Chapter 15 AcidBase Equilibria Buffer A Definition Solutions

Chapter 15 Acid-Base Equilibria

Buffer - A Definition • Solutions that contain components that resist changes in p. H – Imperative to living things which require specific p. H ranges (ex: blood = 7. 4 even during creation of lactic acid…)

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

Example • Which way will the equilibrium shift if NH 4 Cl is added to a 1. 0 M NH 3 solution? How will this affect the p. H? NH 4 Cl -> NH 4+ + Cl. NH 3 + H 2 O <-> NH 4+ + OH- • The creation of NH 4+ ions in the first reaction will shift the second reaction to the LEFT. This will lower the [OH-] and increase the [H+] and lowering p. H.

Example #2 • Calculate the p. H and the percent dissociation of the acid – 0. 200 M solution of HC 2 H 3 O 2 (Ka = 1. 8 X 10 -5) • Answer: p. H = 2. 7, 0. 95% – 0. 200 M HC 2 H 3 O 2 in the presence of 0. 500 M Na. C 2 H 3 O 2 • Answer: p. H = 5. 14, 3. 6 X 10 -3%

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 Na. F. – Answer: [H+] = 7. 2 X 10 -4, 0. 072% dissociation

Buffered Solutions • A buffered solution is one that resists a change in its p. H with addition of OH- ions or protons – BLOOD (HCO 3 - and H 2 CO 3) – Weak acid and its salt (ex HF and Na. F) – Weak base and its salt (ex NH 3 and NH 4 Cl) • Same calculation approach as chapter 14

Example #1 • A buffered solution contains a 0. 50 M acetic acid (HC 2 H 3 O 2, Ka = 1. 8 X 10 -5) and 0. 50 M sodium acetate (Na. C 2 H 3 O 2). Calculate the p. H of this solution. – Answer: 4. 74

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

Example #2 • Calculate the change in p. H that occurs when 0. 010 mol solid Na. OH is added to 1. 0 L of the buffered solution described in the last example. • Compare this p. H change with that which occurs when 0. 010 mol solid Na. OH is added to 1. 0 L water.

Example #3 • Calculate the p. H 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 p. H of the buffer after the addition of 10. 0 m. L of 6. 00 M Na. OH to the original buffered solution volume of 500. 0 m. L

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 p. H very much.

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

• 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- + H 2 O – 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 p. H. The reason why it doesn’t affect the p. H 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 p. H much.

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

Example • Calculate the p. H of a solution containing 0. 75 M lactic acid (Ka = 1. 4 X 10 -4) and 0. 25 M sodium lactate. Lactic acid (HC 3 H 5 O 3) 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

Example #2 • A solution is prepared by adding 31. 56 g Na. CN and 22. 30 g HCN to 600. 0 m. L of water (Ka for HCN = 6. 2 X 10 -10). – What is the p. H of this solution? • Answer: 9. 100 – What is the p. H after the addition of 50. 0 m. L of 3. 00 M HCl? • Answer: 8. 91 – What is the p. H after a further addition of 80. 0 m. L of 4. 00 M Na. OH? • Answer: 9. 30

Example #3 • The Ka of propionic acid, HC 3 H 5 O 2, is 1. 34 X 10 -5 (p. Ka = 4. 87). What is the p. H when [HC 3 H 5 O 2] = [C 3 H 5 O 2 -]? – Answer: 4. 87

Buffering Capacity • Definition: The amount of protons or hydroxide ions a buffer can absorb without a significant change in p. H. • 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 p. H change • NOTE: The p. H of a buffered solution is determined by [A-]/[HA]. The capacity is determined by how large [HA] and [A-] are

Example • Calculate the p. H 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 p. H of the solution after the addition of 10. 0 m. L of 12. 0 M Na. OH – Answer: 3. 87, 12. 95 • When the strong base was added, the p. H changed drastically. When a buffer is used, the goal is for this NOT to happen.

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

Example • We wish to buffer a solution at p. H = 10. 07. Which one of the following bases (and conjugate acid salts) would be most useful? – NH 3 (Kb = 1. 8 X 10 -5) – C 6 H 5 NH 2 (Kb = 4. 2 X 10 -10) – N 2 H 4 (Kb = 9. 6 X 10 -7)

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 p. H and depends on indicator used • p. H curve = titration curve is a plot of the p. H of the solution being analyzed as a function of the amount of titrant added

Titrations and p. H 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/m. L = M

Strong Acid - Strong Base Titration • p. H changes gradually until the titration is close to the equilvalence point where a dramatic change occurs • p. H = 7. 00 at equivalence point • Curve points right/left based on beginning solution • Polyprotic acids have multiple curves

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 1. Stoichiometry Problem: OH- reacts with weak acid. [ ] of acid remaining and conjugate base formed are determined 2. Equilibrium Problem: Position of the weak acid equilibrium is determined, and p. H can be calculated

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 m. L sample of 0. 100 M HCN is titrated with 0. 100 M Na. OH, calculate the p. H of the solution a. After 8. 00 m. L of 0. 100 M Na. OH has been added. b. At the halfway point of the titration. c. At the equivalence point of the titration.

Important Notes • p. H at the equivalence point of a titration of a weak acid with a strong base is always greater than 7. 00 (basic) • p. H 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 p. H at the equivalence point, however. This p. H affects the titration curve.

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 m. L of water and is titrated with 0. 0500 M Na. OH. After 20. 0 m. L Na. OH has been added, the p. H is 6. 00. What is the Ka value for the acid? – Answer: 1. 0 X 10 -6

Weak Base - Strong Acid Titration • **Always think about the major species in solution, then use stoichiometry, then choose the dominant equilibrium and find p. H • The p. H at the equivalence point will be less than 7. 00 (acid)

Example • Calculate the p. H at each of the following points in the titration of 50. 00 m. L of a 0. 01000 M sodium phenolate (Na. OC 6 H 5) solution with 1. 000 M HCl solution (Ka for HOC 6 H 5 = 1. 05 X 10 -10)… - Initially - Answers: -Midpoint 10. 99 - Equivalence Point 9. 99 5. 99

Determining the Equivalence Point 1. p. H meter can be used and a plot of the titration curve can be made. 2. 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).

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 p. H at the equivalence point – Use p. H range chart on page 732