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Chapter 18 Equilibria Involving Acids & bases

Chapter 18 Equilibria Involving Acids & bases. ARRHENIUS THEORY for ACIDS and BASES. ACIDS : produce hydrogen ions (protons), H + , in solution BASES : produce hydroxide ions, OH - ,in solution, NEUTRALIZATION : H + + OH -  H 2 O Problems with Arrhenius Theory

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Chapter 18 Equilibria Involving Acids & bases

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  1. Chapter 18 Equilibria Involving Acids & bases

  2. ARRHENIUS THEORY for ACIDS and BASES ACIDS: produce hydrogen ions (protons), H+, in solution BASES: produce hydroxide ions, OH-,in solution, NEUTRALIZATION: H+ + OH- H2O Problems with Arrhenius Theory * H3O+: Hydronium ion rather than H+ * OH(H2O)3- present in solution, not OH- * Other substances also have acidic or basic properties

  3. H+ surrounded by four H-bonded H2O molecules H9O4+

  4. OH- surrounded by three H-bonded H2O molecules OH(H2O)3-

  5. Bronsted-Lowery Theory of Acids and Bases Acid– any substance donating a proton, H+ Base– any substance accepting a proton Conjugate Acid-Base Pairs: e.g. HF + NH3NH4+ + F- acid 1 base 2 acid 2 base 1 AMPHOTERIC substances have both acidic and basic properties. Mono-, di-, tri-,………. to polyprotic acids. Acidic versus nonacidic H atoms in compounds.

  6. For each of the following reactions, identify the acid, the base, the conjugate base, and the conjugate acid H2O + H2O  H3O+ + OH- H2PO4- + H2PO4- H3PO4 + HPO42- H2SO4 + H2O  H3O+ + HSO4- H2PO4- + H2O  H3PO4 + OH- CO2 + 2H2O  HCO3- + H3O+ H2PO4- + H2O  HPO42- + H3O+ Fe(H2O)63+ + H2O  Fe(H2O)5(OH) 2+ + H3O+ HCN + CO32- CN- + HCO3-

  7. Graphic representations of strong and weak acid equilibria Strong Acid: 100% Dissociation into ions HA(aq) + H2O(l) H3O+(aq) + A- (aq) Equilibrium Favors undissociated acid Weak Acid: Very little Dissociation into ions

  8. Acid strength versus conjugate base strength

  9. Bronsted-Lowery Theory: Acid and Base Strengths Proton transfers occur from a Strong acid to a strong base e.g. HCl + NaOH  H2O + NaCl Weak acid to strong base e.g. CH3COOH + NaOH  ?? Weak base to a stronger base e.g. HSO41- + HSO31- ??? Will a Reaction occur between….. a. HS1- and F 1-?? b. HCl and ClO2 1- ?? c. HCl and ClO4 1- ?? d. HCl and HNO3??

  10. Relative strengths of some Bronsted-Lowry acids and their conjugate bases AcidBase StrongestHClO4 ClO4-Weakest Acids H2SO4 HSO4-bases HI I- HBr Br- HCl Cl- HNO3 NO3- H3O+ H2O HSO4- SO42- H2SO3 HSO3- H3PO4 H2PO4- HNO2 NO2- HF F- CH3CO2H CH3CO2- H2CO3 HCO3- H2S HS- NH4+ NH3 HCN CN- HCO3- CO32- HS- S2- H2O OH- Weakest NH3 NH2- Strongest Acid OH- O2-bases

  11. Bronsted-Lowery Theory: Acid and Base Strengths LEVELING EFFECT of SOLVENTS: The strongest acid in a solvent is the conjugate acid of the solvent. The strongest base is the conjugate base. Acid H3O+ in water Base OH- in water

