Chemistry 12 Unit 4: Acids, Bases & Salts 4.10 – The Relative Strengths of Acids and Bases

Remember: There are 2 conjugate pairs in solution Ex: H2CO3 + SO32-   HCO3- + HSO3-  SO32- can ONLY act as a base (has no protons to give) - We will never have 2 proton transfers in Chem 12!

Let’s look at a “Proton Competition”! Ex: CO32- + H2PO4-   HCO3- + HPO42- When CO32- and H2PO4- are mixed,  The products H2PO4- and HCO3- are both acids at equilibrium - Each can donate a proton  The reactants, therefore, can both act as bases - Each can accept a proton *** We need to identify the STRONGER Acid that will be more successful at donating a proton. - Use the Table of Relative Strengths again

H2PO4- Ka = 6. 2x10-8 (stronger acid) HCO3- Ka = 5 H2PO4- Ka = 6.2x10-8 (stronger acid) HCO3- Ka = 5.6x10-11 (weaker acid) H2PO4- has a greater tendency to donate a proton than HCO3-  Therefore, products are favored (there will be more products than reactants) Summary: In a Bronsted-Lowry acid-base equilibrium, the side of the equilibrium which has the WEAKER ACID will be “favored”

H2PO4-  HPO42- + H+ (stronger) CO32- + H+  HCO3- CO32- + H2PO4-  HCO3- + HPO42- Keq = [HCO3- ] [HPO42-]  [HPO42-] [H+] x [HCO3-] [CO32-] [H2PO4-] [H2PO4-] [CO32-][H+] Keq = Ka(H2PO4-) x 1____  Keq = Ka(H2PO4-) Ka(HCO3-) Ka(HCO3-)

Keq = [products] > 1 More products than reactants [reactants] Keq = Ka(H2PO4-) = 6.2x10-8 = 1.1x103 Ka(HCO3-) 5.6x10-11 Keq = [products] > 1 More products than reactants [reactants] So, the general expression for calculating the equilibrium value is: HReact + Prod-   React- + HProd Keq = Ka (reactant acid) Ka (product acid) Ka (reactant acid) = Ka value of acid on reactant side (HReact) Ka (product acid) = Ka value of acid on product side (HProd)

Note: Keq = the ratio of [products] to [reactants] Note: Keq = the ratio of [products] to [reactants]. This value can also be calculated using the ratio of the Ka value for the reactant acid to the Ka value for the product side [products] = Keq = Ka (reactant acid) [reactants] Ka (product acid)

Example: When HS- and HCO3- are mixed, does the resulting equilibrium favor the reactants or products? Text ex p.132

Let’s Try! Write the Bronsted-Lowry acid-base equilibria which occur when the following pairs of substances are mixed in solution. Identify the conjugate pairs formed. HNO2 and NH3 b) CO32- and HF c) HS- and H3PO4 d) HCO3- and S2- 38

2) In the following equilibria, predict whether reactants or products are favored. H2S + NH3   HS- + NH4+ H2PO4- + HS-   HPO42- + H2S NH4+ + OH-   NH3 + H2O 39

HSO4- and NO2- b) H3PO4 and HPO42- 3) Write the major equilibrium reactions which occur when the following substances are put into water. Ignore reactions between ions and water. All salts are 100% dissociated in water. Do the resulting equilibria favor reactants or products? HSO4- and NO2- b) H3PO4 and HPO42- c) HCO3- and HSO3- d) NH4F e) HSO3- and HC2O4- f) H2O2 and HS- 40

4) Keq = 14 for the equilibrium: H2Te + HSe-   HTe- + H2Se Which acid is stronger: H2Te or H2Se ? Which base is stronger? Based on your answers, complete the equation using STRONGER ACID, weaker acid, STRONGER BASE, weaker base _________ + ________   ________ + _________ Homework: 38efgh, 39de, 40ghij, 42, 43, 44,45,46