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More about equilibria

More about equilibria. Predicting the Position of Acid-Base Equilibria. Acidity, Basicity and the Periodic Table Lewis Acids and Bases Equilibria Between Complexes The Solubility Product Solubility Calculations. [H 3 O + ] [A - ] [HA]. [OH - ] [BH + ] [B].

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More about equilibria

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  1. More about equilibria • Predicting the Position of Acid-Base Equilibria. • Acidity, Basicity and the Periodic Table • Lewis Acids and Bases • Equilibria Between Complexes • The Solubility Product • Solubility Calculations

  2. [H3O+] [A-] [HA] [OH-] [BH+] [B] Dissociation of acids and bases • Weak acids • HA + H2O H3O+ + A- • Ka = • Weak bases • B + H2O BH+ + OH- • Kb = Water is not included in the expressions because its concentration is constant.

  3. Weak acid example • Example • Determine the [H+] of a 0.1M acetic acid solution. • HAc + H2O H3O+ + Ac- • KA = 2.24 x 10-5 at 25oC. • HAc = acetic acid • Ac- = acetate ion

  4. Weak acid example • [ H+ ] [ Ac- ] • [ HAc ] • Since both a H+ and a Ac- is produced for each HAc that dissociates: • [ H+ ] = [ Ac- ] • [ HAx ] = 0.1 M - [ H+ ] • Lets set X = [ H+ ] KA = = 2.24 x 10-5

  5. Weak acid example • KA = 2.24 x 10-5 = X2 / (0.1 - X) • Rearranging give us: • X2 + 2.24 x 10-5 X - 2.24 x 10-6 = 0 • We can solve this quadratic or possibly assume that the amount of acid that dissociates is insignificant (compared to the undissociated form)

  6. Weak acid example • When to round off • If you are willing to accept an error of one percent or less, then you can round off when: • KA / CTOTAL< 10-3 • Realistically, with the common use of calculators, its just as easy to do the full quadratic solution. • Lets give it a try anyway.

  7. Weak acid example • KA = 2.24 x 10-5 • [ H+ ] = [ Ac- ] • [ HAc ] = 0.1 M - [ H+ ] • KA / 0.1 < 10-3 - can assume that [HAc] = 0.1 M • 2.24 x 10-5 = X2 / 0.1M • X = (2.24x10-6)1/2 • = 0.00150

  8. -b + b2 - 4ac 2a Weak acid example • Now, lets go for the exact solution • Earlier, we found that • X2 + 2.24 x 10-5 X - 2.24 x 10-6 = 0 • X =

  9. Weak acid example • X = 0.00149 • In this case, there is no significant difference between our two answers. -2.24 x 10-5 + [(2.24 x 10-5)2 +4 x 2.24 x 10-6]1/2 2 X =

  10. Acid-base equilibria • Earlier we saw that an conjugate acid-base pair can be related by: • KaKb = Kw = 1.0 x 1014 • or • pKa + pKb = pKw = 14 • We can predict the position of an acid-base equilibrium by combining the Ka and Kb expressions.

  11. Acid-base equilibria • Acetic acid example • HAc(aq) + H2O(l) H3O+(aq) + Ac-(aq)Ka • + Ac-(aq) + H2O(l) HAc(aq) + OH-(aq)Kb • H2O(l) + H2O(l) H3O+(aq) + OH-(aq)Kw • In general, equilibrium constants for the sum of equations is the product of the constants for the original equations.

  12. Acid-base equilibria • Acid strength • In general, for Brønsted-Lowry acid-base reactions, stronger acids and bases react to form weaker ones. • The Ka indicates the relative strength of an acid. As the acid becomes stronger, it increases.

  13. Acid Ka Base Chloroacetic 1.4 x 10-3 Chloroacetate Nitrous 5.6 x 10-4 Nitrite Hydrofluoric 6.3 x 10-4 Fluoride Formic 1.8 x 10-4 Formate Benzoic 6.5 x 10-5 Benzoate Acetic 1.7 x 10-5 Acetate Acid-base equilibria • Ka and Kb are closely related. • As one increases, the other must decrease. That enables us to compare the relative strengths of conjugate acid-base pairs. Acid Strength Base Strength

  14. Acidity, basicity andthe periodic table • Strengths of binary acids. • Down a group, acid strength increases. • H2O < H2S < H2Se < H2Te • HF < HCl < HBr < HI • Binary acids. • Acids that consist of just two elements.

