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ANALYTICAL CHEMISTRY

Lecture Note. ANALYTICAL CHEMISTRY. Chem. 243. Acids, Bases and Buffers. Chapter 5. Chapter 5 — ACIDS DONATE HYDROGEN IONS TO BASES. By controlling the acid-base conditions, the clinician can ensure that medications stay in solution.

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ANALYTICAL CHEMISTRY

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  1. Lecture Note ANALYTICAL CHEMISTRY Chem. 243 Acids, Bases and Buffers Chapter 5

  2. Chapter 5 —ACIDS DONATE HYDROGEN IONS TO BASES • By controlling the acid-base conditions, the clinician can ensure that medications stay in solution. • If acid content of blood changes slightly, the patient dies

  3. Dynamic Equilibrium (Exists in a system made up of at least 2 states of the matter when the populations of the 2 states are constant, even though the members of the system are constantly changing from one state to another; e.g., vapor pressure: From Chapter 7) • Most chemical reactions are reversible • Reactants combine to give products • Products can fall apart to give reactants (products can break back down into starting materials) reactants products These two competing processes occur simultaneously • This process is called a dynamic equilibrium: forward and reverse reactions continue to take place but the [reactants] and [products] remain constant

  4. Example: Hemoglobin (Hb loads up on O2 in the lungs and dumps O2 into the cells) Le Chatelier’s principle (Nature strives to attain/maintain equilibrium) • Hemoglobin (Hb) is the protein in blood that carries oxygen from the lungs to the rest of the body. Hb is a quaternary aggregate of four polypeptide (i.e., protein) chains. Each polypeptide carries an organic heme group (the cyclic structure that carries a Fe2+ ion). Each iron (II) ion can carry one O2 molecule, so each Hb molecule can carry 4 O2 molecules. • Hb combines reversibly with oxygen in the lungs • In the lungs the concentration of O2 is high. The system (i.e., your body) “perceives” the increased O2 concentration as adding reactants; therefore, the equilibrium shifts towards the product: oxyhemoglobin, Hb(O2)4 • The opposite happens in cellular tissue (low [O2] is “perceived as removing a reactant: equilibrium shifts towards reactants)

  5. The Equilibrium Constant • A system is in a state of equilibrium when there is a balance between reactants and products • This balance is defined by thermodynamic parameters (e.g., bond strengths and intermolecular forces between all the molecules in the system) • The equilibrium constant (K) is the numerical description of that balance

  6. Equilibrium Constant Expression • For a model reaction: a A + b B ⇌c C + d D The equilibrium constant is given by: • The molar concentrations of the products, raisedto their stoichiometric coefficients, are in the numerator • The molar concentrations of the reactants, raisedto their stoichiometric coefficients, are in the denominator Stoichiometry: calculation of measurable (quantitative) relationships of the reactants and products in a balanced chemical reaction

  7. A Word About Subscripts • The equilibrium constant (K) is often appended with a subscript • The subscript denotes the type of equilibrium reaction • Keq is a generic equilibrium constant • Ka is the equilibrium constant for the inonization of weak acids • Kb is the equilibrium constant for the inonization of weak bases • Ksp is the equilibrium constant for the solubility of sparingly soluble compounds

  8. Meaning of K • As K increases (K > 1), the reaction tends to increasingly favor products (the forward reaction becomes more favorable) a A + b B c C + d D • As K decreases (K < 1), the reaction tends to increasingly favor starting materials (the reverse reaction becomes more favorable) a A + b B c C + d D

  9. Example • K for the reaction below is 56: H2 + I2⇌ 2 HI • If equal amounts of hydrogen and iodine are placed into a reaction vessel and allowed to come to equilibrium, will there be more HI (hydrogen iodide) or H2? Since K >1, the reaction is “product-favored”, so at equilibrium, there will be more HI

  10. Units of K • The units on K simply cancel out the units on the other side of the equation • You have been reminded for the last few weeks about always including units • However, the units on K are meaningless • You do not need to include units with K

  11. Carbon Monoxide • Carbon monoxide (CO) kills by “tying up” the Hb molecule so that not enough O2 can be delivered to the tissues in the body • The K for Hb and CO is about 200 times larger that the K for Hg and O2 (the system favors the forward reaction towards the formation of carboxyhemoglobin: Hb(CO)4 Hb + 4 CO  Hb(CO)4 • There is less free Hb available to bind O2

  12. Solids and Liquids • The concentration of a pure solid or liquid is a constant value (or very nearly so) • Pure solids or liquids comprise a different phase from where the reaction occurs. Physiologicreactions usually occur in aqueous media. • The concentration of a liquid solvent (e.g., water) is also nearly constant (because the concentration of water is so much greater than the concentrations of any of the reacting species in aqueous solutions) so these constant values for concentration are included with the equilibrium constant value. • Therefore, the concentrations of solids, liquids, and water (as a solvent) do not appear explicitly in the equilibrium constant expression

