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The Chemistry of Acids and Bases

Explore the properties of acids and bases, their reactions, and calculating pH levels. Learn about their nomenclature and definitions.

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The Chemistry of Acids and Bases

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  1. The Chemistry of Acids and Bases Chemistry I – Chapter 19

  2. Acids Have a sour taste. Vinegar is a solution of acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas And salt. React with carbonates and bicarbonates to produce carbon dioxide gas Bases Have a bitter taste. Feel slippery. Many soaps contain bases.

  3. Example:Mg + 2HCl  MgCl2 + H2

  4. Some Properties of Acids • Produce H+ (as H3O+) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) • Taste sour • Corrode metals • Electrolytes • React with bases to form a salt and water • pH is less than 7 • Turns blue litmus paper to red “Blue to Red A-CID”

  5. Acid Nomenclature Review No Oxygen w/Oxygen An easy way to remember which goes with which… “In the cafeteria, you ATE something ICky”

  6. Acid Nomenclature Review • HBr(aq) • H2CO3 • H2SO3 hydrobromicacid  carbonicacid  sulfurousacid

  7. Name ‘Em! • HI (aq)Hydroiodic Acid • HCl(aq) Hydrochloric Acid • H2SO4 Sulfuric Acid • HNO3 Nitric Acid • HPO4 Phosphoric Acid • HCO3 Carbonic Acid • HOH Water! (Dihydrogen Monoxide)

  8. Some Properties of Bases • Produce OH- ions in water • Taste bitter, chalky • Are electrolytes • Feel soapy, slippery • React with acids to form salts and water • pH greater than 7 • Turns red litmus paper to blue “Basic Blue”

  9. Some Common Bases NaOH sodium hydroxide lye KOH potassium hydroxide liquid soap Ba(OH)2 barium hydroxide stabilizer for plastics Mg(OH)2 magnesium hydroxide Milk of magnesia Al(OH)3 aluminum hydroxide Maalox (antacid)

  10. Acid/Base definitions • Definition #1: Arrhenius (traditional) Acids – produce H+ ions (or hydronium ions H3O+) Bases – produce OH- ions (problem: some bases don’t have hydroxide ions!)

  11. Arrhenius acid is a substance that produces H+ (H3O+) in water Arrhenius base is a substance that produces OH- in water

  12. A Brønsted-Lowryacidis a proton donor A Brønsted-Lowrybaseis a proton acceptor • A “proton” is really just a hydrogen atom that has lost it’s electron!

  13. A Brønsted-Lowryacidis a proton donor A Brønsted-Lowrybaseis a proton acceptor base acid

  14. Conjugate Pairs

  15. A Brønsted-Lowryacidis a proton donor A Brønsted-Lowrybaseis a proton acceptor conjugatebase conjugateacid base acid

  16. Acids & Base Definitions Definition #3 – Lewis Lewis acid - a substance that accepts an electron pair Lewis base - a substance that donates an electron pair

  17. Lewis Acid/Base Reaction

  18. Neutralization • Acid+Base=Salt+Water • Ex: • HCl + NaOH  NaCl + H2O • Ca(OH)2 + HCl  CaCl2 + H2O

  19. Complete Acid/Base Worksheet on page 10 and 11

  20. The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H+ (or OH-) ion.Under 7 = acid 7 = neutralOver 7 = base

  21. pH of Common Substances

  22. Calculating the pH pH = - log [H+] (Remember that the [ ] means Molarity) Example: If [H+] = 1 X 10-10pH = - log 1 X 10-10 pH = - (- 10) pH = 10 Example: If [H+] = 1.8 X 10-5pH = - log 1.8 X 10-5 pH = - (- 4.74) pH = 4.74

  23. Try These! Find the pH of these: 1) A 0.15 M solution of Hydrochloric acid 2) A 3.00 X 10-7 M solution of Nitric acid pH = - log(0.15) = 0.82 pH = - log(3.00X10-7) = 6.52

  24. pH calculations – Solving for H+ If the pH of Coke is 3.12, [H+] = ??? 10-pH =[H+] [H+] = 10-3.12 = 7.6 x 10-4 M

  25. pH calculations – Solving for H+ • A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution? 10-8.5 = [H+] 3.16 X 10-9 M = [H+]

  26. pOH • Since acids and bases are opposites, pH and pOH are opposites! • pOH does not really exist, but it is useful for changing bases to pH. • pOH looks at the perspective of a base pOH = - log [OH-] Since pH and pOH are on opposite ends, pH + pOH = 14

  27. =10 0 =10 0 pH [H+] [OH-] pOH

  28. [H+], [OH-] and pH What is the pH of the 0.0010 M NaOH solution? [OH-] = 0.0010 M (or 1.0 X 10-3 M) pOH = - log[OH-] = -log 0.0010 pOH = 3 pH = 14 – 3 = 11

  29. The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater? [H+] = 10 -4.82 [H+] =1.51X10 -5 M The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood? pOH = -log [2.5 x 10-7 ] pOH = 6.6 pOH + pH = 14 pH = 14 – pOH = 7.4

  30. [OH-] 1.0 x 10-14 [OH-] 10-pOH 1.0 x 10-14 [H+] -Log[OH-] [H+] pOH 10-pH 14 - pOH -Log[H+] 14 - pH pH

