Precipitation Equilibrium Precipitation reactions reach a position of equilibrium – even the most insoluble electrolyte dissolves to at least a slight extent, establishing an equilibrium with it’s ions in solution. For example, a solution of lead II chloride.

Ion-Product Equilibrium Systems There are 2 types of precipitation equilibria: between a precipitate and its ions. A precipitate (ppt.) forms when a cation from one solution combines with another forming an insoluble ionic solid. For example: Sr(NO3)2 (aq) + K2CrO4 (aq) ⇌ 2KNO3 (aq) + SrCrO4 (s) (2) between a precipitate and the species used to dissolve it. AgCl (s) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq) + Cl- (aq)

The Solubility Product Constant The Ksp expression – this is the equilibrium constant expression for the dissolving of a solid. The smaller the Ksp value the less soluble the precipitate. And like the any equilibrium constant value, the Ksp value is at a fixed temperature.

Writing Ksp Expression: Write the ion-product expression for each compound: (a) magnesium carbonate (b) iron(II) hydroxide (c) calcium phosphate (d) silver chromate (e) silver sulfide *(be careful with sulfide ions, they are so reactive with water that they will produce HS- and OH-)

Calculating Ion Concentration Calcium phosphate is a water-insoluble mineral, large quantities of which are used to make commercial fertilizers. Taking it’s equilibrium constant value of 1 x 10-33, calculate: The concentration of the phosphate ion in equilibrium with the solid if the [Ca2+] = 1 x 10-9 M Ans = 1 x 10-3 M The concentration of the calcium ions in equilibrium with the solid if the [PO43-] = 1 x 10-5 M Ans = 2 x 10-8 M

Determining Precipitate Formation: Ksp values can be used to predict when a precipitate will form or not. To do this we work with the reaction quotient once more: If Q > Ksp - the reaction will shift to the left so ppt. forms If Q < Ksp - the reaction will shift to the right so no ppt. forms If Q = Ksp - then the solution is saturated with ions at the point of ppt.

Determining if a ppt. will form… Sodium chromate is added to a solution in which the original concentration of Sr2+ is 0.0060 M (a) assuming the [Sr2+] stays constant, will a precipitate of strontium chromate form when the chromate ion concentration becomes 0.0030 M? ans: no precipitate will form, Q < K (b) will a precipitate of strontium chromate form if 0.200 L of 0.0060 M of strontium nitrate solution is mixed with 0.800 L of 0.040 M potassium chromate? ans: a precipitate will just barely form, Q > K (just slightly)

Ksp & Water Solubility: One way to establish equilibrium between a slightly soluble solid is to stir the solid with water to form a saturated solution. Solubility of the solid, in moles per liter, is related to the solubility product constant. For example, determine the solubility of barium sulfate. ans: s = 1.0 x 10-5 M

Example: Determine Solubility Calculate the solubility of barium fluoride in moles/liter and grams/liter. ans: 3.6 x 10-3 M & 0.63 g/L

Ksp & The Common Ion Effect The solubility of an insoluble substance decreases when adding a solution with a common ion in comparison to adding water (similar to Le Chatelier’s Principle). Which will have the higher solubility: barium sulfate in water OR barium sulfate in a 0.1M solution of sodium sulfate Why?

Applying the Common-Ion Effect Taking the equilibrium constant value of barium sulfate into account, estimate its solubility in a 0.10 M solution of sodium sulfate (hint: you will need an equilibrium table). ans: 1.1 x 10-9 M

Selective Precipitation A way to separate 2 cations in water solution is to add an ion that precipitates only one of the cations. For example: A flask contains a solution 0.10 M Cl- and 0.010 M CrO42-. When AgNO3 is added: (a) which anion, chloride or chromate, precipitates first? ans: The chloride ion will ppt. first since it requires the lowest silver ion concentration to ppt. (b) what percentage of the first anion has been precipitated when the second anion starts to precipitate? ans: 99.98% of chloride ions precipitated.

Dissolving Precipitates Water insoluble ionic solids can be brought into solution by adding a reagent to react with either the anion or the cation. The 2 most useful reagents for this are: 1. Strong Acids – the H+ will react with the basic anions 2. NH3 or OH- to react with the metal cations

Dissolving Zinc Hydroxide Write the equation that represents the dissolving of zinc hydroxide by a strong acid (hint: two-step process). Determine the equilibrium constant for this reaction (hint: rule of multiple equilibria). The Kw for 1 mole of water is 1 x 10-14.

Strong Acids Strong acids can be used to dissolve water-insoluble salts in which the anion is a weak base: Almost all carbonates Many sulfides Example: Write balanced equations to explain why each of the following precipitates dissolve in a strong acid (assume the acid is in excess): Aluminum hydroxide Calcium carbonate Cobalt II sulfide

Complex Ion Formation Ammonia and hydroxides are commonly used to dissolve precipitates containing a cation that forms a stable complex with NH3 or OH- (reference table 16.2 – page 434). Write the reaction by which zinc hydroxide dissolves in ammoina and solve for the equilibrium constant of this reaction using: Step 1: Zn(OH)2 (s) ⇌ Zn2+ + 2OH- Ksp = 4.0 x 10-17 Step 2: Zn2+ + 4NH3 (aq) ⇌ Zn(NH3)42+ + 2OH- Kf = 3.6x108 Kf = equilibrium constant for the formation of complex ions (page 416)

Dissolving AgCl with Ammonia Consider the reaction by which silver chloride dissolves in ammonia: Taking the Ksp AgCl = 1.8 x 10-10 and the Kf Ag(NH3)2+ = 1.7 x 107, calculate the K for this reaction. ans: 3.1 x 10-3 Calculate the number of moles of AgCl that dissolves in one liter of 6.0 M ammonia. ans: 0.30 moles/liter