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Acids and Bases (Chapter 16):

Acids and Bases (Chapter 16):. Properties of acids:. Taste sour React with metals to form hydrogen gas React with carbonates to form CO 2 gas Form electrolyte solutions. Properties of bases:. Feel slippery Taste bitter Will react with some metals Form electrolyte solutions. 0.

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Acids and Bases (Chapter 16):

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  1. Acids and Bases (Chapter 16): Properties of acids: • Taste sour • React with metals to form hydrogen gas • React with carbonates to form CO2 gas • Form electrolyte solutions Properties of bases: • Feel slippery • Taste bitter • Will react with some metals • Form electrolyte solutions

  2. 0 acidic 7 basic 14 pH scale [H+] > [OH] [H+] = [OH] [H+] < [OH] pOH = log[OH] Calculating pH: pH = -log[H+] pH + pOH = 14 Ex 1: What is the pH of a 0.15 M solution of HCl? pH = log(0.15 M) = 0.82 Ex 2: What is the pH of a 0.2 M solution of NaOH? x x pOH = log(0.2 M) = 0.70 NaOH  Na+ + OH pH = 14  0.70 = 13.30 0.2 M 0.2 M

  3. Arrhenius model of acids and bases: H2SO4(aq) 2 H+(aq) + SO42 (aq) An acid contains acidic protons that will ionize in water to form hydrogen ions. pH of 0.1M soln? x x HCl (aq) H+(aq) + Cl(aq) pH = log(0.1 M) = 1 0.1 M 0.1 M x 2x pH = log(0.2 M) = 0.7 0.1 M 0.2 M A base contains hydroxide ions that will ionize in water and produce aqueous hydroxide ions. pH of 0.1M soln? NaOH (s) Na+(aq) + OH(aq) pH = 14  pOH pOH = log(0.1 M) = 1 pH = 14 1 = 13 Mg(OH)2(s) Mg2+(aq) + 2 OH(aq) pH = 14  0.7 = 13.3 pOH = log(0.2 M) = 0.7

  4. Brønsted-Lowry model of acids and bases: • Acids increase the hydrogen ion concentration in a solution by donating a hydrogen ion. • Bases increase the hydroxide ion concentration in a solution by accepting a hydrogen-ion. HX (aq) + H2O (l)⇋ H3O+(aq) + X-(aq) Acid H+ donor Conjugate base H+ acceptor Base H+ acceptor Conjugate acid H+ donor

  5. Substances that can act as either an acid or a base (e.g. H2O) are called amphoteric. • Substances that can accept or donate a proton (e.g. HSO4) are amphiprotic. Another example: NH3(aq) + H2O (l)⇋ NH4+(aq) + OH-(aq) base acid Conjugate acid Conjugate base • The stronger an acid is, the weaker its conjugate base will be. • The stronger a base is, the weaker its conjugate acid will be. • Strong acids/bases have conjugates of negligible strength.

  6. Strong acids ionize completely in solution. There is no equilibrium between the reactants and the products. HX  H+ + X- Hydrochloric acid HCl Hydrobromic acid HBr Hydroiodic acid HI Perchloric acid HClO4 Nitric acid HNO3 Sulfuric acid H2SO4 Examples: The pH can be found directly from the concentration of the acid in the solution. Ex: What is the pH of a 1 x 10-3 M HCl solution? [HCl] = [H+] = 1 x 10-3 M pH = 3

  7. Ka = [H+][A-] [HA] Weak acids only partially ionize in solution. They establish an equilibrium between the acid and its ions. HA ⇋ H+ + A- The acid dissociation equilibrium constant is given its own designation: Ka Note: [H+] = [A-] Ex: What is the pH of a weak acid if [HA]= 1 x 10-4 M at equilibrium and it has a Ka = 1 x 10-5? [H+] = 3.16 x 10-5 M pH = 4.5

  8. Kb = [CH3NH3+][OH-] [CH3NH2] Like strong acids, strong bases ionize completely in solution. NaOH  Na+ + OH- Examples: Group 1 and Group 2 hydroxides Weak bases only partially ionize in solution, and will establish an equilibrium with its conjugate acid and the conjugate base of water. CH3NH2(aq) + H2O (l)⇋ CH3NH3+(aq) + OH-(aq) Base dissociation constant

  9. Suffixes: -ate -ic -ite -ous Naming Acids Binary acids: contain hydrogen and one other element • First part of name is hydro- • Second part is the root of the second element with the suffix –ic • Ex: HCl is hydrochloric acid Oxyacids: acid form of a polyatomic anion that contains oxygen Format: Root of anion + suffix acid HNO3 HNO2 Nitric acid Nitrous acid

  10. Acid examples: Acid name HC2H3O2 HNO3 H2CO3 HCl H3PO4 H2SO4 H2S Acetic acid (vinegar) Nitric acid (acid rain) Carbonic acid (in sodas) Hydrochloric acid (stomach acid) Phosphoric acid (in colas) Sulfuric acid (battery acid) Hydrosulfuric acid (rotten egg smelling toxin) The hydrogens that appear first in the formula are called ACIDIC PROTONS, or just PROTONS for short. Monoprotic acid: an acid with only one acidic proton Polyprotic acid: an acid with more than one acidic proton (diprotic = 2, triprotic = 3)