  13. Example • Write the equilibrium constant expression for the partial ionization of a weak base: K = [NH4+] [OH-] [NH3] [H2O] this can be rearranged to: K [H2O] = [NH4+] [OH-] [NH3] this can be rearranged to: Kb = [NH4+] [OH-] [NH3] *Water is the solvent and does not appear in the equilibrium constant expression

  14. Reversing a Reaction • Since equilibrating reactions are, by definition, reversible, what happens to K when the chemical equation is reversed? Kforward is the reciprocal of Kreverse Kforward = 1 Kreverse

  15. Le Chatelier’s Principle • When a system in a state of dynamic equilibrium is disturbed, it will react to re-establish the equilibrium condition • Nature likes being in an equilibrium state • Changing reaction conditions (concentrations, temperature, etc.) moves the system away from being at equilibrium • The system will adjust (go forwards or backwards) until it again is in an equilibrium state

  16. Changing Concentration • If you add products, the equilibrium will shift towards reactants • If you remove products, the equilibrium will shift towards products • The system will adjust to counteract whatever change you make on it

  17. Example • In which direction will the equilibrium shift if morecalcium hydroxide is added? Increasing reactants will shift the equilibrium to the products (to the right)

  18. Example • In which direction will the equilibrium shift if calcium hydroxide is removed? Decreasing reactants will shift the equilibrium to the reactants (to the left)

  19. Example • In which direction will the equilibrium shift if more calcium carbonate is added? • Increasing volume will decrease pressure; therefore, not favoring the smaller number of gas particles. Since the gas (CO2) is in the reactants, an increase in volume will shift the equilibrium to the products (shift to the right)

  20. Self-Ionization of Water • The acid-base chemistry we will encounter occurs in aqueous media • A tiny fraction of water molecules fragment or ionize into a hydrogen ion and a hydroxide ion H2O ➔ H+ + OH- • Thus, the is H+ represents the acid and OH-represents the base

  21. Conjugate Acid-Base Pairs When an acid donates a proton, it is converted into its conjugate base HA ➔ H+ + A— acid conjugate base يتحد مؤقتاً When a base accepts a proton, it is converted into its conjugate acid B + H+➔BH+ base conjugate acid Lecture 3

  22. Lecture 3 Conjugate acid-base pairs differ by a proton. When an acid donates a proton it becomes the conjugate base. → HCl(g) Cl-(aq) acid Conjugate base

  23. Lecture 3 Conjugate acid-base pairs differ by a proton. When a base accepts a proton it becomes the conjugate acid. → H2O (l) H3O+(aq) base Conjugate acid

  24. Lecture 3 → HCl(g) + H2O (l) Cl-(aq) + H3O+(aq) acid base base acid Conjugate acid-base pairs differ by a proton.

  25. Examples • Give the conjugate bases for each of these acids (the charge of the conjugate base is always one lower than the charge of its conjugate acid) HCl H2SO4 HSO4- H2O Cl- HSO4- SO42- OH-

  26. Examples • Give the conjugate acids for each of these bases (the charge of the conjugate acid is always one greater than the charge of its conjugate base) OH- NH3 HPO42- H2O H2O NH4+ H2PO4− H3O+

  27. conjugate base acid conjugate acid base Example • Identify the conjugate acid-base pairs in this reaction: NH3 + HC2H3O2 ➔ NH4+ + C2H3O22- acetic acid (gives vinegar its sour taste and pungent smell) acetate ion

  28. Amphoteric (Amphoprotic) Species • An amphoteric species can behave as either an acid or a base • H2O is an amphoteric species: • Here water behaves as a(n) base (water accepts a proton) H+ + H2O  H3O+ • Here, water behaves as a(n) acid (water donates a proton) H2O  H+ + OH-

  29. Diprotic Acids • A diprotic acid has 2 hydrogen ions to donate. A diprotic acid can behave as an acid twice. H2SO3➔ H+ + HSO3— HSO3- ➔H+ + SO32— • The terms triprotic acid and polyproticacid are self-explanatory

  30. Acid and Base Strength • A stronger acid is more determined to give its proton to some base • A stronger base is more determined to take a proton from some acid

  31. Acid Strengths

  32. Ionization Equilibria of Weak Acids • Weak acids do not ionize 100% in water • Weak acids establish an equilibrium: HA ➔H+ + A- • And have a corresponding equilibrium constant expression:

  33. Ionization Equilibria of Weak Bases • Weak bases establish an equilibrium by accepting a proton from water: B + H2O = BH+ + OH- • And have a corresponding equilibrium constant expression:

