1. pK a

2. Strengths of Conjugate Acid-Base Pairs strong medium weak very weak Acid strength increases HCl H 2 SO 4 HNO 3 H 3 O + HSO 4 - H 3 PO 4 HC 2 H 3 O 2 H 2 CO 3 H 2 S H 2 PO 4 - NH 4 + HCO 3 - HPO 4 2- H 2 O negligible very weak weak medium strong Base strength increases Cl - HSO 4 - NO 3 H 2 O SO 4 2- H 2 PO 4 - C 2 H 3 O 2 - HCO 3 - HS - HPO 4 2- NH 3 CO 3 2- PO 4 3- OH -

3. Conjugate Acid Strength Very strong Strong Weak Very weak Relative acid strength Relative acid strength Relative conjugate base strength Relative conjugate base strength Very weak Very strong Weak Strong HA H + + A - pK a = [H + ] [A - ] [HA] Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 508

4. Solutions of Acids and Bases: The Leveling Effect No acid stronger than H 3 O + 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 (H 3 O + ) is leveled to the strength of H 3 O + in aqueous solution because H 3 O + 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

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

6. HC 2 H 3 O 2(aq) H + (aq) + C 2 H 3 O 2 1- (aq) Equilibrium constant K eq = 1.8 x 10 -5 = = K a = Acid dissociation constant K a = 1.8 x 10 -5 @ 25 o C for acetic acid K a =

7. HC 2 H 3 O 2 H + + C 2 H 3 O 2 1- HCl H + + Cl 1- very large HNO 3 H + + NO 3 1- very large H 2 SO 4 H + + HSO 4 1- large 1.8 x 10 -5 H 2 S H + + HS 1- 9.5 x 10 -8 Ionization Constants for Acids KaKa

8. Ionization of Acids Acid Ionization Equation Ionization Constant, pK a Hydrochloric HCl H 1+ + Cl 1- very large Sulfuric H 2 SO 4 H 1+ + HSO 4 1- large AceticHC 2 H 3 O 2 H 1+ + C 2 H 3 O 2 1- 1.8 x 10 -5

9. Formula NameValue of K a * Values of K a for Some Common Monoprotic Acids HSO 4 - hydrogen sulfate ion1.2 x 10 -2 HClO 2 chlorous acid1.2 x 10 -2 HC 2 H 2 ClO 2 monochloracetic acid1.35 x 10 -3 HF hydrofluoric acid7.2 x 10 -4 HNO 2 nitrous acid4.0 x 10 -4 HC 2 H 3 O 2 acetic acid1.8 x 10 -5 HOCl hypochlorous acid3.5 x 10 -8 HCN hydrocyanic acid6.2 x 10 -10 NH 4 + ammonium ion5.6 x 10 -10 HOC 6 H 5 phenol1.6 x 10 -10 * The units of K a are mol/L but are customarily omitted. Increasing acid strength

10. H 2 SO 4 2 H + + SO 4 2- in dilute solutions...occurs ~100% H 2 SO 4 H + + HSO 4 1- & HSO 4 1- H + + SO 4 2- One gram of concentrated sulfuric acid (H 2 SO 4 ) is diluted to a 1.0 dm 3 volume with water. What is the molar concentration of the hydrogen ion in this solution? What is the pH? x mol H 2 SO 4 = 1 g H 2 SO 4 Solution) First determine the number of moles of H 2 SO 4 Sample 1) = 0.010 mol H 2 SO 4 OVERALL: pH = - log [H + ] pH = 1.69 0.010 M0.020 M substitute into equation pH = - log [0.020 M] 98 g H 2 SO 4 1 mol H 2 SO 4

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

12. Step 3) Find the [H + ] 1.8 x 10 -5 = K a = 1.8 x 10 -5 @ 25 o C for acetic acid K a = Substitute into equation: x 2 = 1.8 x 10 -6 M x = 1.3 x 10 -3 molar = [H + ] HC 2 H 3 O 2 H + + C 2 H 3 O 2 1- 0.1 M pH = - log[H + ] pH = - log [1.3 x10 -3 M] pH = 2.9 ? 0.1 M weak acid How do the concentrations of H + and C 2 H 3 O 2 1- compare?

13. Moles of Acid used to make 1 L of solution H+H+ pH 0.010 mol H 2 SO 4 Strong acid 0.100 mol HC 2 H 3 O 2 Weak acid Note: although the sulfuric acid is 10x less concentrated than the acetic acid... …it produces > 10x more H + H + Concentrations …Strong vs. Weak Acid pH = - log[H + ] 1.7 2.9 0.0200 M 0.0013 M

14. 1a) What is the molar hydrogen ion concentration in a 2.00 dm 3 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 HNO 3 in 250 cm 3 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 HC 2 H 3 O 2 in 500 cm 3 of solution. What is the molar concentration of hydrogen ions? 1a) 0.0500 M2a) 0.203 M3a) 2.1 x 10 -3 M 4) 2.7 x 10 -3 M 1b) pH = 1.32b) pH = 0.73b) pH = 2.7 Practice Problems:

15. 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 H 3 O + is 4.8 x 10 -3 M, calculate K a for cyanic acid. HCN(aq) H + (aq) + CN 1- (aq) H 3 O + (aq) 0.150 M 4.8 x 10 -3 M K a = [Products] [Reactants] K a = [H 3 O + ] [HCN] [CN 1- ] K a = [4.8 x 10 -3 M] [0.150 M] [CN 1- ][4.8 x 10 -3 M] K a = 1.54 x 10 -4 How is [H 3 O + ] determined? 4.8 x 10 -3 M Measure pH of solution and work backwards

16. Weak Acid, pK a KeysKeys http://www.unit5.org/chemistry/AcidBase.html