  15. Acidity, basicity andthe periodic table • In the case of the group VII (17) binary acids, we cannot observe a difference in acid strength between HCl, HBr and HI in aqueous solution. • All three acids are stronger than H3O+. • Any acid that is a stronger acid than H3O+ will instantly and completely react to form H3O+ in aqueous solution. • Water is said to level the acidity of these three acids. • One can use other solvents in order to reveal differences in acidity.

  16. pH HCl HBr HI % titration Binary acids in acetone We can use a solvent other than water to see acidity differences. A solvent that allows us to do this is called a differentiating solvent.

  17. Formula Ka pKa CH4 ~10-49 ~49 NH3 ~10-35 ~35 H2O 2 x 10-16 15.7 HF 6.3 x 10-4 3.20 Binary acid strengths ofelements in the same period. • Across a period, acid strength increases.

  18. H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Lr Strength of binary acids

  19. Metal hydroxides • The hydroxides of metals are bases. The basicity of aqueous solutions of metal hydroxides is dependent on solubility and the value of Kb • Base Solubility Basicity • LiOH High High • NaOH High High • KOH High High • Mg(OH)2 Low Low • Ca(OH)2 Low Low

  20. Nonmetal hydroxides • Hydroxides of nonmetals are molecular compounds. The electronegativities of the nonmetal and oxygen results in a very polar H - O bond. • Example. Hypochlorus acid. • H O Cl • When hypochlorous acid ionizes, the H-O bond breaks and H3O+ is formed. HOCl(aq) + H2O(l) H3O+(aq) + OCl-(aq)

  21. Oxo acids • Many important inorganic and organic acids are nonmetal hydroxides. Nitric acid Sulfuric acid Acetic acid Phosphoric acid

  22. Oxo acids • In general, the larger the electronegativity difference between the oxygen and the other (central) atom, the stronger the acid. • In addition, the acidities of oxo acids is also related to the number of additional oxygen attached to the central species. • Acid strength increases as the number of oxygen increases.

  23. Oxo acids Perchloric acid Three extra oxygen Ka = very large Chloric acid Two extra oxygen Ka = large Chlorous acid One extra oxygen Ka = 1.1 x 10-2 Hypochlorous acid No extra oxygen Ka = 4.0 x 10-8

  24. Acidic and basic oxides • Metal oxides • Most are basic. O2- is the conjugate base of the very weak acid OH-. • Na2O(s) + H2O(l) 2NaOH(aq) • MgO(s) + 2HCl(aq) MgCl2 (aq) + H2O(l) • Nonmetal oxides • Most nonmetal oxides are acidic. Acid strength increases with the number of oxygen. • CO2 (g) + H2O (l) H2CO3 (aq) • SO2 (g) + H2O (g) H2SO3 (aq) • SO3 (g) + H2O (g) H2SO4 (aq)

  25. Lewis acids and bases • An even more general acid-base definition. All Brønsted-Lowry and Arrhenius acids and bases are included -- and then some. • Acid • Any species that can attach an electron pair to form a covalent bond. • Base • Any species that has an electron pair that can form a covalent bond.

  26. : Lewis acids and bases • A Lewis acid must have an empty valence orbital for an electron pair to go into. • A Lewis base must have an unshared pair of valence electrons. Coordinate covalent bond A shared electron pair of which both electrons originally belonged to the same atom.

  27. Complexes • Highly charged, small cations attract electrons strongly. • Example. Hydrated cations. • This attraction is so strong that the cation shares the unpaired electrons of oxygen. • The overall species is called a complex ion. Al(H2O)63+

  28. L M L L L Complexes • General structure m+ L - ligand M - central species X - counter ion Xn+

  29. Complexes • Central species • Must have the ability to accept one or more pairs of electrons in available d orbitals - typically a metal ion. • Ligand • Anion, cation or neutral species with ability to donate electron pair to form a coordinate covalent bond. • Counter ion • Required if complex is charged.

