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 SWBAT Learn about The Common Ion Effect.  SWBAT Work out Common Ion Effect icebox problems.

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Presentation on theme: " SWBAT Learn about The Common Ion Effect.  SWBAT Work out Common Ion Effect icebox problems."— Presentation transcript:

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2  SWBAT Learn about The Common Ion Effect.  SWBAT Work out Common Ion Effect icebox problems.

3  The Common Ion effect views the effect of adding a strong electrolyte to a weak electrolyte solution, where both have an ion in common, from the perspective of Le Chatelier’s principle.  The common-ion effect: The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.

4  When a soluble salt, ex: A + C - is added to a solution of another salt (ex: A + B - ) containing a common ion (A + ), the dissociation of AB is suppressed.  AB  A - + B +  AC  A - + C+  By the addition of the salt, AC, the concentration of A + increases. Therefore, according to Le Chatelier's principle, the equillibrium will shift to the left, thereby decreasing the concentration of A + ions.  Of that, the degree of dissociation of AB will be reduced.

5  Potassium nitrite (KNO 2 ) is a soluble ionic compound; a strong electrolyte. It dissociates completely in an aqueous solution:  KNO 2 (aq)  K + (aq) + NO 2 - (aq)  o In contrast, HNO 2 is a weak electrolyte that ionizes like so:  HNO 2 (aq)  H + (aq) + NO 2 - (aq)   The addition of NO 2 from KNO 2 causes this equilibrium to shift to the left, thereby decreasing the equilibrium concentration of H + (aq)  HNO 2 (aq)  H + (aq) + NO 2 - (aq)  This decrease in the dissociation of the weak acid HNO 2 is the common-ion effect.

6  When a strong electrolyte supplies the common ion NO 2 -, the equilibrium shifts to form more HNO 2 (aq).   HNO 2 (aq)  H + (aq) + NO 2 - (aq) Added NO 2 - Equilibrium shifts to form less NO 2 (aq)

7  The ionization of a weak base is also decreased by the addition of a common ion. For example, the addition of NH 4 + (from the strong electrolyte NH 4 Cl) causes the base-dissociation equilibrium of NH 3 to shift to the left, decreasing the equilibrium concentration of OH - and lowering the pH:  NH 3 (aq) + H 2 O(l)  NH 4 + (aq)+OH - (aq)   addition of NH 4 + shifts equilibrium, reducing [OH - ]

8  When, in a problem, the strong electrolyte is dissolved in a solution that is already a certain molarity solution of weak electrolyte, the source of this molarity is unimportant. Let’s just disregard math and assume that the molarity of the solution came a random outside source.

9  Calculate the pH of a solution that is 0.10M NH 4 Cl and.25M NH 3 (aq)

10  pH = 9.65

11  Calculate the fluoride ion concentration and pH of a solution that is 0.20M in HF and 0.10M in HCl

12  Fluoride ion concentration: 1.4x10 -3 M  pH = 1.00

13  Calculate the pH of a solution containing 0.085M nitrous acid(HNO 2 ; K a = 4.5 x ) and 0.10M potassium nitrite(KNO 2 )  Calculate the formate ion concentration and pH of a solution that is 0.050M in formic acid(HCHO 2 ; K a =1.8x10 -4 )and 0.10M in HNO 3.

14  #1: pH = 3.42  #2: [CHO 2 - ] = 9.0x10 -5 ; pH = 1.00

15  The End!


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