Lecture 14 Chemistry

Lecture 14 • Review: Knowing does not fix individual concentrations. • Disturbing equilibrium (Le Chatelier) Being ch14: Acid-base equilibria in aqueous solutions • The pH of water Next: Composition of water. Weak acids and strong acids. Getting weak acid Ka values; pKa = −log( a); Using a to get [H3O+]. Titration: What happens when some OH− is added to an acid [TP] The reaction 3 A( ) ⇌ 2 B( ) is at equilibrium, with [A]e = 0.10 M and [B]e = 2.0 M and so = 4000. Then 0.05 M of A is added. At the moment of this change, the value of [A] will be … 33% 1. 0.05 M 33% 2. 0.10 M 33% 3. 0.15 M Practice question Knowing does not fix individual concentrations At 25 ℃ the equilibrium constant for the reaction 3 A + B ⇌ 2 C is 0.145. In one experiment, a student measured at equilibrium that the concentrations of A and B are each 2.45 M. In a second experiment, the student measured at equilibrium that [A]e = 2.45 M but [B]e = 1.94 M. Must there be errors in one or both of the measurements? If errors, show why. If no errors, show why. Disturbing equilibrium Essential idea: A system at equilibrium responds to a disturbance by … [TP] The reaction 3 A( ) ⇌ 2 B( ) is at equilibrium, with [A]e = 0.10 M and [B]e = 2.0 M and so = 4000. Then 0.05 M of A is added. At the moment of this change, the value of [A] will be … partially offsetting the disturbance. 33% 1. 0.05 M 33% 2. 0.10 M 33% 3. 0.15 M [TP] The reaction 3 A( ) ⇌ 2 B( ) is at equilibrium, with [A]e = 0.10 M and [B]e = 2.0 M and so = 4000. Then 0.05 M of A is added. At the moment of this change, the value of Q will be … [TP] The reaction 3 A( ) ⇌ 2 B( ) is at equilibrium, with [A]e = 0.10 M and [B]e = 2.0 M and so = 4000. Then 0.05 M of A is added. At the moment of this change, the system will … 33% 1. > 4000 33% 2. = 4000 33% 3. < 4000 33% 1. have too much reactant 33% 2. still be at equilibrium 33% 3. have too much product [TP] The reaction 3 A( ) ⇌ 2 B( ) is at equilibrium, with [A]e = 0.10 M and [B]e = 2.0 M and so = 4000. Then 0.05 M of A is added. After the system is once again at equilibrium, [A]e must be … [Quiz] The reaction 3 A( ) ⇌ 2 B( ) is at equilibrium, with [A]e = 0.10 M and [B]e = 2.0 M and so = 4000. Then 0.05 M of A is added. After the system is once again at equilibrium, [B]e must be … 33% 1. < 0.10 M 33% 2. = 0.10 M 33% 3. > 0.10 M 33% 1. < 2.0 M 33% 2. = 2.0 M 33% 3. > 2.0 M Disturbing equilibrium Essential idea: A system at equilibrium responds to a disturbance by partially offsetting the disturbance. This behavior is called Le Chatelier’s principle. Begin ch14: Acid-base equilibria in aqueous solutions The pH of water The pH of water Write down the chemical equilibrium that accounts for the pH of water. H2O( ) + H2O( ) ⇌ H3O+( ) + OH−( ) The pH of water Write down the value of H2O( ) + H2O( ) ⇌ H3O+( O+] w = [H3 w = [H3O+]e [OH−]e ) + OH−( ) [OH−] The pH of water Pure water at 50 ℃ is measured to have a pH of 6.63. [H3O+]e Write down the reaction quotient for the water autoionization equilibrium, at 50 [H3O+]e = 10−6.63 = 2.34 × 10−7 oC. Pure water at 50 ℃ is measured to have a pH of 6.63. Write down the value of [OH−]e at 50 oC. [OH−]e = [H3O+]e = 10−6.63 = 2.34 × 10−7 The pH of water Pure water at 50 ℃ is measured to have a pH of 6.63. Write down the value of the reaction quotient of the water autoionization at equilibrium at 50 oC. w,e = w [TP] Pure water at 50 ℃ is measured to have a pH of 6.63. This means that pure water at 50 ℃ is … 0% 1. acidic 0% 2. neutral 0% 3. basic = [H3O+]e [OH−]e = (10−6.63 )2 = 5.48 × 10−14 [TP] Pure water at 10 ℃ has [H3O+] = 5.39 × 10−8. This means that pure water at 10 ℃ is … 0% 1. acidic 0% 2. neutral 0% 3. basic [Quiz] The pH of pure water is different at different temperatures. This means that as temperature changes … 0% 1. the proportions of H3O+( ) and OH−( ) to one another in pure water change 0% 2. the acidity of pure water changes 0% 3. the value of the equilibrium constant changes 0% 4. All of the above 0% 5. None of the above Equilibrium constants change with temperature Equilibrium constants change with temperature The pH of pure water at 50 ℃ is 6.63. If A ⇌ B is endothermic (gains energy from it surroundings), its increases with temperature. The pH of pure water at 25 ℃ is 7.00. If A ⇌ B is exothermic (loses energy to its surroundings), its decreases with temperature. The pH of pure water at 10 ℃ is 7.27. Sketch versus temperature for H2O( ) + H2O( ) ⇌ H3O+( [TP] The pH of pure water at 50 ℃ is 6.63. The pH of pure water at 25 ℃ is 7.00. This means the autoionization of water, H2O( ) + H2O( ) ⇌ H3O+( ) + OH−( ) is ... 0% 1. Exothermic 0% 2. Endothermic 0% 3. More information needed to know ) + OH−( ). Later, we’ll learn why is related to ∆ in this way.