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Acid-Base Equilibria and Buffer Solutions

Learn about solutions that resist changes in pH and their importance in maintaining specific pH ranges in living organisms. Understand how weak acids and salts interact to determine pH, and discover the Henderson-Hasselbalch equation for calculating pH in buffered solutions.

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Acid-Base Equilibria and Buffer Solutions

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  1. Chapter 15 Acid-Base Equilibria

  2. Buffer - A Definition • Solutions that contain components that resist changes in pH • Imperative to living things which require specific pH ranges (ex: blood = 7.4 even during creation of lactic acid…)

  3. Solutions of Acids/Bases Containing a Common Ion • Weak acid (HA) is present in solution along with a salt (NaA --> Na+ and A-) • Salt will break up (strong electrolyte) • HA will determine the pH (ex: HF and NaF) HF(aq) <-> H+(aq) + F-(aq) NaF -> Na+ + F- • The common ion (in this case F-) will shift equilibrium LEFT (Le Chatelier’s principle) • Fewer H+ ions present, higher/more basic pH

  4. Example • Which way will the equilibrium shift if NH4Cl is added to a 1.0 M NH3 solution? How will this affect the pH? NH4Cl -> NH4+ + Cl- NH3 + H2O <-> NH4+ + OH- • The creation of NH4+ ions in the first reaction will shift the second reaction to the LEFT. This will lower the [OH-] and increase the [H+] and lowering pH.

  5. Example #2 • Calculate the pH and the percent dissociation of the acid • 0.200 M solution of HC2H3O2 (Ka = 1.8 X 10-5) • Answer: pH = 2.7, 0.95% • 0.200 M HC2H3O2 in the presence of 0.500 M NaC2H3O2 • Answer: pH = 5.14, 3.6 X 10-3%

  6. Change From Chapter 14 Problems • The initial concentration of the products will not be zero in the presence of the salt **ALWAYS write the major species in solution!! • Example: The equilibrium [H+] in a 1.0 M HF solution is 2.7 X 10-2 M, and the percent dissociation of HF is 2.7%. Calculate the [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2 X 10-4) and 1.0 M NaF. • Answer: [H+] = 7.2 X 10-4, 0.072% dissociation

  7. Buffered Solutions • A buffered solution is one that resists a change in its pH with addition of OH- ions or protons • BLOOD (HCO3- and H2CO3) • Weak acid and its salt (ex HF and NaF) • Weak base and its salt (ex NH3 and NH4Cl) • Same calculation approach as chapter 14

  8. Example #1 • A buffered solution contains a 0.50 M acetic acid (HC2H3O2, Ka = 1.8 X 10-5) and 0.50 M sodium acetate (NaC2H3O2). Calculate the pH of this solution. • Answer: 4.74

  9. Flow • When dealing with a strong acid or base, rxn stoichiometry comes first, then use ICE table.

  10. Example #2 • Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the buffered solution described in the last example. • Compare this pH change with that which occurs when 0.010 mol solid NaOH is added to 1.0 L water.

  11. Example #3 • Calculate the pH of a solution that contains 0.250 M formic acid, HCOOH (Ka = 1.8 X 10-4), and 0.100 M sodium formate, HCOONa • Calculate the pH of the buffer after the addition of 10.0 mL of 6.00 M NaOH to the original buffered solution volume of 500.0 mL

  12. Adding Acid to a Solution (HA and A-) • If acid (H+) is added instead of OH-, the reaction that will occur is H+ + A- --> HA • The H+ ions are replaced by HA, which won’t change the pH very much.

  13. But WHY? • When a weak acid (HA) dissociates, HA --> H+ + A- the corresponding equilibrium expression is Ka = [H+][A-]/[HA]. In order to determine pH, the equation should be rearranged: • [H+] = Ka [HA]/[A-]

  14. A buffered solution may contain weak acid (HA) and it’s conjugate base (A-). When a base (OH-) is added, the reaction that occurs is OH- + HA --> A- + H2O • The added OH- ions react with HA to make A- ions. With less HA and more A-, the equilibrium expression from the last slide is affected. The [H+] is reduced which will change the pH. The reason why it doesn’t affect the pH much is because [HA] and [A-] are very large compared to the amount of OH- added. The change in [HA]/[A-] will be small and won’t affect the pH much.

  15. Henderson-Hasselbalch Equation • [H+] = Ka [HA]/[A-] will be very helpful in determining [H+] and pH in buffered solutions…can be manipulated to make: pH = pKa + log ([A-]/[HA]) or pH = pKa + log ([base]/[acid]) • NOTE: In a particular buffering system, all solutions that have the same ratio of [A-] to [HA] will have the same pH. • ICE table assumption for x is generally accepted when using this equation

  16. Example • Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 X 10-4) and 0.25 M sodium lactate. Lactic acid (HC3H5O3) is a common constituent of biologic systems. For example, it is found in milk and is present in human muscle tissue during exertion. • Answer: 3.38

