1. Edward Wen, PhD Acids & Bases

2. 2 Learning Outcomes Properties of acids and bases and definitions pH scale and calculation of pH Completing and balancing Neutralization reactions Titration calculations for neutralization reactions Defining Weak vs. Strong electrolytes (using the concept of equilibrium) Buffers – recognition of a buffer system, how a buffer works

3. 3 Types of Ionic Compounds Acids = form H + ions in water solution Bases = combine with H + ions in water solution increases the OH - concentration  may either directly release OH - or pull H + off H 2 O Salts = Ionic compounds formed from Acid and Base. all strong electrolytes Cation: except H + Anion: except OH -

4. 4 Properties of Acids Sour taste react with “active” Metals i.e. Al, Zn, Fe, but NOT w/ Ag, Au Zn + 2 HCl  ZnCl 2 + H 2 react with Carbonates, producing CO 2 marble, baking soda, limestone CaCO 3 + 2 HCl  CaCl 2 + CO 2 + H 2 O change color of vegetable dyes blue litmus turns red react with Bases to form ionic salts

5. 5 Most food contains acids Citric acid (HO 2 CCH(CO 2 H)COHCO 2 H): citrus fruits, tomato Malic acid (HO 2 CCH 2 CHOHCO 2 H): green apple, tomato, grape Ascorbic acid (aka Vitamin C) Folic acid

6. 6 Common Acids

7. 7 Binary acids (H m X): acid hydrogens attached to a nonmetal atom HCl, HF, HBr, HI H 2 S, H 2 Se Hydrofluoric acid

8. 8 Oxyacids acid hydrogens (H + ) attached to an oxygen atom H 2 SO 4, HNO 3, H 3 PO 4 HClO 4

9. 9 Carboxylic acids Many exist in food like vinegar, tomato, citrus fruit -COOH group HC 2 H 3 O 2, H 3 C 6 H 5 O 3 only the first H in the formula is acidic the H is on the COOH

10. 10 Properties of Bases also known as alkalis taste bitter solutions feel slippery change color of vegetable dyes different color than acid red litmus turns blue react with acids to form ionic salts neutralization

11. 11 Common Bases

12. 12 Structure of Bases most ionic bases contain OH - ions Drano clog-remover: NaOH, Ca(OH) 2 some contain CO 3 2- ion: it produces OH - with water Baking soda: CaCO 3 Alka-Seltzer: NaHCO 3 molecular bases that react with H + Windex: Ammonia (NH 3 )

13. 13 Acid-Base Reactions (Neutralization, Double Displacement Reaction) H + (from the acid) + OH - (from the base)  H 2 O it is often helpful to think of H 2 O as H-OH Cation (from base) + Anion (from acid)  Salt acid + base → salt + water HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)

14. 14 Acid Reactions. I. Reaction with Metals Reaction with many metals: Al, Zn, Fe, Mg but not all!! Not for Cu, Au, Ag, etc. Producing a Salt and hydrogen gas H 2 3 H 2 SO 4 (aq) + 2 Al(s) → Al 2 (SO 4 ) 3 (aq) + 3 H 2 (g)

15. 15 Acid Reactions. II Reaction with Metal Oxides when acids react with metal oxides, they produce a salt and water 3 H 2 SO 4 + Al 2 O 3 → Al 2 (SO 4 ) 3 + 3 H 2 O

16. 16 Acid Reactions. III Gas-evolving Reaction with Salts when acids react with metal carbonate, bicarbonate, sulfide, sulfite, and bisulfite, gas will be produced along with other products 2 HNO 3 + FeCO 3 → Fe(NO 3 ) 2 + CO 2 + H 2 O HCl + NaHCO 3 → NaCl + CO 2 + H 2 O ZnS + 2 HBr → ZnBr 2 + H 2 S CaSO 3 + 2 HI → CaI 2 + SO 2 + H 2 O H 2 SO 4 + 2 NH 4 HSO 3 → (NH 4 ) 2 SO 4 + 2SO 2 + 2H 2 O

