1 / 26

Acids, Bases, and Salts

Acids, Bases, and Salts. Characteristics of Acids and Bases. Acids. Strong. Weak. ionizes completely in water. ionizes partially in water. hydrochloric acid, HCl. hydrofluoric acid, HF. hydrobromic acid, HBr. phosphoric acid, H 3 PO 4. hydroiodic acid, HI.

arlen
Download Presentation

Acids, Bases, and Salts

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Acids, Bases, and Salts CHM 1010 PGCC Barbara A. Gage

  2. Characteristics of Acids and Bases CHM 1010 PGCC Barbara A. Gage

  3. Acids Strong Weak ionizes completely in water ionizes partially in water hydrochloric acid, HCl hydrofluoric acid, HF hydrobromic acid, HBr phosphoric acid, H3PO4 hydroiodic acid, HI acetic acid, CH3COOH (or HC2H3O2) nitric acid, HNO3 sulfuric acid, H2SO4 carbonic acid, H2CO3 perchloric acid, HClO4 CHM 1010 PGCC Barbara A. Gage

  4. Strong acid: HA(g or l) + H2O(l) H3O+(aq) + A-(aq) The extent of dissociation for strong acids. Figure 18.2 H+ and H2O  H3O+ (hydronium ion) CHM 1010 PGCC Barbara A. Gage

  5. Weak acid: HA(aq) + H2O(l) H3O+(aq) + A-(aq) The extent of dissociation for weak acids. Figure 18.3 CHM 1010 PGCC Barbara A. Gage

  6. Bases (or alkalis) Strong Moderate Dissociates completely Dissociates completely but is not very soluble sodium hydroxide, NaOH magnesium hydroxide, Mg(OH)2 potassium hydroxide, KOH aluminum hydroxide, Al(OH)3 calcium hydroxide, Ca(OH)2 Weak strontium hydroxide, Sr(OH)2 barium hydroxide, Ba(OH)2 Dissociates partially ammonia, NH3 (NH4OH) carbonates, CO32- bicarbonates, HCO31- CHM 1010 PGCC Barbara A. Gage

  7. Figure 4.8 An aqueous strong acid-strong base reaction on the atomic scale. CHM 1010 PGCC Barbara A. Gage MX is a “salt” – an electrolyte that is not an acid or base

  8. Acid and Base Definitions Arrhenius Acid = compound that forms hydrogen (H+) ions in water Base = compound that forms hydroxide (OH-) ions in water CHM 1010 PGCC Barbara A. Gage

  9. Acid and Base Definitions An acid-base reaction can now be viewed from the standpoint of the reactants AND the products. An acid reactant will produce a base product and the two will constitute an acid-base conjugate pair. Bronsted-Lowry Acid = proton donor (H+ is a proton) Base = proton acceptor CHM 1010 PGCC Barbara A. Gage

  10. Lone pair binds H+ + + HCl H2O Cl- H3O+ Lone pair binds H+ + + NH3 H2O NH4+ OH- Proton transfer as the essential feature of a Brønsted-Lowry acid-base reaction. Figure 18.8 (acid, H+ donor) (base, H+ acceptor) (base, H+ acceptor) (acid, H+ donor) CHM 1010 PGCC Barbara A. Gage

  11. Acid + Base Base + Acid Reaction 1 HF + H2O F- + H3O+ Reaction 2 HCOOH + CN- HCOO- + HCN Reaction 3 NH4+ + CO32- NH3 + HCO3- Reaction 4 H2PO4- + OH- HPO42- + H2O Reaction 5 H2SO4 + N2H5+ HSO4- + N2H62+ Reaction 6 HPO42- + SO32- PO43- + HSO3- Table 18.2 The Conjugate Pairs in Some Acid-Base Reactions Conjugate Pair Conjugate Pair CHM 1010 PGCC Barbara A. Gage

  12. PROBLEM: The following reactions are important environmental processes. Identify the conjugate acid-base pairs. (a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq) (a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq) PLAN: Identify proton donors (acids) and proton acceptors (bases). (b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq) (b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq) SAMPLE PROBLEM 18.4 Identifying Conjugate Acid-Base Pairs conjugate pair2 conjugate pair1 SOLUTION: proton donor proton acceptor proton acceptor proton donor conjugate pair2 conjugate pair1 proton donor proton acceptor proton acceptor proton donor CHM 1010 PGCC Barbara A. Gage

  13. M2+ H2O(l) Molecules as Lewis Acids An acid is an electron-pair acceptor. A base is an electron-pair donor. acid base adduct M(H2O)42+(aq) adduct CHM 1010 PGCC Barbara A. Gage

  14. PROBLEM: Identify the Lewis acids and Lewis bases in the following reactions: (a) H+ + OH- H2O (a) H+ + OH- H2O (b) Cl- + BCl3 BCl4- (b) Cl- + BCl3 BCl4- PLAN: Look for electron pair acceptors (acids) and donors (bases). (c) K+ + 6H2O K(H2O)6+ (c) K+ + 6H2O K(H2O)6+ SAMPLE PROBLEM 18.12 Identifying Lewis Acids and Bases SOLUTION: acceptor donor donor acceptor acceptor donor CHM 1010 PGCC Barbara A. Gage

