1. 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

2. pH scale 0714acidicbasic [H + ] > [OH  ][H + ] < [OH  ][H + ] = [OH  ] Calculating pH: Ex 1: What is the pH of a 0.15 M solution of HCl? pH =  log(0.15 M) = 0.82 pH = -log[H + ] pOH =  log[OH  ] pH + pOH = 14 Ex 2: What is the pH of a 0.2 M solution of NaOH? pOH =  log(0.2 M) = 0.70 pH = 14  0.70 = 13.30 NaOH  Na + + OH  0.2 M x x

3. Arrhenius model of acids and bases: An acid contains acidic protons that will ionize in water to form hydrogen ions. HCl (aq)  H + (aq) + Cl  (aq) A base contains hydroxide ions that will ionize in water and produce aqueous hydroxide ions. NaOH (s)  Na + (aq) + OH  (aq) H 2 SO 4 (aq)  2 H + (aq) + SO 4 2  (aq) pH of 0.1M soln? pH =  log(0.1 M) = 1 pH =  log(0.2 M) = 0.7 pH of 0.1M soln? pH = 14  pOH pH = 14  1 = 13 Mg(OH) 2 (s)  Mg 2+ (aq) + 2 OH  (aq) pH = 14  0.7 = 13.3 pOH =  log(0.1 M) = 1 pOH =  log(0.2 M) = 0.7 0.1 M x x x2x 0.2 M

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) + H 2 O (l) ⇋ H 3 O + (aq) + X - (aq) Acid H + donor Conjugate acid H + donor Base H + acceptor Conjugate base H + acceptor

5. Another example: NH 3 (aq) + H 2 O (l) ⇋ NH 4 + (aq) + OH - (aq) acidbaseConjugate acid Conjugate base  Substances that can act as either an acid or a base (e.g. H 2 O) are called amphoteric.  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.  Substances that can accept or donate a proton (e.g. HSO 4  ) are amphiprotic.

6. Strong acids ionize completely in solution. There is no equilibrium between the reactants and the products. Examples: Hydrochloric acidHCl Hydrobromic acidHBr Hydroiodic acidHI Perchloric acidHClO 4 Nitric acidHNO 3 Sulfuric acidH 2 SO 4 HX  H + + X - 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. 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: K a K a = [H + ][A - ] [HA] Ex: What is the pH of a weak acid if [HA]= 1 x 10 -4 M at equilibrium and it has a K a = 1 x 10 -5 ? [H+] = 3.16 x 10 -5 M pH = 4.5 Note: [H + ] = [A - ]

8. 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. CH 3 NH 2 (aq) + H 2 O (l) ⇋ CH 3 NH 3 + (aq) + OH - (aq) K b = [CH 3 NH 3 + ][OH - ] [CH 3 NH 2 ] Base dissociation constant

9. Naming Acids Binary acids: contain hydrogen and one other element 1.First part of name is hydro- 2.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 Suffixes: -ate  -ic-ite  -ous HNO 3 Nitric acid HNO 2 Nitrous acid

10. HC 2 H 3 O 2 HNO 3 H 2 CO 3 HCl H 3 PO 4 H 2 SO 4 H 2 S 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) 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) Acid nameAcid examples:

11. Ionization of polyprotic acids proceeds stepwise: First ionization: H 3 PO 4 (aq) ⇋ H 2 PO 4 - (aq) + H + (aq) Second ionization: H 2 PO 4 - (aq) ⇋ HPO 4 -2 (aq) + H + (aq) Third ionization: HPO 4 -2 (aq) ⇋ PO 4 -3 (aq) + H + (aq) Anhydrides: oxides that can act like acids or bases in water by reacting with the water. CaO (s) + H 2 O (l) ⇋ Ca 2+ (aq) + 2OH - (aq) CO 2 (g) + H 2 O (l) ⇋ H 2 CO 3 (aq) ⇋ H + (aq) + HCO 3 - (aq)

12. Position of Equilibrium and Acid Strength HSO 4  + CO 3 2  ⇋ SO 4 2  + HCO 3  Ex: What is the relative position (products or reactants) of the equilibrium below? 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 K a (HSO 4  ) = 1.2 x 10 -2 K a (HCO 3  ) = 5.6 x 10 -11 K C > 1 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: H 2 O (l) ⇋ H + (aq) + OH - (aq) The equilibrium constant for this reaction is given a special symbol, K w, which is: In neutral water [H + ] = [OH - ]= 1 x 10 -7 M The pH for any solution can be found by: pH = -log [H + ] The pH of a neutral solution is therefore = 7 2H 2 O (l) ⇋ H 3 O + (aq) + OH - (aq) Or…

14. Acid-Base reactions The H + from the acid reacts with the OH -1 from the base to form water. H + + OH - H2OH2O This is called areaction

15. Examples: What would the products be if HCl and NaOH reacted? HCl + NaOHH 2 O +NaCl If HCl and KOH reacted? HCl + KOHH 2 O +KCl If HCl and Mg(OH) 2 reacted? HCl + Mg(OH) 2 H 2 O +MgCl 2 2 2