  17. Example #2 • A solution is prepared by adding 31.56 g NaCN and 22.30 g HCN to 600.0 mL of water (Ka for HCN = 6.2 X 10-10). • What is the pH of this solution? • Answer: 9.100 • What is the pH after the addition of 50.0 mL of 3.00 M HCl? • Answer: 8.91 • What is the pH after a further addition of 80.0 mL of 4.00 M NaOH? • Answer: 9.30

  18. Example #3 • The Ka of propionic acid, HC3H5O2, is 1.34 X 10-5 (pKa = 4.87). What is the pH when [HC3H5O2] = [C3H5O2-]? • Answer: 4.87

  19. Buffering Capacity • Definition: The amount of protons or hydroxide ions a buffer can absorb without a significant change in pH. • Large buffering capacity means it contains a large amount of buffering components and it can absorb a lot of protons/hydroxide ions and show little pH change • NOTE: The pH of a buffered solution is determined by [A-]/[HA]. The capacity is determined by how large [HA] and [A-] are

  20. Example • Calculate the pH of a 0.500 L solution that contains 0.15 M HCOOH (Ka = 1.8 X 10-4) and 0.20 M HCOONa. • Then calculate the pH of the solution after the addition of 10.0 mL of 12.0 M NaOH • Answer: 3.87, 12.95 • When the strong base was added, the pH changed drastically. When a buffer is used, the goal is for this NOT to happen.

  21. Optimal Buffering • We want to avoid large changes in the [A-]/[HA] ratio • Best buffered solution will have [HA]=[A-], or [A-]/[HA] = 1

  22. Example • We wish to buffer a solution at pH = 10.07. Which one of the following bases (and conjugate acid salts) would be most useful? • NH3 (Kb = 1.8 X 10-5) • C6H5NH2 (Kb = 4.2 X 10-10) • N2H4 (Kb = 9.6 X 10-7)

  23. Titration Term Review • Titrant in buret = solution of known concentration • Equivalence point = stoichiometric point = moles of H+ equal moles of OH- • Endpoint = color changes due to pH and depends on indicator used • pH curve = titration curve is a plot of the pH of the solution being analyzed as a function of the amount of titrant added

  24. Titrations and pH Curves • Strong Acid - Strong Base (ch. 4) • Weak Acid with Strong Base (ch. 14) • Weak Base with Strong Acid • Mole is too large of a unit when working with mililiters, so generally a millimole (mmol) is used…1000 mmol = 1 mol and mmol/mL = M

  25. Strong Acid - Strong Base Titration • pH changes gradually until the titration is close to the equilvalence point where a dramatic change occurs • pH = 7.00 at equivalence point • Curve points right/left based on beginning solution • Polyprotic acids have multiple curves

  26. Weak Acid with Strong Base Titration • Essentially a set of buffer problems • **Even though it is a weak acid, it reacts essentially to completion with the strong base’s hydroxide ion • Stoichiometry Problem: OH- reacts with weak acid. [ ] of acid remaining and conjugate base formed are determined • Equilibrium Problem: Position of the weak acid equilibrium is determined, and pH can be calculated

  27. Example • Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (Ka = 6.2 X 10-10) when dissolved in water. If a 50.0 mL sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solution • After 8.00 mL of 0.100 M NaOH has been added. • At the halfway point of the titration. • At the equivalence point of the titration.

  28. Important Notes • pH at the equivalence point of a titration of a weak acid with a strong base is always greater than 7.00 (basic) • pH is determined by the amount of excess OH- present • Curve looks different before and the same after the equivalence point • The AMOUNT, not strength of acid determines the equivalence point. The STRENGTH affects the pH at the equivalence point, however. This pH affects the titration curve.

  29. Titrations vs. Ka • Titration curves can be used to determine equilibrium constant values • EX: 2.00 mmol of a monoprotic weak acid is dissolved in 100.0 mL of water and is titrated with 0.0500 M NaOH. After 20.0 mL NaOH has been added, the pH is 6.00. What is the Ka value for the acid? • Answer: 1.0 X 10-6

  30. Weak Base - Strong Acid Titration • **Always think about the major species in solution, then use stoichiometry, then choose the dominant equilibrium and find pH • The pH at the equivalence point will be less than 7.00 (acid)

  31. Example • Calculate the pH at each of the following points in the titration of 50.00 mL of a 0.01000 M sodium phenolate (NaOC6H5) solution with 1.000 M HCl solution (Ka for HOC6H5 = 1.05 X 10-10)… • Initially -Midpoint - Equivalence Point • Answers: 10.99 9.99 5.99

  32. Determining the Equivalence Point • pH meter can be used and a plot of the titration curve can be made. • Acid-base indicator can be used to see the endpoint (NOT SAME AS EQUIVALENCE POINT, however various indicators can be used so this error won’t be a big deal).

  33. Acid-Base Indicators • Indicators are represented by HIn. As the H+ ions react with OH- ions from the basic titrant (are removed from the HIn), In- ions remain. These In- ions cause the color to change based on the indicator present. • Indicators can be chosen where the endpoint and equivalence point are very close. • Determine the pH at the equivalence point • Use pH range chart on page 732

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