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Chemistry 1011

Chemistry 1011. TOPIC Acids and Bases TEXT REFERENCE Masterton and Hurley Chapter 4.2 (Review), 13, 14.1, 15.1 (page 427), 21.2 (page589). 14.1 Buffers. YOU ARE EXPECTED TO BE ABLE TO: Explain why a buffer solution is resistant to changes in pH. Calculate the pH of a pure buffer solution.

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Chemistry 1011

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  1. Chemistry 1011 TOPIC Acids and Bases TEXT REFERENCE Masterton and Hurley Chapter 4.2 (Review), 13, 14.1, 15.1 (page 427), 21.2 (page589) Chemistry 1011 Slot 5

  2. 14.1 Buffers YOU ARE EXPECTED TO BE ABLE TO: • Explain why a buffer solution is resistant to changes in pH. • Calculate the pH of a pure buffer solution. • Calculate the pH of a buffer solution to which a known quantity of a strong acid or base has been added. Chemistry 1011 Slot 5

  3. What is a Buffer Solution? • A buffer solution is able to maintain a stable pH when quantities of acid or base are added • If one drop (0.1mL) of 0.1mol/L HCl is added to 10mL pure water, the pH will change from 7 to 3 • If one drop (0.1mL) of 0.1mol/L HCl is added to 10mL of a buffer solution, the pH will not change appreciably Chemistry 1011 Slot 5

  4. What is a Buffer Solution? • A buffer solution is an aqueous mixture of a weak acid or a weak base with its salt • An aqueous mixture of ammonia and ammonium chloride is a buffer solution • An aqueous mixture of acetic acid and sodium acetate is a buffer solution • In each case, small quantities of strong acid or strong base can be added without altering the pH Chemistry 1011 Slot 5

  5. The Acetic Acid/Acetate Ion Buffer • A solution is prepared that contains 1.0 x 10-1 mol/L acetic acid and 1.0 x 10-1 mol/L sodium acetate CH3COOH(aq) CH3COO-(aq) + H+(aq) CH3COONa(aq) CH3COO-(aq) + Na+(aq) • The sodium acetate is a salt and is fully ionized • The [CH3COO-] is 1.0 x 10-1 mol/L (Ignore the acetate ion produced by the acetic acid. This concentration is << 1.0 x 10-1 mol/L ) • The [CH3COOH] is 1.0 x 10-1 mol/L • (Ignore the acetic acid that has dissociated. This concentration is << 1.0 x 10-1 mol/L ) Chemistry 1011 Slot 5

  6. The Acetic Acid/Acetate Ion Buffer For acetic acid: Ka = [CH3COO-]x[H+] = 1.8 x 10-5 [CH3COOH] Ka = (1.0 x 10-1)x[H+] = 1.8 x 10-5 (1.0 x 10-1) [H+] = 1.8 x 10-5 mol/L pH = 4.7 Chemistry 1011 Slot 5

  7. Example – pH of a Buffer Solution • Calculate the [OH-] and pH of a buffer solution prepared so that [CH3NH2] is 0.300mol/L and [CH3NH3+] is 0.400mol/L • CH3NH2 is a weak base and CH3NH3+ is its conjugate acid CH3NH2 + H2O(aq) CH3NH3 +(aq) + OH-(aq) Kb = [CH3NH3 +] x[OH-] = 4.2 x 10-4 [CH3NH2] Kb = (4.00 x 10-1)x [OH-] = 4.2 x 10-4 (3.00 x 10-1) [OH-] = (4.2 x 10-4) x (3.00 x 10-1)= 3.15x 10-4 (4.00 x 10-1) pH = 10.5 Chemistry 1011 Slot 5

  8. Adding a Strong Acid to the Buffer • The buffer is a mixture of a weak acid and its conjugate base CH3COOH(aq) + H2O(aq)CH3COO-(aq) + H3O+(aq) • The concentrations of undissociated acetic acid and acetate ions are quite large. (If they are equal, then pH = pKa) • The concentration of H+ (H3O+) is small • If a strong acid is added (adding more H+ ) the equilibrium system moves to remove the added H+ • Some of the acetate ions react with the added H+ ions to produce undissociated acetic acid. The pH remains largely unaltered Chemistry 1011 Slot 5

  9. Adding a Strong Acid to the Buffer • 10mL of 1.0 x 10-1mol/L HCl is added to 100mL of a buffer solution containing 1.0 x 10-1mol/L CH3COOH and 1.0 x 10-1mol/L CH3COO- • Initial moles CH3COOH = 1.0 x 10-2 mol • Initial moles CH3COO- = 1.0 x 10-2 mol • Added moles H+ = 1.0 x 10-3 mol CH3COOH(aq) CH3COO-(aq) + H+(aq) • The added H+ ions react with CH3COO- ions to form CH3COOH Chemistry 1011 Slot 5

  10. Adding a Strong Acid to the Buffer Resulting moles CH3COOH = 1.0 x 10-2 +1.0 x 10-3 mol = 1.1 x 10-2 mol Resulting moles CH3COO- = 1.0 x 10-2 -1.0 x 10-3 mol = 0.9 x 10-2 mol New [CH3COOH] = 1.1 x 10-2 mol = 1.0 x 10-1mol/L 0.11L New [CH3COO-] = 0.9 x 10-2 mol = 8.0 x 10-2 mol/L 0.11L Ka = [CH3COO-]x[H+] = 1.8 x 10-5 [CH3COOH] Ka = (8.0 x 10-2 )x[H+] = 1.8 x 10-5 (1.0 x 10-1) [H+] = 2.3 x 10-5 mol/L pH = 4.6 (was 4.7) Chemistry 1011 Slot 5

  11. Choosing a Buffer • Different systems require buffer solutions of different pH • The pH depends on • Ka or Kb for the weak acid or base (Greatest impact) • The ratio of amounts of weak acid or base and the salt • If the concentrations of weak acid or base and salt are equal, then pH will equal p Ka or pKb Chemistry 1011 Slot 5

  12. Effectiveness of Buffers • To be effective, there must be sufficient acid or base and salt to react with any added strong acid or base • Usually there will be similar amounts of weak acid or base and salt • This produces a buffer with a pH  pKa or pKbthat is able to absorb either strong acid or strong base • Once so much strong acid or base has been added that all of either the weak acid or its conjugate base have reacted, then the buffer will not be effective Chemistry 1011 Slot 5

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