  12. H2O(l) H2O(l) OH-(aq) H3O+(aq) Autoionization of Water and the pH Scale + +

  13. Autoionization of Water H2O(l) + H2O(l)  H3O+(aq) + OH-(aq) Kw = [H3O+] [OH-] = 1.0 x 10-14 at 25oC At equilibrium [H3O+] = [OH-] = 1.0 x 10-7 Kw changes with temperature but [H3O+] = [OH-]

  14. pH and pOH SCALES pH = - log [H3O+] pOH = - log [OH-]

  15. pH, pOH CALCULATIONS pH = - log [H3O+] pOH = - log [OH-] Kw = [H3O+] [OH-] = 1.0 x 10-14 so p[H3O+] + p[OH-] = 14.00 or pH = 14.00 – pOH CALULATE SOME pH and pOH VALUES

  16. (Household Cleaner) Blood [H+] pH 10-14 14 1 M NaOH 10-13 13 Basic 10-12 12 Ammonia 10-11 11 10-10 10 10-9 9 10-8 8 Neutral 10-7 7 Pure Water 10-6 6 10-5 5 10-4 4 10-3 3 Acidic 10-2 2 10-1 1 1 0 Milk Vinegar Lemon juice Stomach acid 1 M HCl

  17. Calculate the pH of each solution: a. [H+] = 1.4 x 10-3 M e. [OH-] = 8 x 10-11 M b. [H+] = 2.5 x 10-10 M f. [OH-] = 5.0 M c,. [H+] = 6.1 M g. pOH = 10.5 d. [OH-] = 3.5 x 10-2 M h. pOH = 2.3 Calculate [H+] and [OH-] for each solution: a. pH = 7.41 (the normal pH of blood) b. pH = 15.3 c. pH = -1.0 e. pOH = 5.0 d. pH = 3.2 f. pOH = 9.6 How many significant figures are there in the numbers: 10.78, 6.78, 0.78? If these were pH values, to how many significant figures can you express the [H+]? Explain any discrepancies between your answers to the two questions.

  18. Values of Kw as a function of temperature are as follows: Temp (oC)Kw 0 1.14 x 10-15 25 1.00 x 10-14 35 2.09 X 10-14 40 2.92 x 10-14 50 5.47 x 10-14 a. Is the autoionization of water exothermic or endothermic? b. What is the pH of pure water at 50oC?

  19. Values of Kw as a function of temperature are as follows: Temp (oC)Kw 0 1.14 x 10-15 25 1.00 x 10-14 35 2.09 X 10-14 40 2.92 x 10-14 50 5.47 x 10-14 a. Is the autoionization of water exothermic or endothermic? b. What is the pH of pure water at 50oC? c. Restate your answers to water at 50oC. Which of the three criteria for neutrality is most general? d. From a plot of ln(Kw) versus 1/T (using the Kelvin scale), estimate Kw at 37oC, normal physiological temperature. e. What is the pH of a neutral solution at 37oC?

  20. -28 -29 Y = -9.2338 – 6870.6x R^2 = 0.999 -30 ln Kw -31 -32 -33 -34 -35 0.0028 0.0030 0.0032 0.0034 0.0036 1/T

  21. pH MEASUREMENT Indicators: colored weak acids and bases pH Meters: Glass membrane with a voltage (potential) difference across the glass. pH and BODY CHEMISTRY Normal pH 7.3 to 7.5 Acidosis pH < 7.3 Alkalosis pH > 7.45 Body Chemistry is “buffered” with bicarbonates (HCO3-) dihydrogenphosphates (H2PO4-) and proteins which help to maintain a constant pH

  22. WEAK ACIDSIONIZATION CONSTANTS HA(aq) + H2O(l) H3O+(aq) + A- (aq) Ka values at 25oC are known and tabulated for a large number of weak acids.

