Solubility Product Constant Ksp The Terms Solubility a

Solubility Product Constant (Ksp)

The Terms Solubility: - a measure of the ability of a substance to dissolve Saturated Solution: - has the maximum amount of solute dissolved in a given volume of solvent at a given temperature Unsaturated: - has less than the maximum… Supersaturated: - has more than the maximum…

The Terms Solubility: - measured in g of solute per 100 m. L of solution units: (g/100 m. L) Molar Solubility: - measured in mole of solute per litre of solution units: (mol/L)

Solubility Conversions? e. g. Mg. F 2 solubility = 1. 72 x 10 -3 g/100 m. L molar solubility = 2. 76 x 10 -4 mol/L §

Solubility Product Constant For saturated solutions at equilibrium we use the solubility product constant symbolized Ksp Ex: Zn(OH)2(s) ↔ Zn 2+(aq) + 2 OH 1 -(aq) n § Since Ksp is always calculated by just multiplying concentrations, it is called a “solubility product” constant - Ksp.

Solubility Product Constant Ex: Zn(OH)2(s) ↔ Zn 2+(aq) + 2 OH 1 -(aq) § § § Note: Notice that Zn(OH)2(s) is left out of the expression! Why? Physical states matter when you are writing the eqm law equation! See Tables for values of Ksp

Solubility Product Constant (Ksp) Noten Used for saturated solutions. n Used for ionic compounds of low solubility, because under ordinary conditions highly soluble ionic compounds do not form precipitates n Units are omitted from values. n The lower the value of Ksp the lower the solubility of the compound.

Sample Problems n n These problems are similar to K except used for solubility. We seldom require an “ICE” table for Ksp problems.

Sample Problems 1. Zinc carbonate is a hard, slightly soluble salt. Calculate the solubility (in g/100 m. L) given the molar solubility of Zn. CO 3 is 3. 87 x 10 -6 mol/L. (Ans: 4. 85 x 10 -5 g/100 m. L) 2. A chemist finds that the molar solubility of silver carbonate, Ag 2 CO 3, is 1. 3 x 10 -4 mol/L at 25 C. Calculate Ksp for silver carbonate. (Ans: 8. 8 x 10 -12 )

Will a Precipitate Form? Use trial ion product, Qsp, and compare the value to Ksp. n If Qsp < Ksp - rxn is not at eqm - soln is unsaturated, ppt is not present n If Qsp > Ksp - rxn is not at eqm - soln is supersaturated, ppt is present n If Qsp = Ksp - rxn is at eqm - soln is saturated, ppt is not present

Sample Problems 3. Consider an aqueous solution containing Ag 1+ and Cl 1 - ions, each at a concentration of 1. 0 x 10 -5 mol/L. Will precipitation occur? (ans: Qsp < Ksp , no ppt forms) 4. A chemist mixes 100. 0 m. L of 0. 25 mol/L Pb(NO 3) 2 with 200. 0 m. L of 0. 070 mol/L Na. Br. a) Write the balanced chemical equation for the reaction. b) Does a precipitate form? Hint: Recall dilution. (ans: Qsp > Ksp , ppt forms)

Ions in Aqueous Solutions So far we have studied the solubility of ionic compounds (salts) in pure water. n Equilibrium will shift if - Another compound containing a common is added or - Another compound that removes an ion already in solution is added n

Adding a Common Ion The solubility of Mg. F 2 in pure water is 2. 6 x 10 -4 mol/L. What happens to the solubility if we dissolve the Mg. F 2 in a solution of Na. F, instead of pure water? Ans: Solubility is lowered due to the addition of the fluoride ion (F 1 -) common ion

Explaining the Common Ion Effect The presence of a common in a solution will lower the solubility of a salt. n Le Chatelier’s Principle: The addition of the common ion will shift the solubility equilibrium backwards (left). This means that there is more solid salt in the solution and therefore the solubility is lower! Ex: Mg. F 2(s) ↔ Mg 2+(aq) + 2 F 1 -(aq)

Example 1. The solubility of lead (II) chromate (Pb. Cr. O 4(s)) in water is 4. 8 x 10 -7 mol/L. Determine the solubility of Pb. Cr. O 4(s) in a 0. 10 mol/L solution of Na 2 Cr. O 4. For lead chromate, Ksp = 2. 3 x 10 -13 (Ans: 2. 3 x 10 -12 mol/L) 2. Try p 492 #10