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Solubility Product

Solubility Product. SCH4U0. Solubility Equilibrium. We have been investigating the equilibria involved with chemical reactions We can also investigate physical processes Ie : Dissolving and dissociation When ionic compounds dissolve they dissociate to produce solvated ions

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Solubility Product

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  1. Solubility Product SCH4U0

  2. Solubility Equilibrium • We have been investigating the equilibria involved with chemical reactions • We can also investigate physical processes • Ie: Dissolving and dissociation • When ionic compounds dissolve they dissociate to produce solvated ions • This produces an equilibrium between the solid compound and the dissolved ions

  3. Solubility Product • We can generate an equilibrium law equation for these equilibria • We must remember to omit solid concentrations (since they are constant and are incorporated into K) • The equilibrium constant for this type of process is called the “solubility product” • It gives us information about the solubility of the solid compound

  4. Solubility Product • If we compare the solubility product constant of two different solids (that have the same exponents in the equilibrium law), we can draw conclusions about their solubility • Solubility is the concentration of the dissolved compound in a saturated solution • These values are 1 trillion times different • Meaning that the ion concentrations in a saturated solution of silver bromate are much larger than that of silver cyanide • Silver bromate is more soluble than silver cyanide

  5. Solubility Product • We can calculate the solubility of these compounds to compare their concentrations directly. • Lets start with silver bromate • Solubility is the concentration of dissolved compound, or • We can solve these concentrations just like any other equilibrium • With an ICE chart Since 1 silver is produced when 1 compound dissolves =

  6. Solving Solubility • Let’s assume that we have put a large sample of silver bromate in water • Enough that we will produce a saturated solution

  7. Sample Problem • Substitute the values from the ICE chart into the equilibrium law:

  8. Solving Solubility • Let’s try this with silver cyanide instead

  9. Sample Problem • Substitute the values from the ICE chart into the equilibrium law:

  10. Compare Solubility • Lets compare the solubilities of the two compounds • The lower the , the lower the solubility • Silver bromate is approximately 1 million () times more soluble than silver cyanide

  11. Solving Solubility • Let’s try something a little trickier. • What is the solubility of calcium phosphate at RT? In this case, x represents

  12. Sample Problem • Substitute the values from the ICE chart into the equilibrium law:

  13. Predicting Precipitation • We can use the solubility product and the reaction quotient (Q) to predict whether or not a solution will create a precipitate • Example: • If Q < Ksp, the ion concentrations are too low to be at equilibrium, the solution is unsaturated, and no precipitate will form • If Q = Ksp, the solution is at equilibrium and is saturated. No precipitate will form unless more calcium phosphate is added • If Q > Ksp, the ion concentrations are higher than they should be at equilibrium. The rxn will shift left creating a solid precipitate

  14. Sample Problem • If we mix together 100 mL of 0.100 M calcium chloride with 100 mL of 0.0400 M sodium sulfate at 25ºC, will a precipitate form? • What are the potential precipitates? • What are their solubility products? • Since calcium sulfate is the least soluble substance, it is the most likely to precipitate Solutions before being mixed, will be solutions after mixing too

  15. Sample Problem • If we mix together 100 mL of 0.100 M calcium chloride with 100 mL of 0.0400 M sodium sulfate at 25ºC, will a precipitate form? • We need to know the concentration of the calcium and sulfate ions to determine Q • The two solutions were diluted when they were mixed:

  16. Sample Problem • Now that we know the concentrations, we can determine Q and compare it to K. Q > K, therefore the reaction will shift left to reach equilibrium, forming a precipitate.

  17. Practice Problem • If 200 mL of M magnesium chloride were mixed with 300 mL of M copper (II) nitrate at 25ºC, will a precipitate form? • What are the potential precipitates? • What are their solubility products? • Since copper (II) chloride is the least soluble substance, it is the most likely to precipitate Solutions before being mixed, will be solutions after mixing too

  18. Sample Problem • If 200 mL of M magnesium chloride were mixed with 300 mL of M copper (II) nitrate at 25ºC, will a precipitate form? • We need to know the concentration of the copper and chloride ions to determine Q • The two solutions were diluted when they were mixed:

  19. Sample Problem • Now that we know the concentrations, we can determine Q and compare it to K. Q < K, therefore the reaction will NOT shift left to reach equilibrium, and no precipitate will be formed. The solution is unsaturated.

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