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Ksp : The Solubility Product Constant

Ksp : The Solubility Product Constant. Learning Goals: I will understand the solubility product constant, Ksp , and use it to compare to Q, the trial ion product, to determine if a precipitate will form I will understand how the common ion effect affects solubility.

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Ksp : The Solubility Product Constant

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  1. Ksp:The Solubility Product Constant Learning Goals: I will understand the solubility product constant, Ksp, and use it to compare to Q, the trial ion product, to determine if a precipitate will form I will understand how the common ion effect affects solubility

  2. Ksp: The Solubility Product Constant. An equilibrium can exist between a partially soluble substance and its solution(i.e. not all ionic compounds are fully soluble)

  3. For example: BaSO4 (s)  Ba2+ (aq) + SO42- (aq) • When writing the equilibrium constant expression for the dissolution of BaSO4, we remember that the concentration of a solid is constant. The equilibrium expression is therefore: K = [Ba2+][SO42-] K = Ksp, the solubility-product constant. Ksp= [Ba2+][SO42-]

  4. The Solubility Expression AaBb(s)  aAb+ (aq) + bBa- (aq) Ksp = [Ab+]a [Ba-]b Example: PbI2 (s)  Pb2+ + 2 I- Ksp = [Pb2+] [I-]2 • The greater the Ksp the more soluble the solid is in H2O.

  5. Solubility and KspThree important definitions: • Solubility: quantity of a substance that dissolves to form a saturated solution • Molar solubility: the number of moles of the solute that dissolves to form a liter of saturated solution • Ksp (solubility product): the equilibrium constant for the equilibrium between an ionic solid and its saturated solution

  6. Comparing Q and Ksp to Determine Precipitate Formation -The Reaction Quotient, Q, is used to determine if a system is at equilibrium (but is called trial ion product when comparing to Ksp) -If Q=Ksp, then the solution is perfectly saturated and a precipitate will not form -If Q<Ksp, more solid can still dissolve, therefore, no precipitate forms -If Q>Ksp, a precipitate forms

  7. Example #1: Calculating Molar Solubility when given Ksp Calculate the molar solubility of Ag2SO4 in one liter of water. Ksp =1.4 x 10-5 • Write the balanced equation • Write the Ksp equation • Use an ICE table to find [ions]

  8. Calculate the molar solubility of Ag2SO4 in one liter of water. Ksp =1.4 x 10-5 • Ag2SO4  2Ag1+ + SO42- • Ksp = [Ag]2[SO4] • ICE table • 1.4 x 10-5 = (2x)2(x) 1.4 x 10-5 = 4x3 …… = x

  9. Example 2 • What is the Ksp of a saturated solution of zinc hydroxide if the molar solubility is 2.5 x 10-2mol/L? Zn(OH)2 Zn + 2OH Ksp = [Zn][OH]2 Ratio = 1:1:2 Ksp = [2.5 x 10-2] 2[5 x 10-2] Ksp = 1.25 x 10-3

  10. Factors Affecting Solubility • Some of the main factors that affect solubility of a solute are:   (1) Temperature   (2) Nature of solute or solvent   (3) Pressure

  11. Temperature • Generally solubility increases with the rise in temperature and decreases with the fall of temperature but not in all cases. • In endothermic process solubility increases with the increase in temperature and vice versa.    • In exothermic process solubility decrease with the increase in temperature.  • Gases are more soluble in cold solvent than in hot solvent.

  12. Nature of Solute and Solvent • A polar solute dissolved in polar solvent. • A polar solute has low solubility or is insoluble in a non-polar solvent. • In general like dissolves like

  13. Pressure • The effect of pressure is observed only in the case of gases. • An increase in pressure increases of solubility of a gas in a liquid. • For example carbon dioxide is filled in cold drink bottles (such as coca cola, Pepsi 7up etc.) under pressure.

  14. Homework • Pg 471 Q4, 6, 8 (don’t do)

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