Chapter 17 Common Ion Effect Drill l Use

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Chapter 17 Common Ion Effect

Chapter 17 Common Ion Effect

Drill l Use AP Review Drill # 50 -53

Drill l Use AP Review Drill # 50 -53

Objectives SWBAT l Complete Common Ion Effect calculations. l Explain how the Common Ion

Objectives SWBAT l Complete Common Ion Effect calculations. l Explain how the Common Ion Effect is a special case of Le Chatelier’s Principle. l

Common Ion Effect l The Common Ion Effect is the shift in equilibrium caused

Common Ion Effect l The Common Ion Effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.

Uses of Common Ion Effect l The common ion effect plays an important role

Uses of Common Ion Effect l The common ion effect plays an important role in determining: l the p. H of a solution l and the solubility of a slightly soluble salt. l

The Common Ion Effect When a salt, with the anion of a weak acid,

The Common Ion Effect When a salt, with the anion of a weak acid, is added to that acid the addition of the salt reverses the dissociation of the acid. l Acid Dissociation: l CH 3 COOH CH 3 COO -1 + H+1 l Salt that is added: l Na. CH 3 COO -1 + Na+1 l

Common Ion Effect l Addition of the salt will increase the concentration of the

Common Ion Effect l Addition of the salt will increase the concentration of the CH 3 COO-1 ion. l Le Chatelier’s Principle says that the increased ion concentration will reverse the equilibrium reaction. This reversal decreases production of the products and increases production of the reactants.

Equilibrium Special Case l You have probably realized that the Common Ion Effect is

Equilibrium Special Case l You have probably realized that the Common Ion Effect is just a special case of Le Chatelier’s Principle.

Common Ion Effect Lowers the percent dissociation of the acid. (If the equilibrium shifts

Common Ion Effect Lowers the percent dissociation of the acid. (If the equilibrium shifts left, you produce less products. ) l The same principle applies to salts with the cation of a weak base. l This is an “ICE BOX” problem with an exception. Now you have an initial concentration of the anion. l

Common Ion Effect Example l In this situation, the soluble salt of the weak

Common Ion Effect Example l In this situation, the soluble salt of the weak acid's conjugate base simply provides a source of the conjugate base.

Consider acetic acid in water: l HC 2 H 3 O 2 (aq) ↔

Consider acetic acid in water: l HC 2 H 3 O 2 (aq) ↔ H+ (aq) + C 2 H 3 O 2 - (aq) l Now, add Na. C 2 H 3 O 2, which dissociates completely, we increase the concentration of C 2 H 3 O 2 -1 (aq). l l l Le Chatelier's Principle : the equilibrium will shift to the left (towards the reactants), causing the [H+] to decrease, and therefore the p. H increases!

Common Ion Effect l The common ion effect is the shift in equilibrium that

Common Ion Effect l The common ion effect is the shift in equilibrium that occurs when an ion ALREADY PRESENT in the equilibrium reaction is added.

EXAMPLE l If 0. 100 moles of Na. C 2 H 3 O 2

EXAMPLE l If 0. 100 moles of Na. C 2 H 3 O 2 are added to a 1. 00 L of 0. 100 M solution of acetic acid, HC 2 H 3 O 2, what is the resultant p. H ?

STEP 1: Identify the major species in solution l HC 2 H 3 O

STEP 1: Identify the major species in solution l HC 2 H 3 O 2 (weak acid) l Na+1 (neither acidic nor basic which means it is a SPECTATOR ION) l C 2 H 3 O 2 -1 (conjugate base of weak acid) l H 2 O (very weak acid or base, amphoteric)

STEP 2: Identify the important equilibrium reaction(s). In this case it is a reaction

STEP 2: Identify the important equilibrium reaction(s). In this case it is a reaction that involves both the weak acid and its conjugate base: HC 2 H 3 O 2 (aq) ↔ H+ (aq) + C 2 H 3 O 2 -1 (aq)

STEP 3: Make an I. C. E. table to determine the equilibrium concentrations of

STEP 3: Make an I. C. E. table to determine the equilibrium concentrations of substances: HC 2 H 3 O 2 (aq) ↔ H+ (aq) + C 2 H 3 O 2 -1 (aq) http: //www. chem. ubc. ca/courseware/p. H/section 12/index. html

STEP 4: Determine and solve the equilibrium constant expression: l Ka = ([H+][C 2

STEP 4: Determine and solve the equilibrium constant expression: l Ka = ([H+][C 2 H 3 O 2 -1] / [HC 2 H 3 O 2] ) l 1. 8 x 10 -5 = ( x(0. 100 + x) / (0. 100 - x) ) l Use the 5% assumption. l 1. 8 x 10 -5 = ( x(0. 100) / (0. 100) ) x = [H+] = 1. 8 x 10 -5 l

STEP 5: Calculate the p. H! l p. H = -log(1. 8 x 10

STEP 5: Calculate the p. H! l p. H = -log(1. 8 x 10 -5) = 4. 74 l The addition of a common ion can also affect the dissociation of a weak base in a similar manner. l See http: //www. chem. ubc. ca/courseware/p. H/secti on 12/index. html for more information.

Practice Problems l Try B&L #

Practice Problems l Try B&L #

Wrap Up l How is the Common Ion Effect a special case of Le

Wrap Up l How is the Common Ion Effect a special case of Le Chatelier’s Principle?