17. 17 Base Reactions Neutralization of acids Reaction with Nonmetal oxides, CO 2 2 NaOH + CO 2 → Na 2 CO 3 + H 2 O Strong bases will react with Al metal to form sodium aluminate and hydrogen gas Example: Dissolving recycled aluminum can with NaOH solution 2 NaOH + 2 Al + 6 H 2 O → 2 NaAl(OH) 4 + 3 H 2

18. 18 Relative Strength of Acids

19. 19 Strong Acids The stronger the acid, the more willing it is to donate H + use water as the standard base Strong acids donate practically all their H + 100% ionized in water [H 3 O + ] = [strong acid] [ ] = molarity Stomach acid HCl  H + + Cl -

20. 20 Strong Acids Examples: Binary Acid: HCl, HBr, HI Oxyacid: HNO 3, H 2 SO 4, HClO 4, HClO 3 Example: HNO 3 = H + + NO 3 - H 2 SO 4 = 2H + + SO 4 2-

21. 21 Weak Acids Weak acids donate a small fraction of their H + most of the weak acid molecules do not donate H + to water [H 3 O + ] << [weak acid] Vinegar HC 2 H 3 O 2  H + + C 2 H 3 O 2 -

22. 22 Weak Acids Examples: Binary Acid: HF, H 2 S, H 2 Se Oxyacid: HNO 2, H 2 SO 3, H 3 PO 4, HClO Most carboxylic acids, such as acetic acid

23. 23 Strong Bases The stronger the base, the more willing it is to accept H + use water as the standard acid Strong bases: practically all molecules are dissociated into OH – or accept H + 1 mol NaOH = 1 mol OH - 1 mol Ca(OH) 2 = 2 mol OH - [OH – ] = [strong base] x (# OH) Drano TM NaOH  Na + + OH -

24. 24 Weak Bases Definition: a small fraction of molecules accept H + most of the weak base molecules do not take H + from water [HO – ] << [weak base] Windex TM NH 3 + H 2 O  NH 4 + + OH -

25. 25 Autoionization of Water Water: extremely Weak electrolyte therefore there must be a few ions present about 2 out of every 1 billion water molecules form Ions: Autoionization H 2 O + H 2 O  H 3 O + + OH – H 2 O  H + + OH – ALL aqueous solutions contain both H + and OH – the concentration of H + and OH – are equal in water @ 25°C: [H + ] = [OH – ] = 10 -7 M

26. 26 Ion Product of Water [H + ] x [OH – ] = constant: Ion Product of water, K w At 25°C, [H + ] x [OH – ] = 1 x 10 -14 = K w as [H + ] increases, [OH – ] must decrease so the product stays constant

27. 27 Acidic and Basic Solutions Neutral solutions have equal [H + ] and [OH – ] [H + ] = [OH – ] = 1 x 10 -7 M Acidic solutions : [H + ] > [OH – ] [H + ] > 1 x 10 -7 M [OH – ] < 1 x 10 -7 M Basic solutions: [OH – ] > [H + ] [H + ] 1 x 10 -7 M

28. 28 Practice - Determine the [H + ] concentration and whether the solution is acidic, basic or neutral for the following All [H + ] compared to 1 x 10 -7 M [OH – ] = 3.50 x 10 -8 M [NaOH] = 0.000250 M [HCl] = 0.50 M

29. Practice - Determine the [H + ] concentration and whether the solution is acidic, basic or neutral for the following [OH – ] = 3.50 x 10 -8 M NaOH = 0.000250 M [HCl] = 0.50 M [H + ] = 1 x 10 -14 3.50 x 10 -8 = 2.86 x 10 -7 M [H + ] >[OH - ], therefore acidic [H + ] = 1 x 10 -14 0.000250 = 4.00 x 10 -11 M [H + ] < [OH - ], therefore basic [H + ] > 1.0 x 10 -7 M therefore acidic [H + ] = 0.50 M

