1. Chapter 14 Arrhenius –Acid – create H + in water –Base – create OH - in water Bronsted-Lowery –Acid – donates proton (H + ) –Base – accepts proton (H + ) Hydronium ion - H 3 O +

2. Conjugate pairs –HCN/CN - HCl/Cl - NH 3 /NH 4 + Acid dissociation constant –Equilibrium expression for its dissociation –K a –HF  H + + F - –Ka = [H + ][F - ] / [HF] Strong vs weak

3. Oxyacids –HNO 3, HClO 4, H 2 SO 4 Organic Acids –COOH group HC 2 H 3 O 2 (CH 3 COOH) Monoprotic, diprotic, triprotic –HCl, H 2 SO 4, H 3 PO 4

4. Water as an acid and base –H 2 O + H 2 O  H 3 O + + OH - Amphoteric Dissociation constant for water –K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 –[OH - ] = 1.0 x 10 -5 [H 3 O + ] = ? –[H 3 O + ] = 1.0 x 10 -14 / 1.0 x 10 -5 = 1.0 x 10 -9

5. pH –pH = - log[H + ] pOH –pOH = - log[OH - ] pH + pOH = 14 Calculate the pH and pOH of a 1.0 x10 -9 M HCl solution. –pH = - log(1.0x10 -9 ) = 9.0 pOH = 14 – 9 = 5.0

6. –What is the [H + ] if the pH = 4.4 –pH = -log[H + ] [H + ] = 10 -pH –[H + ] = 10 -4.4 = 3.98 x 10 -5 = 4.0 x 10 -5 M pH of strong acids –Completely dissociates so [acid] = [H + ] pH of weak acids –Have to do an equilibrium problem using K a –Follow the same process

7. –HF  H + + F - –Ka = [H + ][F - ] / [HF] = 7.2 x 10 -4 –Calculate the pH of a 1.0 M HF solution –Chem.InitialEquil. –[H+]0+x –[F-]0+x –[HF]1.01.0 – x = 1.0 (small x) –Ka = [H + ][F - ] / [HF] = 7.2 x 10 -4 –7.2 x 10 -4 = x 2 / 1.0 x = 2.7 x 10 -2 M –pH = -log(2.7 x 10 -2 ) = 1.57

8. pH of a mixture of acids –Focus on the strongest acid –Our HF example also contained H 2 O but we can ignore it since its constant is 1.0 x 10 -14 and HF is much larger, 7.2 x 10 -4 Percent dissociation –%diss. = [H + ] / [HA] o x 100

9. What is the pH of an aqueous solution of 1.00M HCN and 5.00M HNO 2. –HCN Ka = 6.2 x 10 -10 –HNO 2 Ka = 4.0 x 10 -4 –H 2 O Kw = 1.0 x 10 -14 –Strongest acid is HNO 2 so we use it

10. HNO 2 Ka = 4.0 x 10 -4 –HNO 2  H + + NO 2 - Chem.[init.][equil] HNO 2 5.005.00-x = 5.00 (x is small) H + 0+x NO 2 - 0+x –4.0 x 10 -4 = x 2 / 5.00 –x = 4.5 x 10 -2 –pH = -log(4.5 x 10 -2 ) = 1.35

11. Bases –We calculate [OH - ] just like [H + ] but use K b instead of K a pH of strong bases –Completely dissociates so the [Base] = #[OH - ] pH of weak bases –Focus on the strongest base present and do an equilibrium problem

12. Calculate the [OH - ] for a 15.0 M NH3 solution, K b = 1.8 x 10 -5. –NH 3 + H 2 O  NH 4 + + OH - –K b = [NH 4 + ][OH - ] / [NH 3 ] = 1.8 x 10 -5 –Chem.[Init.][Equil] –NH 4 +0+x –OH-0+x –NH 3 15.015.0 – x = 15.0

13. –Chem.[Init.][Equil] –NH 4 +0+x –OH-0+x –NH 3 15.015.0 – x = 15.0 –K b = [NH 4 + ][OH - ] / [NH 3 ] = 1.8 x 10 -5 –1.8 x 10 -5 = x 2 / 15.0 –x = 1.6 x 10 -2 –pOH = -log(1.6 x 10 -2 ) = 1.80 –pH = 14 – 1.80 = 12.20

14. Strong bases make weak adics Strong acids make weak bases So the larger the Ka the smaller the Kb Ka x Kb = Kw

15. Polyprotic acids –Dissociate one proton at a time –Follow the same steps as an equil. problem –Calculate the pH as well as the concentration of all other chemicals present in a 5.0M H 3 PO 4 aqueous solution. –H 3 PO 4 K a1 = 7.5 x 10 -3 –H 2 PO 4 - K a2 = 6.2 x 10 -8 –HPO 4 2- K a3 = 4.8 x 10 -13

