Solubility Equilibria Ksp

Solution A homogeneous mixture (solute + solvent) Solubility - a measure of the maximum of solute that will dissolve in a given amount of solvent at a specified temperature (g/100ml or mol/L=molar solubility) Saturated Solution – contains maximum amount of dissolved solute possible at given temperature Supersaturated Solution – under unusual conditions a solvent will dissolve more than its maximum amount

Solubility Equilibrium When salt dissolves in water to form a saturated solution, an equilibrium is created between the solid and its ions Salts Insoluble Soluble Slightly Soluble In Water

Dissolving a salt... http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/molvie1.swf A salt is an ionic compound - usually a metal cation bonded to a non-metal anion. The cations and anions are attracted to each other in the salt. They are also attracted to the water molecules. The water molecules will start to pull out some of the ions from the salt crystal.

At first, the only process occurring is the dissolving of the salt - the dissociation of the salt into its ions. However, soon the ions floating in the water begin to collide with the salt crystal and are “pulled back in” to the salt. (precipitation) Eventually the rate of dissociation is equal to the rate of precipitation. The solution is now “saturated”. It has reached equilibrium.

Solubility Equilibrium: Dissociation = Precipitation In a saturated solution, there is no change in amount of solid precipitate at the bottom of the beaker. Concentration of the solution is constant. The rate at which the salt is dissolving into solution equals the rate of precipitation. Na+ and Cl - ions surrounded by water molecules NaCl Crystal Dissolving NaCl in water

Examples of Solubility of Salts and Solubility Equilibrium Kidney/bladder stones (Story Time) http://www2.canada.com/montrealgazette/news/insight/story.html?id=7c04b6f5-56c4-479a-ade3-1491175551eb Barium sulfate Stalagmites and Stalactites

Barium Sulfate Colon cancer

Stalagmites and Stalactites

Carbonate Ion Equilibria Reactions CO2(g) ↔ CO2(aq) CO2(aq) + 2H2O(l) ↔ H3O+(l) + HCO3-(aq) (2) CaCO3(s)+CO2(aq)+H2O(l)↔Ca2+(aq) + 2HCO3(aq)

Ksp - Solubility Product Constant Measures how far to the right, the dissolution proceeds at equilibrium (saturated) Ex. NaCl(s)  Na+ (aq) + Cl- (aq) Ksp = [Na+ (aq) ][Cl- (aq) ] * Solids are not included Since K is always calculated by just multiplying concentrations, it is called a “solubility product” constant - Ksp.

Dissolving silver sulfate, Ag2SO4, in water When silver sulfate dissolves it dissociates into ions. When the solution is saturated, the following equilibrium exists: Ag2SO4 (s)  2 Ag+ (aq) + SO42- (aq) Since this is an equilibrium, we can write an equilibrium expression for the reaction: Ksp = [Ag+]2[SO42-] Notice that the Ag2SO4 is left out of the expression! Why? Since K is always calculated by just multiplying concentrations, it is called a “solubility product” constant - Ksp.

Writing solubility product expressions... For each salt below, write a balanced equation showing its dissociation in water. Then write the Ksp expression for the salt. Iron (III) hydroxide, Fe(OH)3 Nickel sulfide, NiS Silver chromate, Ag2CrO4 Zinc carbonate, ZnCO3 Calcium fluoride, CaF2

Some Ksp Values Note: These are experimentally determined, and may be slightly different on a different Ksp table.

Determining Solubility (Equilibrium Concentrations) of Ions Example: Calcium fluoride has a Ksp of 3.9x10-11 at 25°C. What is the solubility of the fluoride ion at equilibrium? 1) Write equation: CaF2(s) ↔ Ca2+(aq) + 2F-(aq) 2) Write Ksp expression: Ksp = [Ca2+] [F-] 2 = 3.9x10-11 3) Create a RICE table to represent the equilibrium. 4) Ksp = (x)(2x)2 = 3.9x10-11 4x3 = 3.9x10-11 x3 = 9.8x10-12 x = 2.1x10-4 M for [Ca2+] and [F-] is 2x = 4.2x10-4 M R CaF2(s) ↔ Ca2+ (aq) + 2F- (aq) I C +x +2x E x 2x

Determining the Ksp Constant Example 2: The solubility of silver carbonate (Ag2CO3) is 3.6x10 -3 g/100 mL of solvent at 25˚C. Calculate the value of the solubility product constant of silver carbonate. 1) Convert g/100mL into units of molar solubility: n = m = 3.6x10 -3 g x 1 mol = 1.3 x 10 -5 mol MM 275.8 g Molar Solubility = n = 1.3 x 10 -5 mol x 1000 mL = 1.3 x 10 -4 mol/L v 100 mL 1 L 2) Find the concentration of each ion. Ag2CO3 (s) ↔ 2 Ag+ (aq) + CO3 2- (aq) 3) Create a RICE table. 4) Ksp = [Ag +]2 [CO3 2-] = (2.6 x 10 -4 mol/L) 2 ∙ 1.3 x 10 -4 mol/L = 8.8 x 10 -12 R Ag2CO3 (s) ↔ 2 Ag+ (aq) + CO3 2- (aq) I C +2.6 x 10 -4 mol/L +1.3 x 10 -4 mol/L E 2.6 x 10 -4 mol/L 1.3 x 10 -4 mol/L

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