Chapter+19 - Chapter 19 Ionic Equilibria in Aqueous Systems...

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Click to edit Master subtitle style 6/6/11 Chapter 19 Ionic Equilibria in Aqueous Systems Dr. Shamindri M. Arachchige [email protected] Davidson 11 1 Office Hours: Location: Davidson 11 Monday 1:45-5:00 PM Wednesday 1:45-4:15 PM Other hours by appointment 11
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6/6/11 Chapter 19 Ionic Equilibria in Aqueous Systems Buffer Solutions Demonstration: What happens when Strong acid is added to: Strong base is added to: Water Water Water with Alka-Seltzer Water with Alka-Seltzer The Alka-Seltzer provides a buffer solution : Buffer: 22
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6/6/11 Chapter 19 Ionic Equilibria in Aqueous Systems Examples of buffers : HC 2 H 3 O 2 /NaC 2 H 3 O 2 NH 4 Cl/NH 3 H 2 CO 3 /NaHCO 3 (buffer in blood) K a = 1.8 x 10 –5 K a = 5.6 x 10 –10 K a = 4.5 x 10 –7 pK a = 4.75 pK a = 9.25 pK a = 6.35 The pH of a buffer is approximately equal to the pK a of the weak acid: pH pK a Buffers range = pK a ± 1. Some Buffer Systems at Different pH Values Desired pH Weak Acid Weak Base K a pK a 4 HC 3 H 5 O 3 C 3 H 5 O 3 8.4 x 10 –4 3.08 5 HC 2 H 3 O 2 C 2 H 3 O 2 1.8 x 10 –5 4.75 6 H 2 CO 3 HCO 3 4.5 x 10 –7 6.35 7 H 2 PO 4 HPO 4 2– 6.3 x 10 –8 7.20 8 HClO ClO 2.9 x 10 –8 7.54 9 NH 4 + NH 3 5.6 x 10 –10 9.25 10 HCO 3 - CO 3 2- 4.7 x 10 -11 10.33 33
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6/6/11 Chapter 19 Ionic Equilibria in Aqueous Systems How does a buffer work? Example: HC 2 H 3 O 2 / C 2 H 3 O 2 The acid component of the buffer reacts with added base: HC 2 H 3 O 2 (aq) + OH (aq) The base component of the buffer reacts with added acid: C 2 H 3 O 2 (aq) + H 3 O + (aq) 44
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6/6/11 Chapter 19 Ionic Equilibria in Aqueous Systems How a buffer works. 55
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6/6/11 Chapter 19 Ionic Equilibria in Aqueous Systems Buffer Calculations HA(aq) + H 2 O H 3 O + (aq) + A - K a = [ ] 3 H O A HA + -       Henderson-Hasselbalch Equation for buffers: [acid] [base] log pK pH a ± = 66 pH = p K a + log [bas e] [aci d]
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6/6/11 Example Problem : What mass of NaCHO 2 needs to be added to 0.10 M HCHO 2 to prepare 1.0 L of a buffer with a pH of 3.4? (K a = 1.8 x 10 –4 ) Acid = Base = pK a = Is this a suitable buffer for this application? Chapter 19 Ionic Equilibria in Aqueous Systems 77
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6/6/11 Chapter 19 Ionic Equilibria in Aqueous Systems 88 Example Problem 1. Calculate the pH of a buffer solution that is 0.90 M HC2H3O2 and 0.80 M NaC2H3O2 (Ka =1.8 × 10-5 ). What kind of a solution are you taking the pH ?
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6/6/11 99
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6/6/11 Chapter 19 Ionic Equilibria in Aqueous Systems 2. Calculate the pH of the buffer in Part 1 after 0.10 mol HCl is added to 1.0 L of the buffer. Write the reaction occurring and determine composition of the solution after reaction: Solve the buffer equation: pH = 1010
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6/6/11 Chapter 19 Ionic Equilibria in Aqueous Systems 3. Calculate the pH of the solution that results when 0.15 mol of NaOH is added to 1.0 L of the buffer in Part 1: Write the reaction that occurs and determine the composition of the solution: Solve the buffer equation: pH = 1111
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6/6/11 1212 Chapter 19 Ionic Equilibria in Aqueous Systems What happens when 0.90 mol of NaOH is added to the original buffer described in Part 1 of the problem above? Buffering capacity: a measure of the ability to resist change in pH.
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This note was uploaded on 06/06/2011 for the course CHEM 1036 taught by Professor Amateis during the Spring '08 term at Virginia Tech.

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Chapter+19 - Chapter 19 Ionic Equilibria in Aqueous Systems...

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