Presentation on theme: " SWBAT Learn about The Common Ion Effect. SWBAT Work out Common Ion Effect icebox problems."— Presentation transcript:
SWBAT Learn about The Common Ion Effect. SWBAT Work out Common Ion Effect icebox problems.
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.
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.
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.
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)
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 - ]
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.
Calculate the pH of a solution that is 0.10M NH 4 Cl and.25M NH 3 (aq)
pH = 9.65
Calculate the fluoride ion concentration and pH of a solution that is 0.20M in HF and 0.10M in HCl
Fluoride ion concentration: 1.4x10 -3 M pH = 1.00
Calculate the pH of a solution containing 0.085M nitrous acid(HNO 2 ; K a = 4.5 x 10 -4 ) 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.