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Precipitation Equilibrium

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.

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Precipitation Equilibrium

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  1. 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.

  2. 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)

  3. 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.

  4. 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-)

  5. 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

  6. 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.

  7. 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)

  8. 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

  9. 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

  10. 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?

  11. 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

  12. 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.

  13. 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

  14. 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.

  15. 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

  16. 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)

  17. 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

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