16. H 3 PO 4  H + + H 2 PO 4 - K a1 = 7.5 x 10 -3 = [H+][H 2 PO 4 - ] / [H 3 PO 4 ] –Chem.[init][equil] –H + 0+x –H 2 PO 4 - 0+x –H 3 PO 4 5.05.0 – x = 5.0 –7.5 x 10 -3 = x 2 / 5.0 –x = 0.19 = [H+] = [H 2 PO 4 - ]

17. H 2 PO 4 -  H + + HPO 4 2- K a2 = 6.2 x10 -8 = [H + ][HPO 4 2- ]/[H 2 PO 4 - ] –Chem[init][equil] –H + 0.190.19+x = 0.19 –HPO 4 2- 0+x –H 2 PO 4 - 0.190.19 – x = 0.19 –6.2 x10 -8 = (0.19)x / 0.19 x = 6.2 x10 -8 –[HPO 4 2- ] = 6.2 x 10 -8

18. HPO 4 2-  H + + PO 4 3- K a3 = 4.8 x 10 -13 = [H + ][PO 4 3- ]/[HPO 4 2- ] –Chem [init.] [equil.] –H + 0.190.19 + x = 0.19 –PO 4 3- 0+x –HPO 4 2- 6.2 x 10 -8 6.2 x10 -8 -x = 6.2x10 -8 –4.8 x 10 -13 = (0.19)x / 6.2 x 10 -8 –x = 1.6x10 -19 –[PO 4 3- ] = 1.6 x 10 -19

19. [OH - ] –K w = 1.0 x 10 -14 = [H+][OH-] –1.0 x 10 -14 = (0.19)[OH-] –[OH-] = 5.3 x 10 -14

20. Sulfuric acid –Similar to phosphoric acid except the first dissociation is complete and we can use that info for the second dissociation problem. –Calculate the [SO 4 2- ] in a 1.0M aqueous H 2 SO 4 solution. –H 2 SO 4  H + + HSO 4 - –[H 2 SO 4 ] = [H + ] = [HSO 4 - ] = 1.0

21. HSO 4 -  H + + SO 4 2- K a2 = 1.2 x 10 -2 = [H + ][SO 4 2- ] / [HSO 4 - ] –Chem.[init.][equil.] –H + 1.01.0 + x = 1.0 –SO 4 2- 0+ x –HSO 4 - 1.01.0 – x = 1.0 –1.2 x 10 -2 = (1.0)x / 1.0 x = 1.2 x 10 -2 –[SO 4 2- ] = 1.2 x 10 -2 M

22. Acid/Base properties of salts –Salts from strong acids/bases produce neutral solution, NaCl, KNO 3 etc –Salts from weak acid, strong base produce basic solutions, NaF, KC 2 H 3 O 2 –F - + H 2 0  HF + OH - –Salts from strong acid, weak base produce acidic solutions, NH 4 Cl –NH 4 + + H 2 O  NH 3 + H 3 O +

23. –Highly charged metal ions like Al 3+ produce acidic solutions when hydrated, because of the large charge it is easier to release H+ K a x K b = K w So if you know a chemicals K a or K b you can determine the corresponding K a / K b as needed

24. Calculate the pH for a 0.1 M NH 4 Cl aqueous solution. K b = 1.8 x 10 -5 for NH 3 NH 4 + reacts with water like an acid so we need it’s K a. We get it from the K b –K a = K w / K b = 1.0 x 10 -14 / 1.8 x 10 -5 = 5.6 x 10 -10 –NH 4 + + H 2 O  NH 3 + H 3 O + –Chem[init.][equil.] –NH 4 + 0.10.1 – x = 0.1 –NH 3 0+ x –H 3 O + 0+ x

25. –K a = 5.6 x 10 -10 = [NH 3 ][ H 3 O+] / [NH 4 +] – 5.6 x 10 -10 = x 2 / 0.1 –x = 7.5 x 10 -6 = [H 3 O+] –pH = -log (7.5 x 10 -6 ) = 5.13

26. Structure effect HClO vs HClO 4 –The extra oxygens draw the electrons away from the hydrogen allowing it to be released more easily. Thus HClO 4 is stronger –How do HNO 2 and HNO 3 compare?

27. Acid/Base properties of oxides –Non-metal oxides produce acids when mixed with water –SO 3 + H 2 O  H 2 SO 4 –CO 2 + H 2 O  H 2 CO 3 –Metal oxides produce bases when mixed with water –CaO + H 2 O  Ca(OH) 2 –Na 2 O + H 2 O  NaOH

28. Lewis Acid/Bases –Acids accept electron pairs –Bases donate electron pairs –BH 3 + NH 3  BH 3 NH 3 –BH 3 is the acid –NH 3 is the base