16. 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 pK a Example: Acetic acid has a K a = 1.8 x 10 -5 pK a = -log(K a ) = -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. There are two main buffering equilibria… H 2 CO 3 + H 2 O ⇋ HCO 3 -1 + H 3 O + H 2 PO 4 -1 + H 2 O ⇋ HPO 3 -2 + H 3 O + K a = 4.3 x 10 -7 pK a = 6.4 K a = 6.2 x 10 -7 pK a = 7.2 Example: What is the pH of a solution containing 0.2 M H 2 CO 3 and 0.4 M NaHCO 3 ? K a = [HCO 3 -1 ][H 3 O + ] [H 2 CO 3 ] 4.3 x 10 -7 = [0.4 M][H 3 O + ] [0.2 M] [H 3 O + ] = 2.15 x 10 -7 MpH = 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. M 1 V 1 = M 2 V 2 M 1 = unknown concentration V 1 = volume used for unknown M 2 = known concentration V 2 = 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? M 1 (25 mL) = (1.0 M)(12.5 mL) M 1 V 1 = M 2 V 2 M 1 = 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 H 2 SO 4. What is the concentration of the NaOH? M 1 (10 mL) = 2(1.0 M)(10 mL) M 1 V 1 = M 2 V 2 M 1 = 2.0 M The concentration of the NaOH is 2.0 M There are TWO acidic protons in H 2 SO 4. The [H 3 O + ] will be two times more concentrated than the original acid.

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

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

26. The pH at V eq for a weak acid is greater than 7. pKa = pH at ½ V eq pK a = 4.9 V eq K a = 1.3 x 10 -5

27. The pH at V eq for a weak base is less than 7. pK a = pH at ½ V eq pK a = 9.0 pK b = 5.0 K b = 1.0 x 10 -5 K a K b = K w pK a + pK b = pK w pK a + pK b = 14

28. Buffer region pH = 4.74 Titration of Acetic Acid with NaOH Volume NaOH added (mL) pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 01020304050

29. Acid-Base Properties of Salt Solutions 1. Salt of a weak acid or a weak base. NH 4 Cl(aq) + H 2 O(l) ⇋ NH 3 (aq) + H 3 O+(aq) + Cl  (aq) a)What kind of salt is involved? b)What will happen to the pH of the solution? c)The K b for NH 3 is 1.8 x 10 -5, what is the pH of a 1.0 M solution of ammonium chloride? Weak acid = NH 4 + decrease K a = K w /K b = 1E-14/1.8E-5 = 5.6E-10 [H 3 O + ] = 2.4E-5 M pH = 4.62

30. Ex 2: NaC 2 H 3 O 2 (aq) + H 2 O(l) ⇋ HC 2 H 3 O 2 (aq) + OH  (aq) Weak base = C 2 H 3 O 2  increase K b = K w /K a = 1E-14/1.8E-5 = 5.6E-10 [OH  ] = 2.4E-5 M pOH = 4.62 a)What kind of salt is involved? b)What will happen to the pH of the solution? c)The K a for HC 2 H 3 O 2 is 1.8 x 10 -5, what is the pH of a 1.0 M solution of sodium acetate? pH = 9.38  If your equilibrium results in the formation of OH , use K b and solve for pOH and then pH.  If H 3 O + or H + is produced, use K a and solve for pH directly.  If your equilibrium results in the formation of OH , use K b and solve for pOH and then pH.  If H 3 O + or H + is produced, use K a and solve for pH directly.

31. What if the ion is amphiprotic? HPO 4 2- + H 2 O K a = 4.2 x 10  13 K b = 1.6 x 10  7 H 2 PO 4 - + OH  PO 4 3- + H 3 O + ? HPO 4 2  is a stronger base than it is an acid, so the pH of the solution will increase. Ex 3: Ex 4: What about NaNO 3 and MgBr 2 ? The solution will be neutral because HNO 3 and HBr are strong acids and Group 1 and 2 cations form strong bases.

32. 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. HClO > HBrO > HIO HClO 4 > HBrO 4 > HIO 4 Relative acid strength 2. For oxyacids that have the same central atom, acid strength increases as the number of oxygen atoms attached to the central atom increases. HClO 4 > HClO 3 > HClO 2 > HClO

33. Lewis Acids An Arrhenious acid is one that An Arrhenious base is one that Recall: A Bronsted-Lowry acid is one that A Bronsted-Lowry base is one that contains an acidic proton. contains hydroxide groups. donates a proton. accepts a proton A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor. Ex: AlCl 3 + Cl  ⇋ AlCl 4  LA LB +

34. pH of solutions containing metal ions It is because of Lewis acid/base behavior that some solutions of metal ions are acidic. Fe(H 2 O) 6 3+ ⇋ Fe(H 2 O) 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. NaNO 3 pH = 7 Ca(NO 3 ) 2 pH = 6.9 Zn(NO 3 ) 2 pH = 5.5 Al(NO 3 ) 3 pH = 3.5

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 HX ⇋ H + + X  Calculating pH of a buffer: Henderson-Hasselbalch Eq.

36. Ex: What is the pH of a buffer that is 0.12 M in lactic acid at equilibrium (HC 3 H 5 O 3 ) and 0.10 M in sodium lactate? For lactic acid, K a = 1.4 × 10 −4. pH = 3.85 + log[(0.10 M)/(0.12 M)] pKa = -log(1.4E-4) = 3.85 = 3.77 Ex: How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH is 9.00? NH 3 + H 2 O ⇋ NH 4 + + OH  K b = 1.8E-5 We really want to use the reverse reaction, since this will give us pH directly NH 4 + + OH  ⇋ NH 3 + H 2 O K a = 5.6E-10 pK a = 9.26

37. [NH 4 + ] = 0.18 M Trying to determine how many moles of NH 4 + are needed to prepare 2.0 L of the buffer. 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