  15. Acid Anhydrides Non-metal oxides react with water to form acidic solutions CO2 (g) + H2O (l)  H2CO3 (aq) N2O5 (s) + H2O (l)  2 HNO3 (aq) SO3 (g) + H2O (l)  H2SO4 (aq) Dissolved non-metal oxides cause acid rain. CHM 1010 PGCC Barbara A. Gage

  16. Basic Anhydrides Metal oxides react with water to form alkaline solutions Na2O (s) + H2O (l)  2 NaOH(aq) CaO (s) + H2O (l)  Ca(OH)2 (aq) Al2O3 (s) + 3 H2O (l)  2 Al(OH)3 (aq) Lime (CaO) is used on lawns and is converted to Ca(OH)2 when it rains. CaO is less hazardous to handle. CHM 1010 PGCC Barbara A. Gage

  17. An acid-base titration. Figure 4.7 Start of titration Excess of acid Point of neutralization Slight excess of base CHM 1010 PGCC Barbara A. Gage

  18. PROBLEM: You perform an acid-base titration to standardize an HCl solution by placing 50.00 mL of HCl in a flask with a few drops of indicator solution. You put 0.1524 M NaOH into the buret, and the initial reading is 0.55 mL. At the end point, the buret reading is 33.87 mL. What is the concentration of the HCl solution? NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) 1L 103 mL 5.078x10-3 mol HCl 0.05000 L Sample Problem 4.5 Finding the Concentration of Acid from an Acid-Base Titration PLAN: SOLUTION: volume(L) of base multiply by M of base (33.87-0.55) mL x = 0.03332 L mol of base molar ratio 0.03332 L X 0.1524 M = 5.078x10-3 mol NaOH mol of acid Molar ratio is 1:1 divide by L of acid = 0.1016 M HCl M of acid CHM 1010 PGCC Barbara A. Gage

  19. H2O(l) + H2O(l) H3O+(aq) + OH-(aq) [H3O+][OH-] Kc = [H2O]2 The Ion-Product Constant for Water Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0 x 10-14 at 250C A change in [H3O+] causes an inverse change in [OH-]. In an acidic solution, [H3O+] > [OH-] In a basic solution, [H3O+] < [OH-] In a neutral solution, [H3O+] = [OH-] CHM 1010 PGCC Barbara A. Gage

  20. Divide into Kw [H3O+] > [OH-] [H3O+] = [OH-] [H3O+] < [OH-] The relationship between [H3O+] and [OH-] and the relative acidity of solutions. Figure 18.4 [H3O+] [OH-] ACIDIC SOLUTION BASIC SOLUTION NEUTRAL SOLUTION CHM 1010 PGCC Barbara A. Gage

  21. SAMPLE PROBLEM 18.2 Calculating [H3O+] and [OH-] in an Aqueous Solution PROBLEM: A research chemist adds a measured amount of HCl gas to pure water at 250C and obtains a solution with [H3O+] = 3.0x10-4M. Calculate [OH-]. Is the solution neutral, acidic, or basic? PLAN: Use the Kw at 250C and the [H3O+] to find the corresponding [OH-]. SOLUTION: Kw = 1.0x10-14 = [H3O+] [OH-] so [OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 = 3.3x10-11M [H3O+] is > [OH-] and the solution is acidic. CHM 1010 PGCC Barbara A. Gage

  22. Figure 18.5 The pH values of some familiar aqueous solutions. pH = -log [H3O+] pOH = -log [OH-] pH + pOH = 14 CHM 1010 PGCC Barbara A. Gage

  23. Figure 18.6 The relations among [H3O+], pH, [OH-], and pOH. CHM 1010 PGCC Barbara A. Gage

  24. PLAN: HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-] and then convert to pH and pOH. SAMPLE PROBLEM 18.3 Calculating [H3O+], pH, [OH-], and pOH PROBLEM: In an art restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated HNO3 to 2.0M, 0.30M, and 0.0063M HNO3. Calculate [H3O+], pH, [OH-], and pOH of the three solutions at 250C. SOLUTION: For 2.0M HNO3, [H3O+] = 2.0M and -log [H3O+] = -0.30 = pH [OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15M; pOH = 14.30 For 0.3M HNO3, [H3O+] = 0.30M and -log [H3O+] = 0.52 = pH [OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; pOH = 13.48 For 0.0063M HNO3, [H3O+] = 0.0063M and -log [H3O+] = 2.20 = pH [OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12M; pOH = 11.80 CHM 1010 PGCC Barbara A. Gage

  25. Buffers Solutions that resist change in pH Can maintain any pH value between 0 and 14 (not just neutral pH 7) Composed of a weak acid and a salt made from the weak acid or weak base and salt made from the weak base Examples: HC2H3O2 and NaC2H3O2 NH4OH and NH4Cl CHM 1010 PGCC Barbara A. Gage

  26. Buffers Reaction with acid: HC2H3O2 + C2H3O2- + H+ HC2H3O2 + HC2H3O2 Reaction with base: HC2H3O2 + C2H3O2- + OH- C2H3O2- + C2H3O2- + HOH A buffer regenerates it’s own components. The pH it maintains depends on the ratio of salt to acid (or base) and the nature of the acid (or base). CHM 1010 PGCC Barbara A. Gage

More Related