30. 30 Acidic/Basic: [H + ] vs. [OH - ] OH - H+H+ H+H+ H+H+ H+H+ H+H+ [OH - ]10 -14 10 -13 10 -11 10 -9 10 -7 10 -5 10 -3 10 -1 10 0 [H + ] 10 0 10 -1 10 -3 10 -5 10 -7 10 -9 10 -11 10 -13 10 -14 even though it may look like it, neither H + of OH - will ever be 0 the sizes of the H + and OH - are not to scale because the divisions are powers of 10 rather than units Acid Base

31. 31 pH The measure of the acidity/basicity of a solution pH = -log[H + ], [H + ] = 10 -pH exponent on 10 with a positive sign pH water = -log[10 -7 ] = 7 need to know the [H + ] concentration to find pH pH 7 : Basic pH = 7 : Neutral

32. 32 pH scale pH↓, Acidity↑ pH↑, basicity↑ 1 pH unit corresponds to a factor of 10 difference in acidity normal range 0 to 14 pH 0 is [H + ] = 1 M, pH 14 is [OH – ] = 1 M

33. 33 pH measurement pH can be measured by pH meter: The change in [H + ] affects the voltage of a standard cell

34. 34 pH of Common Substances SubstancepH 1.0 M HCl0.0 0.1 M HCl1.0 stomach acid1.0 to 3.0 lemons2.2 to 2.4 soft drinks2.0 to 4.0 plums2.8 to 3.0 apples2.9 to 3.3 cherries3.2 to 4.0 unpolluted rainwater5.6 human blood7.3 to 7.4 egg whites7.6 to 8.0 milk of magnesia (sat’d Mg(OH) 2 )10.5 household ammonia10.5 to 11.5 1.0 M NaOH14

35. 35 Example - Calculate the pH of the following strong acid or base solutions 0.0020 M HCl 0.010 M NaOH

36. 36 Example - Calculate the pH of the following strong acid or base solutions 0.0020 M HCl HCl as strong acid, so [H + ] = 0.0020 M pH = - log (2.0 x 10 -3 ) = 2.7 pH = - log (1.0 x 10 -12 ) = 12 [H + ] = 1 x 10 -14 1 x 10 -2 = 1 x 10 -12 M NaOH as strong base, so [OH - ] = 0.010 M 0.010 M NaOH

37. 37 pH in everyday life OH - H+H+ H+H+ H+H+ H+H+ H+H+ [OH - ]10 -14 10 -13 10 -11 10 -9 10 -7 10 -5 10 -3 10 -1 10 0 [H + ] 10 0 10 -1 10 -3 10 -5 10 -7 10 -9 10 -11 10 -13 10 -14 pH 0 1 3 5 7 9 11 13 14 Acid Base Stomach acid Vinegar Pure water Windex Drano

38. 38 Example - Calculate the concentration of [H + ] for a solution with pH 3.7 [H + ] = 10 -pH [H + ] = 10 -3.7 = 2 x 10 -4 M = 0.0002 M

39. 39 Find concentration of Acid or Base? Titration Purpose: using Reaction Stoichiometry to determine the Concentration of an unknown solution Titrant (solution of known concentration) added from a Buret Indicators: chemicals added to help determine when a reaction is complete the Endpoint of the titration occurs when the reaction is complete

40. 40 TitrationTitration: Color change w/ Indicator

41. 41 Titration Start: The base solution as titrant in the buret. Titrating: As the Base is added to the Acid, H + + OH –  HOH. But still excess Acid present so the color does not change. Endpoint: just enough Base to neutralize all the acid. The indicator changes color.

42. Calculations in Titration At the Endpoint of the titration, acid base neutralization reaction is complete. The mole ratio between acid and base in the reaction mixture is the same as in the balanced equation. Given the concentration of titrant, the mole of titrant can be calculated as: mole = Molarity x Volume (L) Then the mole of the other reactant can be calculated from the mole of titrant and the mole ratio in the equation (review stoichiometry: mole-to-mole). Finally, the molarity of other reactant can be determined.