  11. Ionization of polyprotic acids proceeds stepwise: First ionization: H3PO4(aq) ⇋ H2PO4-(aq) + H+(aq) Second ionization: H2PO4-(aq) ⇋ HPO4-2(aq) + H+(aq) Third ionization: HPO4-2(aq) ⇋ PO4-3(aq) + H+(aq) Anhydrides: oxides that can act like acids or bases in water by reacting with the water. CaO (s) + H2O (l) ⇋ Ca2+(aq) + 2OH-(aq) CO2(g) + H2O (l) ⇋ H2CO3(aq) ⇋ H+(aq) + HCO3-(aq)

  12. Position of Equilibrium and Acid Strength The position of the equilibrium favors the reaction of the stronger acid and the stronger base to form the weaker acid and the weaker base Ex: What is the relative position (products or reactants) of the equilibrium below? KC > 1 HSO4 + CO32 ⇋ SO42 + HCO3 Ka(HSO4) = 1.2 x 10-2 Ka(HCO3) = 5.6 x 10-11 The equilibrium favors product formation because the hydrogen carbonate ion is a weaker acid than the hydrogen sulfate ion.

  13. Self-ionization of water: Water will self-ionize according to the following reaction: H2O (l)⇋ H+(aq) + OH-(aq) Or… 2H2O (l)⇋ H3O+(aq) + OH-(aq) The equilibrium constant for this reaction is given a special symbol, Kw, which is: In neutral water [H+] = [OH-]= 1 x 10-7 M pH = -log [H+] The pH for any solution can be found by: The pH of a neutral solution is therefore = 7

  14. H2O H+ + OH- NEUTRALIZATION This is called a reaction Acid-Base reactions The H+ from the acid reacts with the OH-1 from the base to form water.

  15. HCl + NaOH HCl + KOH HCl + Mg(OH)2 Examples: What would the products be if HCl and NaOH reacted? H2O + NaCl If HCl and KOH reacted? H2O + KCl If HCl and Mg(OH)2 reacted? 2 H2O + MgCl2 2

  16. Acid + Base Salt + Water In general... *A salt is any ionic compound that is NOT an acid or a base.

  17. Buffers: solutions that resist changes in pH Buffers are mixtures of either: • A weak acid and the salt of its conjugate base OR • A weak base and the salt of its conjugate acid Buffers will maintain a pH that is ± 1 pH unit of their pKa Example: Acetic acid has a Ka = 1.8 x 10-5 pKa = -log(Ka) = -log(1.8 x 10-5) = 4.74 An acetic acid/sodium acetate mixture will buffer a solution at a pH of 4.74.

  18. An important buffer example: The pH of blood must be maintained at 7.4 ± 0.2 or death may occur. Ka = [HCO3-1][H3O+] 4.3 x 10-7 = [0.4 M][H3O+] [0.2 M] [H2CO3] There are two main buffering equilibria… H2CO3 + H2O ⇋ HCO3-1 + H3O+ Ka= 4.3 x 10-7 pKa = 6.4 H2PO4-1 + H2O ⇋ HPO3-2 + H3O+ Ka = 6.2 x 10-7 pKa = 7.2 Example: What is the pH of a solution containing 0.2 M H2CO3 and 0.4 M NaHCO3? [H3O+] = 2.15 x 10-7 M pH = 6.67

  19. Titrations Titrations allow the concentration of an acid or base to be determined using an acid-base reaction and an indicator. 1. Measure out a volume of the acid or base that has the unknown concentration. 2. Add small volumes of the other reactant of a known concentration until the indicator changes color. 3. Use the ‘magic equation’ to calculate the unknown concentration.

  20. Indicators: organic dyes whose color depends upon the pH of the solution.

  21. The Magic Equation!! M1V1 = M2V2 M1 = unknown concentration V1 = volume used for unknown M2 = known concentration V2 = total volume added of known concentration

  22. Examples: 25 mL of HCl are titrated with 12.5 mL of 1.0 M NaOH. What is the concentration of the HCl? M1V1 = M2V2 M1(25 mL) = (1.0 M)(12.5 mL) M1 = 0.5 M The concentration of the HCl is 0.5 M

  23. 10 mL of NaOH are titrated with 10 mL of 1.0 M H2SO4. What is the concentration of the NaOH? There are TWO acidic protons in H2SO4. The [H3O+] will be two times more concentrated than the original acid. M1V1 = M2V2 M1(10 mL) = 2(1.0 M)(10 mL) M1 = 2.0 M The concentration of the NaOH is 2.0 M

  24. Moles acid = Moles base • This volume is called the EQUIVALENCE POINT for the neutralization reaction. Why does the ‘Magic Equation’ work? Recall that Molarity = moles/volume of soln. MAVA = MBVB • The ‘Magic Equation’ gives the volume at which the number of moles of H+exactly equal the number of moles of OH-.

