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Chapter 14 Acids and Bases

Chapter 14 Acids and Bases. 14.10 Buffers. Buffers. When an acid or base is added to water, the pH changes drastically to a buffer solution, the pH does not change very much; pH is maintained. Buffers. Buffers

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Chapter 14 Acids and Bases

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  1. Chapter 14 Acids and Bases 14.10 Buffers

  2. Buffers When an acid or base is added • to water, the pH changes drastically • to a buffer solution, the pH does not change very much; pH is maintained

  3. Buffers Buffers • in the body absorb H3O+ or OH from foods and cellular processes to maintain pH • are important in the proper functioning of cells and blood • in blood maintain a pH close to 7.4. A change in the pH of the blood affects the uptake of oxygen and cellular processes

  4. Components of a Buffer The components of abuffer solution • are acid-base conjugate pairs • can be a weak acid and a salt of its conjugate base • can be a weak base and a salt of its conjugate acid • typically has equal concentrations of a weak acid (or base) and its salt

  5. Learning Check Does each of the following combinations produce a buffer solution or not? Explain. A. HCl and KCl B. H2CO3 and NaHCO3 C. H3PO4 and NaCl D. CH3COOHand KCH3COO

  6. Solution A. HCl + KCl No; HCl is a strong acid. B. H2CO3 + NaHCO3 Yes; this is a weak acid and its salt. C. H3PO4 + NaCl No; NaCl does not contain a conjugate base of H3PO4. D. CH3COOHand KCH3COOYes; this is a weak acid and its salt.

  7. Buffer Action An acetic acid/acetate buffer with acetic acid (CH3COOH) and the salt of its conjugate base sodium acetate (NaCH3COO). • Acid dissociation in water produces acetate ion and hydronium ion: CH3COOH(aq) + H2O(l) CH3COO(aq) + H3O+(aq) Acetic acid Acetate ion Hydronium ion

  8. The sodium acetate salt produces sodium ions and acetate ions. NaCH3COO(aq) Na+(aq) + CH3COO(aq) The buffer solution with sodium acetate provides a higher concentration of the conjugate base CH3COO than the weak acid alone. CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO(aq) Large amount Large amount Buffer Action(continued) 8

  9. Function of the Weak Acid • The function of the acetic (weak) acid in a buffer is to neutralize added base. • The acetate ion produced by neutralization becomes part of the available acetate ion. CH3COOH(aq) + OH−(aq) CH3COO−(aq) + H2O(l) Acetic acid Base Acetate ion Water

  10. Function of the Conjugate Base • The function of the acetate ion, CH3COO−, is to neutralize added H3O+. • The acetic acid produced by the neutralization becomes part of the available acetic (weak) acid. CH3COOH(aq) + H2O(l)CH3COO−(aq) + H3O+(aq) Acetic acid Water Acetate ion Acid

  11. Summary of Buffer Action Buffer action occurs • as the weak acid in a buffer neutralizes base • as the conjugate base in the buffer neutralizes acid • to maintain the pH of the solution

  12. Learning Check Which combination makes a buffer solution? A. HCl and KCl B. H2CO3 and NaHCO3 C. H3PO4 and NaCl D. CH3COOH and LiCH3COO

  13. Solution B. H2CO3 + NaHCO3 A weak acid and its salt D. CH3COOH + LiCH3COO A weak acid and its salt

  14. pH of a Buffer The [H3O+] in the Ka expression is used to determine the pH of a buffer. weak acid(aq) + H2O(l) H3O+(aq) + conjugate base(aq) Ka = [H3O+][conjugate base] [weak acid] [H3O+] = Ka x [weak acid] l [conjugate base] pH = -log [H3O+]

  15. Calculating the pH of a Buffer

  16. Example of Calculating a Buffer pH The weak acid H2PO4- in a buffer H2PO4−/HPO42− has Ka = 6.2 x 10−8. What is the pH of the buffer if it is 0.20 M in both H2PO4−and HPO42−? STEP 1 Write the Ka or Kb expression. H2PO4(aq) + H2O(l) HPO42(aq) + H3O+(aq) Ka = [H3O+] x [HPO42] [H2PO4−] STEP 2 Rearrange the Ka or Kb for [H3O+]. [H3O+] = Ka x [H2PO4−] [HPO42−]

  17. Example of Calculating a Buffer pH(continued) STEP 3 Substitute in [HA] and [A−]. [H3O+] = 6.2 x 10−8 x [0.20] = 6.2 x 10−8 M [0.20] STEP 4 Use [H3O+] to calculate pH. pH = -log [6.2 x 10−8] = 7.21

  18. Learning Check What is the pH of a H2CO3 buffer that is 0.20 M H2CO3 and 0.10 M HCO3−? Ka of H2CO3 = 4.3 x 107

  19. Solution STEP 1 Write the Ka or Kb expression. H2CO3(aq) + H2O(l) HCO3(aq) + H3O+(aq) Ka = [H3O+][HCO3] [H2CO3] STEP 2 Rearrange the Ka or Kb for [H3O+]. [H3O+] = Ka x [H2CO3] [HCO3−]

  20. Solution (continued) STEP 3 Substitute in [HA] and [A−]. [H3O+] = 4.3 x 107 x [0.20] = 8.6 x 107 M [0.10] STEP 4 Use [H3O+] to calculate pH. pH = -log [8.6 x 107] = 6.07

  21. Concept Map

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