  30. [ M(L)xn+ ] [ Mn+ ] [ L ]m Complex formation • In complex formation, two or more species will join, forming a single, new species. • Mn+ + mL M(L)mn+ • Kf =

  31. [ Mn+ ] [ L ]m [ M(L)mn+ ] Complex formation • If we are evaluating the dissociation of a complex, we can use a Kd. • M(L)mn+ Mn+ + mL • Kd = • Kd = 1 / Kf

  32. Solubility Products, KSP • KSP expressions are used for ionic materials that are only slightly soluble in water. • Their only means of dissolving is by dissociation. • AgCl(s) Ag+ (aq) + Cl-(aq) • KSP = [ Ag+] [ Cl-]

  33. Solubility Products, KSP • At equilibrium, our system is a saturated solution of silver and chloride ions. • The only way to know that it is saturated it to observe some AgCl at the bottom of the solution. • As such, [AgCl] is a constant and KSP expressions do not include the solid form in the equilibrium expression.

  34. Solubility Products, KSP • Determine the solubility of AgCl in water at 20 oC in grams / 100 ml. • KSP = [Ag+] [Cl-] = 1.0 x 10-10 • At equilibrium, [Ag+] = [Cl-] so • 1.0 x 10-10 = [Ag+]2 • [Ag+] = 1.0 x 10-5 M • g AgCl = 1.0 x 10-5 mol/l * 0.10 l * 143.32 g/mol • Solubility = 1.4 x 10-4 g / 100 ml

  35. Solubility Products, KSP • Another example • Calculate the silver ion concentration when excess silver chromate is added to a 0.010 M sodium chromate solution. • KSP Ag2CrO4 = 1.1 x 10-12 • Ag2CrO4 (s) 2Ag+ + CrO42-

  36. Solubility Products, KSP • KSP = 1.1 x 10-12 = [ Ag+ ]2 [ CrO42- ] • [CrO42-] = [CrO42-]Ag2CrO4 + [CrO42-]Na2CrO4 • With such a small value for KSP, we can assume that the [CrO42-] from Ag2CrO4 is negligible. • If we’re wrong, our silver concentration will be significant (>1% of the chromate concentration.) • Then you’d use the quadratic approach.

  37. Solubility Products, KSP • KSP = 1.1 x 10-12 = [ Ag+ ]2 [ CrO42- ] • [CrO42-] = 0.010 M • [ Ag+ ] = ( KSP / [ CrO42- ] ) 1/2 • = 1.1 x 10-12 / 0.010 M • = 1.1 x 10-5 M • [ Ag+ ] << [ CrO42- ] • so our assumption was valid.

  38. Factors influencing solubility • Common ion and salt effects. • As with other equilibria we’ve discussed, adding a ‘common’ ion will result in a shift of a solubility equilibrium. • AgCl (s) Ag+(aq) + Cl- (aq) • KSP = [Ag+] [Cl-] • Adding either Ag+ or Cl- to our equilibrium system will result in driving it to the left.

  39. Factors influencing solubility • Complex ion formation. • The solubility of slightly soluble salts can be increased by complex ion formation. • Example. Addition of excess Cl- • AgCl(s) Ag+(aq) + Cl-(aq) • + • 2Cl-(aq) • AgCl2-(aq) A large excess of chloride results in the formation of the complex. More AgCl will dissolve as a result.

  40. Factors influencing solubility • Hydrolysis. • If the anion of a weak acid, or cation of a weak base is part of a KSP, solubilities are greater than expected. • AgCN(s) Ag+(aq) + CN-(aq) • + • H2O(l) • HCN(aq) + OH-(aq) This competing equilibrium causes the CN- to be lower than expected. More AgCN will dissolve as a result.

  41. Qual Schemes • A series of sequential precipitations used to identify ions in solution. • A number of these qualitative procedures exist for both cations and anions. • Qualitative analysis • Determining if a substance is present. • Quantitative analysis • Determining how much of a substance is present.

  42. A qual scheme Sample solution Add HCl Precipitate Group 1 AgCl, Hg2Cl2, PbCl2 Add H2S Precipitate Group 2 Bi2S3, CdS, CuS HgS, PbS, SnS2 Add NH3 + H2S Precipitate Group 3 CoS, FeS, MnS, NiS, ZnS, Al(OH)3, Cr(OH)3 Add Na2CO3 Precipitate Group 4 BaCO3, CaCO3, MgCO3 Solution Group 5 K+, Na+

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