  25. 1. What is the pH of 0.100 M HCl? pH = 1 2. What is the pH of 0.100 M NaOH? pH = 13 3. What is Veq? Veq = 50.0 mL Note: The ONLY time the pH = 7 at the equivalence point is during strong acid : strong base titrations.

  26. pKa = 4.9 Veq Ka = 1.3 x 10-5 The pH at Veq for a weak acid is greater than 7. pKa = pH at ½ Veq

  27. pKa = 9.0 KaKb = Kw pKa + pKb = pKw pKa + pKb = 14 pKb = 5.0 Kb = 1.0 x 10-5 The pH at Veq for a weak base is less than 7. pKa = pH at ½ Veq

  28. Buffer region Titration of Acetic Acid with NaOH 14 13 12 11 10 9 8 pH 7 6 pH = 4.74 5 4 3 2 1 0 0 10 20 30 40 50 Volume NaOH added (mL)

  29. Acid-Base Properties of Salt Solutions 1. Salt of a weak acid or a weak base. NH4Cl(aq) + H2O(l)⇋ NH3(aq) + H3O+(aq) + Cl(aq) Weak acid = NH4+ • What kind of salt is involved? • What will happen to the pH of the solution? • The Kb for NH3 is 1.8 x 10-5, what is the pH of a 1.0 M solution of ammonium chloride? decrease 5.6E-10 Ka = Kw/Kb = 1E-14/1.8E-5 = [H3O+] = 2.4E-5 M pH = 4.62

  30. Ex 2: NaC2H3O2(aq) + H2O(l)⇋ HC2H3O2(aq) + OH(aq) • What kind of salt is involved? • What will happen to the pH of the solution? • The Ka for HC2H3O2 is 1.8 x 10-5, what is the pH of a 1.0 M solution of sodium acetate? Weak base = C2H3O2 increase 5.6E-10 Kb = Kw/Ka = 1E-14/1.8E-5 = [OH] = 2.4E-5 M pOH = 4.62 • If your equilibrium results in the formation of OH, use Kb and solve for pOH and then pH. • If H3O+ or H+ is produced, use Ka and solve for pH directly. pH = 9.38

  31. What if the ion is amphiprotic? PO43- + H3O+ Ka = 4.2 x 10 13 ? Ex 3: HPO42- + H2O H2PO4- + OH Kb = 1.6 x 107 HPO42 is a stronger base than it is an acid, so the pH of the solution will increase. Ex 4: What about NaNO3 and MgBr2? The solution will be neutral because HNO3 and HBr are strong acids and Group 1 and 2 cations form strong bases.

  32. HClO4 > HBrO4 > HIO4 HClO > HBrO > HIO Molecular structure and acidity 1. For oxyacids that have the same number of OH groups and the same number of doubly bonded O atoms, acid strength increases with increasing electronegativity of the central atom. 2. For oxyacids that have the same central atom, acid strength increases as the number of oxygen atoms attached to the central atom increases . Relative acid strength HClO4 > HClO3 > HClO2 > HClO

  33. Lewis Acids + Recall: contains an acidic proton. An Arrhenious acid is one that 1 An Arrhenious base is one that contains hydroxide groups. A Bronsted-Lowry acid is one that donates a proton. 2 A Bronsted-Lowry base is one that accepts a proton A Lewis acid is an electron pair acceptor. 3 A Lewis base is an electron pair donor. LA LB Ex: AlCl3 + Cl⇋ AlCl4

  34. pH of solutions containing metal ions It is because of Lewis acid/base behavior that some solutions of metal ions are acidic. Fe(H2O)63+⇋ Fe(H2O)5(OH)2+ + H+ The higher the charge on the metal ion and the smaller the ion’s radius, the greater the hydrolyzing effect that the cation will have and the lower the pH will be. Ca(NO3)2 pH = 6.9 Zn(NO3)2 pH = 5.5 Al(NO3)3 pH = 3.5 NaNO3 pH = 7

  35. Chapter 17 (cont): Additional Aspects of Acid/Base Equilibria Buffers: solutions that resist changes in pH • Buffers are mixtures of either a weak acid and the salt of its conjugate base OR a weak base and the salt of its conjugate acid • Buffers will maintain a pH that is ± 1 pH unit of their pKa Calculating pH of a buffer: HX ⇋ H+ + X Henderson-Hasselbalch Eq.

  36. Ex: What is the pH of a buffer that is 0.12 M in lactic acid at equilibrium (HC3H5O3) and 0.10 M in sodium lactate? For lactic acid, Ka = 1.4 × 10−4. pKa = -log(1.4E-4) = 3.85 = 3.77 pH = 3.85 + log[(0.10 M)/(0.12 M)] Ex: How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00? NH3 + H2O ⇋ NH4+ + OH Kb = 1.8E-5 We really want to use the reverse reaction, since this will give us pH directly Ka = 5.6E-10 NH4+ + OH⇋ NH3 + H2O pKa = 9.26

  37. Trying to determine how many moles of NH4+ are needed to prepare 2.0 L of the buffer. [NH4+] = 0.18 M x 2.0 L 0.36 mol Buffer capacity: the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree (± 1 pH unit). A 1.0 M buffer has 10 x the capacity of a 0.1 M buffer

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