  23. Graphic representations of strong and weak acid equilibria Strong Acid: 100% Dissociation into ions HA(aq) + H2O(l) H3O+(aq) + A- (aq) Equilibrium Favors undissociated acid Weak Acid: Very little Dissociation into ions

  24. Values of Ka for Some Common Monoprotic Acids Formula Name Value of Ka* HSO4- Hydrogen sulfate ion 1.2 x 10-2 HClO2 Chlorous acid 1.2 x 10-2 HC2H2ClO2 Monochloroacetic acid 1.35 x 10-3 HF Hydrofluoric acid 7.2 x 10-4 HNO2 Nitrous acid 4.0 x 10-4 HC2H3O2 Acetic acid 1.8 x 10-5 [Al(H2O)6]3+ Hydrated aluminum (III) ion 1.4 x10-5 HOCl Hypochlorous acid 3.5 x 10-8 HCN Hydrocyanic acid 6.2 x 10-10 NH 4+ Ammonium ion 5.6 x 10-10 HOC6H5 Phenol 1.6 x 10-10 *The units of Ka are mol/L, but are customarily omitted. Increasing acid strength

  25. Write the dissociation reaction and the corresponding equilibrium expression for each of the following acids in water. a. H3PO4 b. H2PO41- c. HCO31- d. HCN e. Glycine, H2NCH2COOH f. Acetic acid, CH3COOH (HC2H3O2) g. Phenol, C6H5OH h. Benzoic acid, C6H5COOH

  26. Write the reaction and the corresponding Kb equilibrium expression for each of the following substances acting as bases in water. a. PO43- g. Glycine, NH2CH2COOH b. HPO42- h. Ethylamine, CH3CH2NH2 c. H2PO4- I. Aniline, C6H5NH2 d. NH3 j. Dimethylamine, (CH3)2NH e. CN- f. Pyridine, C5H5N

  27. WEAK ACID CALCULATIONS HA(aq) + H2O(l)  H3O+(aq) + A-(aq) 2H2O(l)  H3O+(aq) + OH-(aq) To simplify calculations, if % ionization is < 5%, then CHA [HA] OR [HA] = CHA – [H3O+], Set up pH equilibria calculations in tables as in previous equilibria problems.

  28. Solving Weak Acid Equilibrium Problems List the major species in the solution Choose the species that can produce H+, and write balanced equations for the reactions producing H+ Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing H+ Write the equilibrium expression for the dominant equilibrium. List the initial concentrations of the species participating in the dominant equilibrium. Define the change needed to achieve equilibrium; that is, define x.. Write the equilibrium concentrations in terms of x. Substitute the equilibrium concentrations into the equilibrium expression. Solve for x the “easy” way; that is , by assuming that [HA]0-x [HA]0. Use the 5% rule to verify whether the approximation is valid. Calculate [H+] and pH.

  29. Ka Problems For trichlorophenol (HC6H2Cl3O), Ka = 1 x 10-6, Calculate the concentrations of all species and the pH of a 0.05 M solution of trichlorophenol in water. A solution is prepared by dissolving 0.56 g benzoic acid (C6H5CO2H), Ka = 6.4 x 10-5) in enough water to make 1.0 L of solution. Calculate [C6H5CO2H]. [C6H5CO2-], [H+], [OH-], and the pH in this solution. Calculate the pH of a solution containing a mixture of 0.050 M HNO3 and 0.50M HC2H3O2.

  30. WEAK BASES IONIZATION CONSTANTS B(aq) + H2O(l) BH+ (aq) + OH- (aq) Kb values at 25oC are tabulated or may be calculated from Kw and Ka Kw = (Ka)(Kb) so Kb = Kw/Ka Where Ka is the conjugate acid constant

  31. Kb Problems Thallium (Tl) hydroxide is a strong base used in the synthesis of some organic compounds. Calculate the pH of a solution containing 2.48 g TlOH per liter. For the reaction of hydrazine (N2H4) in water. H2NNH2 + H2O  H2NNH3+ + OH- Kb is 3.0 x 10-6. Calculate the concentrations of all species and the pH of a 2.0 M solution of hydrazine in water.

