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Chemistry 12 Unit 4: Acids, Bases & Salts
4.10 – The Relative Strengths of Acids and Bases
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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!
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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
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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”
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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-)
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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)
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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)
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Example: When HS- and HCO3- are mixed, does the resulting equilibrium favor the reactants or products? Text ex p.132
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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
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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
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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
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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
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