Solubility%20equilibria_Powerpoint[1]

Solubility%20equilibria_Powerpoint[1] - Solubility...

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Solubility Equilibria
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Solubility of ionic compounds in water Solubility of ionic compounds Exceptions Group 1(1A) compounds are soluble Example: Li + , Na + ,K + All ammonium compounds (NH 4 + ) are soluble All nitrates (NO 3 - ) and acetate (CH 3 COO - ) are soluble Chlorides (Cl - ), bromides (Br - ) iodides (I - ) are soluble Pb 2+ , Hg 2 2+ , Cu + , Ag + ,
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Solubility of ionic compounds Exceptions All sulfates are soluble All hydroxides are insoluble Ca 2+ , Ba 2+ Sr 2+ , Pb 2+ Li + , Na + ,K + , NH 4 + Group 2 compounds xample: Ca 2+ Ba 2+ Example: Ca , Ba All phosphates are insoluble Li + , Na + ,K + , NH 4 + All carbonates are insoluble Li + , Na + ,K + , NH 4 + Li + , Na + ,K + , NH 4 + All sulfides are insoluble
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For slightly soluble ionic compounds , equilibrium exists between the solid solute and the dissolved ions. This type of equilibrium is heterogeneous because it involves two phases – solid and liquid. Saturated solution of BaSO 4 Undissolved BaSO4 BaSO 4 (s) a Ba 2+ (aq) + SO 4 2- (aq)
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Solubility Product Constant If the solution is saturated, an equilibrium is reached: BaSO 4 (s) a Ba 2+ (aq) + SO 4 2- (aq) [Ba 2+ [SO 2- K sp , is the solubility-product constant. K sp = [Ba ] [SO 4 ] ** The concept of solubility product does not apply to highly soluble ionic compounds due to inter-ionic interactions For example: NaCl, KNO 3
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Calculate the K sp value for bismuth sulfide (Bi 2 S 3 ) which has a solubility of 1.0 x 10 -15 mol / L at 25 o C. Bi 2 S 3 (s) a 2Bi 3+ (aq) + 3S 2- (aq) K sp = [Bi 3+ ] 2 [S 2- ] 3 Bi 2 S 3 (s) [Bi 3+ ] [S 2- ] I C E Excess 0 0 - 1.0 x 10 -15 Excess - 1.0 x 10 -15 K sp = (2.0 x 10 -15 ) 2 x (3.0 x 10 -15 ) 3 = 1.1 x 10 -73 + 2 x 1.0 x 10 -15 + 3 x 1.0 x 10 -15 2.0 x 10 -15 3.0 x 10 -15
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The K sp for Cu(IO 3 ) 2 is 1.4 x 10 -7 at 25 o C. Find the solubility of Cu(IO 3 ) 2 at 25 o C Cu(IO 3 ) 2 (s) a Cu 2+ (aq) + 2IO 3 2- (aq) K sp = [Cu 2+ ] [IO 3 2- ] 2 = 1.4 x 10 -7 Cu(IO 3 ) 2 [Cu 2+ ] [IO 3 2- ] I C E Excess 0 0 -x +x +2x Excess - x x 2x x(2x) 2 = 1.4 x 10 -7 4x 3 = 1.4 x 10 -7 x = (1.4 x 10 -7 / 4) 1/3 = 3.3 x 10 -3 Therefore, concentration (solubility) is 3.3 x 10 -3 M
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Factors that affect the solubility The common ion effect What happens when water contains an ion in common with
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Solubility%20equilibria_Powerpoint[1] - Solubility...

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