oC. For barium sulfate, Ksp= 1.1 x 10-10.**Uses of Solubility Product Constants**• The solubility product constant for magnesium hydroxide, Mg(OH)2, is 1.5 x 10-11. Calculate the molar solubility of magnesium hydroxide and the pH of a saturated magnesium hydroxide solution at 25oC. • Be careful, do not forget the stoichiometric coefficient of 2! • Substitute the algebraic expressions into the solubility product expression. • Solve for the pOH and pH.**The Common Ion Effect in Solubility Calculations**• Calculate the molar solubility of barium sulfate, BaSO4, in 0.010 M sodium sulfate, Na2SO4, solution at 25oC. Compare this to the solubility of BaSO4 in pure water. (Example 20-3). (What is the common ion? How was a common ion problem solved in Chapter 19?) Write equations to represent the equilibria. Substitute the algebraic representations of the concentrations into the Ksp expression and solve for x. • The molar solubility of BaSO4 in 0.010 M Na2SO4 solution is 1.1 x 10-8M. • The molar solubility of BaSO4 in pure water is 1.0 x 10-5M. • BaSO4 is 900 times more soluble in pure water than in 0.010 M sodium sulfate! • Adding sodium sulfate to a solution is a fantastic method to remove Ba2+ ions from solution! • If your drinking water were suspected to have lead ions in it, suggest a method to prove or disprove this suspicion.**The Reaction Quotient in Precipitation Reactions**• The reaction quotient, Q, and the Ksp of a compound are used to calculate the concentration of ions in a solution and whether or not a precipitate will form. • We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form? Write out the solubility expressions. • Calculate the Qsp for PbSO4. • Assume that the solution volumes are additive. • Concentrations of the important ions are: • Finally, calculate Qsp for PbSO4 and compare it to the Ksp.**The Reaction Quotient in Precipitation Reactions**• Suppose we wish to remove mercury from an aqueous solution that contains a soluble mercury compound such as Hg(NO3)2. We can do this by precipitating mercury (II) ions as the insoluble compound HgS. What concentration of sulfide ions, from a soluble compound such as Na2S, is required to reduce the Hg2+ concentration to 1.0 x 10-8M? • For HgS, Ksp=3.0 x 10-53. • Equlibria that simultaneously involve two or more different equilibrium constant expressions are simultaneous equilibria.**Simultaneous Equilibria Involving Slightly Soluble Compounds**• If 0.10 mole of ammonia and 0.010 mole of magnesium nitrate, Mg(NO3)2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution? • For Mg(OH)2, Ksp = 1.5 x 10-11. Kb for NH3 = 1.8 x 10-5. • Calculate Qsp for Mg(OH)2 and compare it to Ksp. • Mg(NO3)2 is a soluble ionic compound so [Mg2+] = 0.010 M. • Aqueous ammonia is a weak base that we can calculate [OH-]. • Once the concentrations of both the magnesium and hydroxide ions are determined, the Qsp can be calculated and compared to the Ksp.**Simultaneous Equilibria Involving Slightly Soluble Compounds**• Use the ion product for water to calculate the [H+] and the pH of the solution. • How many moles of solid ammonium chloride, NH4Cl, must be used to prevent precipitation of Mg(OH)2 in one liter of solution that is 0.10 M in aqueous ammonia and 0.010 M in magnesium nitrate, Mg(NO3)2 ? • Calculate the maximum [OH-] that can exist in a solution that is 0.010 M in Mg2+. • Using the maximum [OH-] that can exist in solution, determine the number of moles of NH4Cl required to buffer 0.10 M aqueous ammonia so that the [OH-] does not exceed 3.9 x 10-5M. • Check these values by calculating Qsp for Mg(OH)2.**Dissolving Precipitates**• For example, look at the dissolution of Mg(OH)2 in HCl.**In General How to Solve and Predict Reversible Reactions**1. Are the compounds involved strong or weak electrolytes a. If strong then there is a 100% dissociation and no way back use a single headed arrow and when adding the simultaneous equations together only and add the products. i.e. NaOH → Na+ + OH- b. if a weak electrolyte use a double headed arrow and when adding simultaneous equations together add both reactants and products i.e. CH3COOH ↔ CH3COO- + H+ 2. Determine if reactions are neutralization, dissociation, or equilibrium reactions I. Neutralization a. Class I – strong electrolyte/strong electrolyte: produces salt and water. Use ICE table with units of amount (mmol or moles), cancel terms, determine amounts of products. Stop. b. Class II – weak electrolyte/weak electrolyte: Compare Kas or Kbs to determine which predominates products or reactants. Use the equlibrium equation and ICE table with units of concentration (mmol/mL or mol/L) to determine equilibrium concentrations. Stop. c. Class III – strong electrolyte/weak electrolyte: produces salt of the conjugate and water and further actions, Dissociation or Equilibrium reactions must be determined continue to 2 II II. Dissociate any products and begin again at repeat step 1 then continue to 2 III. Products of the dissociation though by name appear to be the same compounds they are not identical in that they have a different source and are treated separately.**III. Equilibrium – write new stoichiometrically balanced**chemical equilibrium equations. • Write the equilibrium equation and solve for x and determine [H+] or [OH-] • Find pH or pOH • Determine the pOH from pH or pH from pOH • Determine the [H+] or [OH-] not found in step 3 Ksp Write a stoichiometrically balanced equation for the limited dissociation of the solid If provided with the Ksp and looking for molar solubility of products Write the stoichiometric ratio, i.e. 1:2 Multiply the ratio through by x, i.e. x:2x Write the equilibrium equation, i.e. Ksp = [M2+][Y-]2 Substitute in the stoichiometrically determined x values for the [ ]s, Ksp = (x)(2x)2 Solve for x If provided a mass determined to be in solution and looking for Ksp convert mass to moles convert moles to molarity Write the equilibrium equation, i.e. Ksp = [M2+][Y-]2 Solve for Ksp