43. Example: Acid-Base Titration

44. 44 Example: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.51 mL of 0.100 M Ba(OH) 2 solution to reach the endpoint. What is the concentration of the unknown HCl solution?

45. 45 First, write balanced equation: 2 HCl(aq) + Ba(OH) 2 (aq) → BaCl 2 (aq) + 2H 2 O(l)  2 mole HCl = 1 mole Ba(OH) 2  0.100 M Ba(OH) 2  0.100 mol Ba(OH) 2  1 L sol’n Information Given:10.00 mL HCl 12.54 mL 0.100 M Ba(OH) 2 Find:M HCl Example: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M Ba(OH) 2 solution to reach the end point. What is the concentration of the unknown HCl solution?

46. 46 Write a Solution Map: mL Ba(OH) 2 L Ba(OH) 2 mol Ba(OH) 2 mol HCl Information Given:10.00 mL HCl 12.51 mL Ba(OH) 2 Find:M HCl CF: 2 mol HCl = 1 mol Ba(OH) 2 0.100 mol Ba(OH) 2 = 1 L M = mol/L Example: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.51 mL of 0.100 M Ba(OH) 2 solution to reach the end point. What is the concentration of the unknown HCl solution? mL HCl L HCl

47. 47 = 2.50 x 10 -3 mol HCl Information Given:10.00 mL HCl 12.51 mL Ba(OH) 2 Find:M HCl CF: 2 mol HCl = 1 mol Ba(OH) 2 0.100 mol Ba(OH) 2 = 1 L M = mol/L SM: mL Ba(OH) 2 → L Ba(OH) 2 → mol Ba(OH) 2 → mol HCl; mL HCl → L HCl & mol  M Example: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.51 mL of 0.100 M Ba(OH) 2 solution to reach the end point. What is the concentration of the unknown HCl solution?

48. 48 Information Given:10.00 mL HCl 12.51 mL Ba(OH) 2 Find:M HCl CF: 2 mol HCl = 1 mol Ba(OH) 2 0.100 mol Ba(OH) 2 = 1 L M = mol/L SM: mL Ba(OH) 2 → L Ba(OH) 2 → mol Ba(OH) 2 → mol HCl; mL HCl → L HCl & mol  M Example: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.51 mL of 0.100 M Ba(OH) 2 solution to reach the end point. What is the concentration of the unknown HCl solution?

49. 49 How does pH change? Initial pHpH after adding 1 mL 1 M HCl pH after adding 1 mL 1 M NaOH 1 L Pure water 7.00 4.0010.00 1 L 0.14 M K 2 HPO 4 + 0.10 M KH 2 PO 4 7.00 6.997.01

50. 50 Buffers Definition: solutions that resist changing pH when small amounts of acid or base are added The mixture of 0.14 M K 2 HPO 4 + 0.10 M KH 2 PO 4 solution has much smaller pH change when strong acid or base is added, thus is called Buffer. Ingredient: mixing together a weak acid and its conjugate base or weak base and it conjugate acid Online demo: https://www.youtube.com/watch?v=P-R-Cqvb5yo https://www.youtube.com/watch?v=P-R-Cqvb5yo Human body fluid as buffer: H 2 CO 3 /HCO 3 -

51. 51 Buffer Composition: a weak acid + its salt; example: HC 2 H 3 O 2 / NaC 2 H 3 O 2, HF/KF When acid is added: C 2 H 3 O 2 - + H +  HC 2 H 3 O 2 When base is added: OH - + HC 2 H 3 O 2  C 2 H 3 O 2 - + H 2 O OR, a weak base + its salt example: NH 3 / NH 4 Cl

52. 52 Acetic Acid/Acetate Buffer

53. 53 Treasure Hunt: Which two can combine into a Buffer? HCl NH 4 + C 2 H 3 O 2 - Cl - HCO 3 - CO 3 2- HC 2 H 3 O 2 NH 3 H 2 CO 3