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. –K sp = 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 g of dissolved AgCl at 25 o C. Calculate the molar solubility of, and K sp 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 gram of CaF 2 at 25 o C. Calculate the molar solubility of, and K sp for, CaF 2. 1.Calculate the molar solubility of CaF 2. 2.From the molar solubility, we can find the ion concentrations in saturated CaF 2. Then use those values to calculate the K sp. 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 25 o C. Calculate the molar solubility of barium sulfate, BaSO 4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25 o C. For barium sulfate, K sp = 1.1 x

Uses of Solubility Product Constants The solubility product constant for magnesium hydroxide, Mg(OH) 2, is 1.5 x Calculate the molar solubility of magnesium hydroxide and the pH of a saturated magnesium hydroxide solution at 25 o C. 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, BaSO 4, in M sodium sulfate, Na 2 SO 4, solution at 25 o C. Compare this to the solubility of BaSO 4 in pure water. (Example 20-3). (What is the common ion? How was a common ion problem solved in Chapter 19?) 1.Write equations to represent the equilibria. 2.Substitute the algebraic representations of the concentrations into the K sp expression and solve for x. The molar solubility of BaSO 4 in M Na 2 SO 4 solution is 1.1 x M. The molar solubility of BaSO 4 in pure water is 1.0 x M. –BaSO 4 is 900 times more soluble in pure water than in M sodium sulfate! –Adding sodium sulfate to a solution is a fantastic method to remove Ba 2+ 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 K sp 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 M potassium sulfate, K 2 SO 4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO 3 ) 2 solutions. Will a precipitate form? 1.Write out the solubility expressions. 2.Calculate the Q sp for PbSO 4. –Assume that the solution volumes are additive. –Concentrations of the important ions are: 3.Finally, calculate Q sp for PbSO 4 and compare it to the K sp.

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(NO 3 ) 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 Na 2 S, is required to reduce the Hg 2+ concentration to 1.0 x M? For HgS, K sp =3.0 x 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 mole of magnesium nitrate, Mg(NO 3 ) 2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution? –For Mg(OH) 2, K sp = 1.5 x K b for NH 3 = 1.8 x Calculate Q sp for Mg(OH) 2 and compare it to K sp. –Mg(NO 3 ) 2 is a soluble ionic compound so [Mg 2+ ] = M. –Aqueous ammonia is a weak base that we can calculate [OH - ]. 2.Once the concentrations of both the magnesium and hydroxide ions are determined, the Q sp can be calculated and compared to the K sp.

Simultaneous Equilibria Involving Slightly Soluble Compounds How many moles of solid ammonium chloride, NH 4 Cl, must be used to prevent precipitation of Mg(OH) 2 in one liter of solution that is 0.10 M in aqueous ammonia and M in magnesium nitrate, Mg(NO 3 ) 2 ? Calculate the maximum [OH - ] that can exist in a solution that is M in Mg 2+. Using the maximum [OH - ] that can exist in solution, determine the number of moles of NH 4 Cl required to buffer 0.10 M aqueous ammonia so that the [OH - ] does not exceed 3.9 x M. Check these values by calculating Q sp for Mg(OH) 2. Use the ion product for water to calculate the [H+] and the pH of the solution.

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. CH 3 COOH ↔ CH 3 COO - + 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 K a s or K b s 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. 3.Write the equilibrium equation and solve for x and determine [H + ] or [OH - ] 4.Find pH or pOH 5.Determine the pOH from pH or pH from pOH 6.Determine the [H + ] or [OH - ] not found in step 3 K sp Write a stoichiometrically balanced equation for the limited dissociation of the solid If provided with the K sp 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. K sp = [M 2+ ][Y - ] 2 Substitute in the stoichiometrically determined x values for the [ ]s, K sp = (x)(2x) 2 Solve for x If provided a mass determined to be in solution and looking for K sp convert mass to moles convert moles to molarity Write the equilibrium equation, i.e. K sp = [M 2+ ][Y - ] 2 Solve for K sp