  32. Two Weak Acids in Solution A solution of 0.100 M HClO, Ka =3.5 x 10-8 and 0.100 M Formic acid, HCO2H, Ka = 1.8 x 10-4 are mixed in equal proportions. What is the resulting solution pH?

  33. Polyprotic Acid Equilibria What is the pH of a solution 0.100 M sulfurous acid, H2SO3, Ka1 = 1.5 x 10-2 and Ka2 = 1.0 x 10-7?

  34. ACID-BASE PROPERTIES OF THE OXIDES (PART I)

  35. ACID BASE PROPERTIES OF THE OXIDES (PART II)

  36. ACID-BASE PROPERTIES OF THE OXIDES (PART III)

  37. ACID BASE PROPERTIES OF THE OXIDES (PART IV)

  38. ACID-BASE PROPERTIES OF THE OXIDES (PART V)

  39. ACID BASE PROPERTIES OF THE OXIDES PART (VI)

  40. ACID BASE PROPERTIES OF THE OXIDES PART (VII)

  41. ACID-BASE PROPERTIES OF THE OXIDES PART (VIII)`

  42. BRONSTED-LOWRY THEORY OXIDES, HYDROXIDES, ANHYDRIDES Acid, base reactions Dehydrations (formation of anhydrides) Hydration of oxides

  43. ACID STRENGTHSof WEAK ACIDS Oxoacids: Ka  up as the central atom oxidation state  up. Ka  up as central atom of same oxidation state moves left to right in the periodic table. Ka  up as the central atom of the same oxidation state moves UP in the same Group or family. Polyprotic acids: Ka decreases by approximately 105 for each successive H+ ionized. Binary acids: (only H and another element) Within a period, Ka  up as electronegativity of the other element. Within a group, Ka  up going down the group to higher mass and larger size.

  44. SnO2 + ? H2O  ? ?HCrO4-  ?Cr2O72- + ? ?HMnO4-  ? Mn (VI) compound + ?

  45. HYDROLOYSIS OF IONIC SALTS The pH of each type salt in solution depends on the Ka or Kb of the hydrolyzing ion(s).

  46. Salt Derived Ions Undergoing From: Hydrolysis pH Examples Strong base Neither Neutral NaCl, KNO3, strong acid pH = 7 BaCl2, CaBr2 Strong base Anion Basic LiCN, KNO2, CaF2 weak acid pH > 7 NaCH3CO2 Weak base Cation Acidic, NH4Cl, Al(NO3)3, strong acid pH < 7 (CH3)3NHBr Weak base, Both Acidic NH4NO2 weak acid if Kb< Ka; Neutral NH4CH3CO2 if Kb = Ka; basic NH4CN if Kb > Ka

  47. Arrange the following 0.10 M solutions in order from most acidic to most basic KOH, KBr, KCN, NH4Br, NH4CN, HCN Given that the Ka value for acetic acid is 1.8 x 10-5 and the Ka value for hypochlorous acid is 3 x 10-8, which is the stronger base, OCl- or C2H3O2-

  48. Acid-Base Equilibria What is the pH of a solution of 0.150 M sodium nitrite, NaNO2? HNO2, Ka = 4.0 x 10-4. What is the pH of a solution of 0.150 M hydrazinnium chloride, H2NNH3+? H2NNH3+, Kb = 3.0 x 10-6

  49. Calculate the pH of each of the following solutions. a. 0.10 M CH3NH3Cl b. 0.050 M NaCN c. 0.20 M Na2CO3 (consider only the reaction ) CO32- + H2O  HCO3- + OH- Sodium azid (NaN3) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a 0.010 M solution of NaN3. The Ka value for hydrazoic acid (HN3) is 1.9 x 10-5.

  50. LEWIS THEORYACIDS and BASES Lewis base - an electron pair donor Lewis acid - an electron pair acceptor Lewis acid + Lewis base  Adduct (coordination compound) e.g. Cu2+(aq) + 4CN-(aq)  Cu(CN)42-(aq) Look at Lewis Dot structures for lone pairs.

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