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Equilibrium

Equilibrium. Equilibrium is a macroscopic event, regardless of the direction of approach the forward rate and the reverse rate are the same and changes in concentration cannot be observed. Products favored 10 3 > K > 10 -3 Reactants favored. Heterogeneous Solutions

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Equilibrium

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  1. Equilibrium Equilibrium is a macroscopic event, regardless of the direction of approach the forward rate and the reverse rate are the same and changes in concentration cannot be observed. Products favored 103 > K > 10-3 Reactants favored Heterogeneous Solutions For mixtures the pure solids, pure liquids, and components that represent large amounts of solvents can be removed from the equilibrium expression and rolled into the equilibrium constant. Mathematical Relationships Reverse the reaction equation K’ = 1/K Scale the reaction equation K’’ = Kn Add equations Krxn = K1∙ K2 ∙ K3 ... Relationship to Kp = K(RT)Dn [C]c[D]d Kc = Predicting reaction direction from non-equilibrium conditions – Q The equilibrium expression for Q is the same as for K. Compare the values of Q and K (the ratios of products to reactants) Q=K the system is at equilibrium Q>K favors reactants Q<K favors products [A]a[B]b To solve for concentrations using K you will need a balanced reaction equation and an ICE table. Factors that affect Equilibrium – [ ], P&V, T For [ ] check Q or use the teeter totter, For gases – P&V, [ ] is to 1/V which is to P, ↑P, ↓V, or ↑[ ] will push to fewer moles of gas. For T check DH is q product or reactant

  2. Buffer Common Ion Effect Titrant Equivalence Point Primary Standard End Point Secondary Standard Titration Solubility Molar solubility Ksp - Solubility Product Constant Ka – Acid dissociation constant Kb – Base dissociation constant Kw – water dissociation constant Le Chatelier’s Principle Lewis Acid Lewis Base Bronsted-Lowry Acid Bronsted-Lowry Base Arrenhius Acid Arrhenius Base Henderson-Hasselbach equation Conjugate acid Conjugate Base pH = -log[H+] pOH = -log[OH-] pKa = -logKa etc. Kw = 1E-14 = [H+][OH-] = 1E-7∙1E-7 pKw = pH + POH = 14 = 7 + 7 Henderson-Hasselbach equation pH = pKa + log([salt]) – log([acid]) pOH = pKb + log([salt]) – log([base]) Determination of strength – or who wins the tug of war for the proton Strong acids Ka > 1 Weak acids Ka < 1 Strong bases Kb > 1 Weak bases Kb < 1

  3. Titration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. 4. Net ionic equation H+ + OH- --> H2O 5. At equivalence point moles H+ = moles OH-

  4. http://www.chem1.com/acad/webtext/abcon/abcon-2.html

  5. In General How to Solve Acid/Base Equilibria and Buffer Problems 1. Are the compounds involved strong or weak electrolytes a. If strong then there is a 100% dissociation and no way back use a single headed arrow and when adding the simultaneous equations together only and add the products. i.e. NaOH → Na+ + OH- b. if a weak electrolyte use a double headed arrow and when adding simultaneous equations together add both reactants and products i.e. CH3COOH ↔ CH3COO- + H+ 2. Determine if reactions are neutralization, dissociation, or equilibrium reactions I. Neutralization a. Class I – strong electrolyte/strong electrolyte: produces salt and water. Use ICE table with units of amount (mmol or moles), cancel terms, determine amounts of products. Stop. b. Class II – weak electrolyte/weak electrolyte: Compare Kas or Kbs to determine which predominates products or reactants. Use the equlibrium equation and ICE table with units of concentration (mmol/mL or mol/L) to determine equilibrium concentrations. Stop. c. Class III – strong electrolyte/weak electrolyte: produces salt of the conjugate and water and further actions, Dissociation or Equilibrium reactions must be determined continue to 2 II II. Dissociate any products and begin again at repeat step 1 then continue to 2 III. Products of the dissociation though by name appear to be the same compounds they are not identical in that they have a different source and are treated separately.

  6. III. Equilibrium – write new stoichiometrically balanced chemical equilibrium equations. • Write the equilibrium equation and solve for x and determine [H+] or [OH-] • Find pH or pOH • Determine the pOH from pH or pH from pOH • Determine the [H+] or [OH-] not found in step 3 Ksp Write a stoichiometrically balanced equation for the limited dissociation of the solid If provided with the Ksp and looking for molar solubility of products Write the stoichiometric ratio, i.e. 1:2 Multiply the ratio through by x, i.e. x:2x Write the equilibrium equation, i.e. Ksp = [M2+][Y-]2 Substitute in the stoichiometrically determined x values for the [ ]s, Ksp = (x)(2x)2 Solve for x If provided a mass determined to be in solution and looking for Ksp convert mass to moles convert moles to molarity Write the equilibrium equation, i.e. Ksp = [M2+][Y-]2 Solve for Ksp

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