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Strengths of Conjugate Acid-Base Pairs

Strengths of Conjugate Acid-Base Pairs. Acid strength increases. strong medium weak very weak.

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Strengths of Conjugate Acid-Base Pairs

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  1. Strengths of Conjugate Acid-Base Pairs Acid strength increases strong medium weak very weak HCl H2SO4 HNO3 H3O+ HSO4- H3PO4 HC2H3O2 H2CO3 H2S H2PO4- NH4+ HCO3- HPO42- H2O Cl- HSO4- NO3 H2O SO42- H2PO4- C2H3O2- HCO3- HS- HPO42- NH3 CO32- PO43- OH- negligible very weak weak medium strong Base strength increases

  2. Conjugate Acid Strength Relative acid strength Relative conjugate base strength Very weak Very strong Strong HA H+ + A- Weak [H+] [A-] [HA] Weak pKa = Strong Very weak Very strong Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 508

  3. Solutions of Acids and Bases: The Leveling Effect • No acid stronger than H3O+ and no base stronger than OH– can exist in aqueous solution, leading to the phenomenon known as the leveling effect. • Any species that is a stronger acid than the conjugate acid of water (H3O+) is leveled to the strength of H3O+ in aqueous solution because H3O+ is the strongest acid that can exist in equilibrium with water. • In aqueous solution, any base stronger than OH– is leveled to the strength of OH– because OH– is the strongest base that can exist in equilibrium with water • Any substance whose anion is the conjugate base of a compound that is a weaker acid than water is a strong base that reacts quantitatively with water to form hydroxide ion

  4. Weak Acids (pKa) Weak Acids – dissociate incompletely (~20%) Strong Acids – dissociate completely (~100%) A(g) + 2 B(g) 3 C(g) + D(g) Equilibrium constant (Keq) = Keq = LeChatelier’s Principle (lu-SHAT-el-YAY’s)

  5. HC2H3O2(aq) H+(aq) + C2H3O21-(aq) Ka = 1.8 x 10-5 = Equilibrium constant Keq = = Ka = Acid dissociation constant Ka = 1.8 x 10-5 @ 25 oC for acetic acid

  6. Ionization Constants for Acids Ka HCl H+ + Cl1- very large HNO3 H+ + NO31- very large H2SO4 H+ + HSO41- large 1.8 x 10-5 HC2H3O2 H+ + C2H3O21- 9.5 x 10-8 H2S H+ + HS1-

  7. Ionization of Acids Acid Ionization Equation Ionization Constant, pKa Hydrochloric HCl H1+ + Cl1- very large Sulfuric H2SO4 H1+ + HSO41- large Acetic HC2H3O2 H1+ + C2H3O21- 1.8 x 10-5

  8. 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 monochloracetic 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 HOCl hypochlorous acid 3.5 x 10-8 HCN hydrocyanic acid 6.2 x 10-10 NH4+ 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

  9. H2SO4 H+ + HSO41- & HSO41- H+ + SO42- Sample 1) One gram of concentrated sulfuric acid (H2SO4) is diluted to a 1.0 dm3 volume with water. What is the molar concentration of the hydrogen ion in this solution? What is the pH? Solution) First determine the number of moles of H2SO4 1 mol H2SO4 x mol H2SO4 = 1 g H2SO4 = 0.010 mol H2SO4 98 g H2SO4 OVERALL: H2SO4 2 H+ + SO42- in dilute solutions...occurs ~100% 0.010 M 0.020 M pH = - log [H+] substitute into equation pH = - log [0.020 M] pH = 1.69

  10. A volume of 5.71 cm3 of pure acetic acid, HC2H3O2, is diluted with water at 25 oC to form a solution with a volume of 1.0 dm3. What is the molar concentration of the hydrogen ion, H+, in this solution? (The density of pure acetic acid is 1.05 g/cm3.) Step 1) Find the mass of the acid Mass of acid = density of acid x volume of acid = 1.05 g/cm3 x 5.71 cm3 = 6.00 g (From the formula of acetic acid, you can calculate that the molar mass of acetic acid is 60 g / mol). Step 2) Find the number of moles of acid. x mol acetic acid = 6.00 g HC2H3O2 = 0.10 mol acetic acid (in 1 L) Molarity: M = mol / L Substitute into equation M = 0.10 mol / 1 L M = 0.1 molar HC2H3O2 Step 3) Find the [H+] Ka =

  11. Ka = 1.8 x 10-5 = HC2H3O2 H+ + C2H3O21- weak acid ? 0.1 M 0.1 M Step 3) Find the [H+] Ka = 1.8 x 10-5 @ 25 oC for acetic acid How do the concentrations of H+ and C2H3O21- compare? Substitute into equation: pH = - log[H+] x2 = 1.8 x 10-6M pH = - log [1.3 x10-3M] x = 1.3 x 10-3 molar = [H+] pH = 2.9

  12. H+ Concentrations …Strong vs. Weak Acid 0.0200 M 1.7 2.9 0.0013 M Note: although the sulfuric acid is 10x less concentrated than the acetic acid... …it produces > 10x more H+ pH = - log[H+]

  13. Practice Problems: 1a) What is the molar hydrogen ion concentration in a 2.00 dm3 solution of hydrogen chloride in which 3.65 g of HCl is dissolved? 1b) pH 2a) What is the molar concentration of hydrogen ions in a solution containing 3.20 g of HNO3 in 250 cm3 of solution? 2b) pH 3a) An acetic acid solution is 0.25 M. What is its molar concentration of hydrogen ions? 3b) pH 4) A solution of acetic acid contains 12.0 g of HC2H3O2 in 500 cm3 of solution. What is the molar concentration of hydrogen ions? 1a) 0.0500 M 2a) 0.203 M 3a) 2.1 x 10-3M 4) 2.7 x 10-3M 1b) pH = 1.3 2b) pH = 0.7 3b) pH = 2.7

  14. [Products] Ka = [H3O+] [CN1-] [Reactants] Ka = [HCN] Weak Acids Cyanic acid is a weak monoprotic acid. If the initial concentration of cyanic acid is 0.150 M and the equilibrium concentration of H3O+ is 4.8 x 10-3M, calculate Ka for cyanic acid. H3O+(aq) + CN1-(aq) H+(aq) HCN(aq) 4.8 x 10-3M 4.8 x 10-3M 0.150 M [4.8 x 10-3M] [4.8 x 10-3M] [CN1-] Ka = 1.54 x 10-4 Ka = [0.150 M] How is [H3O+] determined? Measure pH of solution and work backwards

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