  31. Calculating [H+], pH, [OH-], and pOH Problem 1: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 1.0 M and (b) 0.0024 M. Calculate the pH, [H+], pH, pOH, and [OH-] of the two solutions at 25°C. Solution 1: pH = -log [H+] = -log(1.0) = 0 10-pH =[H+] = 10-0 = 1 pOH = 14 - pH = 14 – 0 =14 10-pOH =[OH-] = 10-14

  32. Calculating [H+], pH, [OH-], and pOH Problem 1: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 1.0 M and (b) 0.00240 M. Calculate the pH, [H+], pH, pOH, and [OH-] of the two solutions at 25°C. Solution 2: pH = -log [H+] = -log(0.0024) = 2.62 10-pH =[H+] = 10-2.6 = 0.00240 pOH = 14 - pH = 14 – 2.62 = 11.4 10-pOH =[OH-] = 10-11.4 = 3.98X10 -12 M

  33. Calculating [H+], pH, [OH-], and pOH Problem 2: What is the [H+], pOH, and [OH-] of a solution with pH = 3.67? Is this an acid, base, or neutral? 10-pH =[H+] = 10-3.67 = 0.000214 M pOH = 14 - pH = 14 – 3.67 = 10.3 10-pOH =[OH-] = 10-10.3 = 5.01X10 -11 M

  34. Complete pH problems on page 15

  35. pH testing • There are several ways to test pH • Blue litmus paper (red = acid) • Red litmus paper (blue = basic) • pH paper (multi-colored) • pH meter (7 is neutral, <7 acid, >7 base) • Universal indicator (multi-colored) • Indicators like phenolphthalein • Natural indicators like red cabbage, radishes

  36. Paper testing • Paper tests like litmus paper and pH paper • Put a stirring rod into the solution and stir. • Take the stirring rod out, and place a drop of the solution from the end of the stirring rod onto a piece of the paper • Read and record the color change. Note what the color indicates. • You should only use a small portion of the paper. You can use one piece of paper for several tests.

  37. pH paper

  38. pH meter • Tests the voltage of the electrolyte • Converts the voltage to pH

  39. pH indicators • Indicators are dyes that can be added that will change color in the presence of an acid or base. • Some indicators only work in a specific range of pH • Once the drops are added, the sample is ruined • Some dyes are natural, like radish skin or red cabbage

  40. ACID-BASE REACTIONSTitrations Setup for titrating an acid with a base

  41. Titration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. • Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base) This is called NEUTRALIZATION.

  42. If it is a 1:1 Ratio between H+ and OH-in their formulas in a neutralization you can use the dilution equation • MbaseX Vbase = MacidX Vacid HCl + NaOH H2O + NaCl 1:1 ratio!

  43. If it is not a 1:1 molar ratio between H+ and OH- in their formulas (ignore coefficients for this!), then you must MULTIPLY the molarity of the acid by the # of H+ ions over the # of OH- ions in the acid and base formulas before you solve the dilution equation. • Ex. 2NaOH + H2SO4 Na2SO4 + 2H2O • MULTIPLY the molarityacid by 2/1 or 2. • Mbase X Vbase = 2Macid X Vacid

  44. LAB PROBLEM #1: Titration Problem. Given: HCl + NaOH H2O + NaCl 35.62 mL of NaOH is neutralized with 25.2 mL of 0.0998 M HCl by titration to an equivalence point. What is the concentration of the NaOH? 1:1 Ratio between acid and base! Mbase X Vbase = Macid X Vacid Mbase X 35.62mL = 0.0998M X 25.2mL Mbase = 0.0998M X 25.2mL = 0.0706 M 35.62mL

  45. LAB PROBLEM #2: Titration Problem. Given: 2H3PO4 + 2KOH  3H2O + 2KPO4 35.62 mL of KOH is neutralized with 25.2 mL of 0.0998 M H3PO4 by titration to an equivalence point. What is the concentration of the NaOH? 3:1 RATIO -- MULTIPLY THE MOLARITY OF THE ACID BY 3 BEFORE SOLVING!! Mbase X Vbase = 3Macidx Vacid Mbase X 35.62mL = 3(0.0998M) X 25.2mL Mbase = 3(0.0998M) X 25.2mL = 0.212 M 35.62mL

  46. COMPLETE TITRATION PROBLEMS on page 12

  47. Chemistry I – Chapter 20 Oxidation - reduction • Oxidation is loss of electrons • Reduction is gain of electrons • Oxidation is always accompanied by reduction • The total number of electrons is kept constant • Oxidizing agents oxidize and are themselves reduced • Reducing agents reduce and are themselves oxidized

  48. Oxidation numbers • Metals are typically considered more 'cation-like' and would possess positive oxidation numbers, while nonmetals are considered more 'anion-like' and would possess negative oxidation numbers. • Oxidation number is the number of electrons gained or lost by the element in making a compound

  49. Memory Device • Loss of Electrons is Oxidation • Gain of Electrons is Reduction • LEO the lion says GER

  50. Predicting oxidation numbers • Oxidation number of atoms in element is zero in all cases • Oxidation number of element in monatomic ion is equal to the charge • Sum of the oxidation numbers in a compound is zero • Sum of oxidation numbers in polyatomic ion is equal to the charge • F has oxidation number –1 • H has oxidation no. +1 • Oxygen is usually –2.

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