Strengths of Conjugate Acid-Base Pairs Acid strength increases strong HCl H2SO4 medium weak very weak HNO3 H3O+ HSO4- H3PO4 HC2H3O2 H2CO3 H2S H2PO4- NH4+ HCO3- HPO42- H2O Cl- HSO4- negligible NO3 H2O SO42- H2PO4- very weak C2H3O2- HCO3- HS- HPO42- weak Base strength increases NH3 medium CO32- PO43- OH- strong Conjugate Acid Strength Relative acid strength Relative conjugate base strength Very strong Very weak Strong HA H+ + A- Weak [H+] [A-] pKa = [HA] Weak Strong Very weak Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 508 Very strong 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 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 = [Products] [Reactants] [C ]3 [D] [ A][B ]2 LeChatelier’s Principle (lu-SHAT-el-YAY’s) H+(aq) + C2H3O21-(aq) HC2H3O2(aq) 1 Equilibrium constant Keq = [H ][C2H3O2 ] [HC2H3O2 ] = Ka = Acid dissociation constant Ka = 1.8 x 10-5 @ 25 oC for acetic acid 1 [H ][C2H3O2 ] Ka = [HC2H3O2 ] 1 1.8 x 10-5 [H ][C2H3O2 ] = [HC2H3O2 ] Ionization Constants for Acids Ka HCl H+ + Cl1- very large HNO3 H+ + NO31- very large H2SO4 H+ + HSO41- HC2H3O2 H2S large H+ + C2H3O21- 1.8 x 10-5 H+ 9.5 x 10-8 + HS1- Ionization of Acids Acid Hydrochloric Ionization Equation HCl Sulfuric H2SO4 Acetic HC2H3O2 Ionization Constant, pKa H1+ + Cl1H1+ + HSO41H1+ + C2H3O21- very large large 1.8 x 10-5 Formula Name Value of Ka* HSO4HClO2 HC2H2ClO2 HF HNO2 HC2H3O2 HOCl HCN NH4+ HOC6H5 hydrogen sulfate ion chlorous acid monochloracetic acid hydrofluoric acid nitrous acid acetic acid hypochlorous acid hydrocyanic acid ammonium ion phenol 1.2 x 10-2 1.2 x 10-2 1.35 x 10-3 7.2 x 10-4 4.0 x 10-4 1.8 x 10-5 3.5 x 10-8 6.2 x 10-10 5.6 x 10-10 1.6 x 10-10 *The units of Ka are mol/L but are customarily omitted. Increasing acid strength Values of Ka for Some Common Monoprotic Acids 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 x mol H2SO4 = 1 g H2SO4 H2SO4 H+ + HSO41- 1 mol H2SO4 98 g H2SO4 & = 0.010 mol H2SO4 HSO41- H+ + SO42- OVERALL: H2SO4 0.010 M 2 H+ + SO42- in dilute solutions...occurs ~100% 0.020 M pH = - log [H+] substitute into equation pH = - log [0.020 M] pH = 1.69 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 Step 2) Find the number of moles of acid. (From the formula of acetic acid, you can calculate that the molar mass of acetic acid is 60 g / mol). 1 mol HC 2H3O 2 = 0.10 mol acetic acid (in 1 L) x mol acetic acid = 6.00 g HC2H3O2 60 g HC H O 2 3 2 Molarity: M = mol / L Substitute into equation M = 0.10 mol / 1 L M = 0.1 molar HC2H3O2 Step 3) Find the [H+] Ka = HC2H3O2 Step 3) Find the 0.1 M [H+] weak acid H+ + C2H3O21- 0.1? M Ka = 1.8 x 10-5 @ 25 oC for acetic acid 1 [H ][C2H3O2 ] Ka = [HC2H3O2 ] 1 1.8 x 10-5 [H ][C2H3O2 ] = [HC2H3O2 ] How do the concentrations of H+ and C2H3O21- compare? [x][x] [HC2H3O2 ] Substitute into equation: 1.8 x 10-5 1.8 x 10 -5 x2 [0.10 M] pH = - log[H+] x2 = 1.8 x 10-6 M x = 1.3 x 10-3 molar pH = - log [1.3 x10-3 M] = [H+] pH = 2.9 H+ Concentrations …Strong vs. Weak Acid Moles of Acid used to make 1 L of solution H+ pH 0.010 mol H2SO4 0.0200 M 1.7 Strong acid 0.100 mol HC2H3O2 0.0013 M 2.9 Weak acid Note: although the sulfuric acid is 10x less concentrated than the acetic acid... …it produces > 10x more H+ pH = - log[H+] 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 1b) pH = 1.3 2a) 0.203 M 2b) pH = 0.7 3a) 2.1 x 10-3 M 3b) pH = 2.7 4) 2.7 x 10-3 M Weak Acids Cyanic acid is a weak monoprotic acid. If the initial concentration of cyanic + is 4.8 x 10-3 M, acid is 0.150 M and the equilibrium concentrationHof3O H3+O(aq) calculate Ka for cyanic acid. 4.8 x 10-3 M 0.150 M Ka = + CN1-(aq) H+(aq) HCN(aq) [Products] [Reactants] Ka = 4.8 x 10-3 M [H3O+] [CN1-] [HCN] [4.8 x 10-3 M][4.8 [CNx1-10 ] -3 M] Ka = [0.150 M] Ka = 1.54 x 10-4 How is [H3O+] determined? Measure pH of solution and work backwards