  34. Acid/Base Strength and Ka/Kb • Stronger acids have larger Ka values • Stronger bases have larger Kb values

  35. Self Ionization of Water : Kw • The ionization of water is an equilibrium: H2O ↔ H+ + OH- • The equilibrium constant is called Kw: Kw = [H+][OH-] = 1.0x10-14 at 25°C

  36. pKa and pKb pKa = -log Ka pKb = -log Kb

  37. The Key Relationships pH = -log [H+] pOH = -log[OH-] [H+][OH-] = 1.00 x10-14 pH + pOH = pKw = 14.00

  38. Calculting pH: Strong Acids • Strong acids are 100% ionized in water • Well, at least as long as the concentration is less than 0.1M, this is mostly true • Therefore, [H+] = cacid (formal concentration or molarity of the acid)

  39. Example: [H+] ➔ pH • What is the pH of a 0.025 M solution of HCl? HCl is a strong acid so it fully dissociates. The [H+] = formal concentration (molarity) of HCl. pH = -log [H+] = -log 0.025 = 1.6

  40. Example: pH➔[H+] • The pH of a dilute nitric acid solution is 4.59. What isthe[H+]? What is the formal concentration of nitric acid? Nitric acid is a strong acid. antilog of -4.59 = 2.6 x 10-5 M To take the “antilog” of -4.59 enter -4.59 then click “inv” and “log” keys in your calculator

  41. Calculating pH: Strong Bases • In a strong base, [OH-] = cbase

  42. Example • What is the pH of a 0.025 M solution of NaOH? NaOH is a strong base so: pOH = -log [OH-] = -log [0.025M] = 1.6 Since pH + pOH i= 14 so pH = 14 – 1.6 = 12.4

  43. Example • The pH of a dilute KOH solution is 9.59. What is the [H+]? Strong base so full dissociation of KOH into K+1 and OH- . K+1 does not matter (it is a pure “spectator”). pH = -log[H+] so [H+] is antilog of pH antilog of -9.59 = 2.57 x 10-10 so [H+] = 2.57 x 10-10

  44. Buffers • A pH buffer is a solution that resists changes in pH • A pH buffer contains a weak acid (HA) and its conjugate base (A-); or a weak base (A-) and its conjugate acid. • If a strong base (OH-) is added to a buffered solution, the weak acid in the buffer (HA) will react with OH- to give H2O and the weak base (A-). This results in converting a strong base (OH-) into a weak base (A-). As a result, the pH will slightly increase. HA + OH-➔ H2O + A- • If a strong acid acid (H+) is added to a buffered solution, the weak base in the buffer (A-) will react with the (H+) ion to give HA. This results in converting a strong acid H+ into a weak acid HA. As a result, the pH will slightly decrease. A- + H+ ➔ HA

  45. Buffer pH: The Henderson-Hasselbach Equation (the Buffer Equation) • The pH of a buffer IS NOT NECESSARILY 7!

  46. Example • Calculate the pH of a buffer solution that is 0.10 M inacetic acid and 0.15 M in sodium acetate. The ka of acetic acid is 1.8 x 10-5. • pka = -log ka = -log 1.8 x 10-5 = 4.74 • [A-] = acetate ion = 0.15 M • [HA] = acetic acid = 0.10 M pH = 4.74 + log 0.15 M 0.10 M pH = 4.74 + log 1.5 = 4.74 + 0.18 = 4.92 pH = 4.92 is slightly higher than pka (on the “basic” side of pka since the concentration of the base (0.15 M) is higher than the concentration of the acid (0.10 M)

  47. Example • Calculate the pH of a solution that is 0.25 molar in ammonia and 0.75 M in ammonium chloride. The pkb for ammonia is 4.74. The Henderson-Hasselbach equation calls for the pka of the acid so 14.00-4.74 = 9.26 is the pka. pH = 9.26 + log 0.25M 0.75 M pH = 9.26 + log 0.33 = 9.26 + (-0.48) = 8.78 pH = 8.78 is slightly lower than pka (on the “acidic” side of pka since the concentration of the base (0.25 M) is smaller than the concentration of the acid (0.75 M)

  48. Special Example • Calculate the pH of a buffer that is 0.10 M each inمهم sodium acetate and acetic acid. The ka of acetic acid is 1.8 x 10-5. pka of acetic acid = 4.74 pH = 4.74 + log 0.10 M 0.10 M pH = 4.74 + log 1 pH = 4.74 + 0 pH = 4.74 pH = pKa (true of any buffer system when the concentrations of the weak acid and its conjugate base are equal) pKa is the pH at which the concentrations of ionized and unionized species in a system are the same

  49. Buffers Example: What is the pH of a buffer containing 0.12 M benzoic acid and 0.20 M sodium benzoate? Ka = 6.3 x 10-5.

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