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Zumdahl’s Chapter 15

Zumdahl’s Chapter 15. Applications of Aqueous Equilibria. Acid-Base Equilibria Common Ion Effect Buffers Titration Curve Indicators. Solubility Solubility Product Common Ion Effect pH and Solubility Complex Equilibria Complexes and Solubilities. Chapter Contents.

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Zumdahl’s Chapter 15

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  1. Zumdahl’s Chapter 15 Applications of Aqueous Equilibria

  2. Acid-Base Equilibria Common Ion Effect Buffers Titration Curve Indicators Solubility Solubility Product Common Ion Effect pH and Solubility Complex Equilibria Complexes and Solubilities Chapter Contents

  3. Acid-Base Titrations • Le Châtlier: restoration of equilibrium replaces species lost. QK • E.g., H2O is a weaker electrolyte than virtually any other weak acid, so … • Titrating weak acid with strong base binds proton in water, removing product! •  such titrations are quantitative.

  4. Common Ion in Acid-Base • Le Châtlier: restoration of equilibrium consumes addends. QK • Addition of an ion already in equilibrium (Common Ion Effect) restores K by consuming the common ion. • NH3 + H2O  NH4+ + OH– Kb=1.810–5 • 0.1 M NH3  [OH–]  [1.810–50.1]½ = 410–3 • Make it 0.1 M NH4+ and [OH–]  1.810–5 !

  5. Buffer Solutions • Kb = [BH+][OH–]/[B] • If [B]=[BH+], then [OH–] = Kb • Ka = [H+][A–]/[HA] • If [HA]=[A–], then [H+] = Ka • Furthermore, in either case, excess H+or OH– finds abundance of its reactant! • Associated robust pH, a buffer hallmark.

  6. Buffer Calculation • 0.1 M ea. [NH4+] & [NH3]; pOH = 4.74 • 100 mL of this buffer contains 10 mmol of each of those species. • React fully 5 mmol OH–(in same 100 mL) • Kb = (0.1–0.05+x) (0+x) / (0.1+0.05–x) • x = [OH–]new (3 Kb)½ or pOHnew = 4.27 • 5% rule OK due to starting point of full reaction! • pOH = 0.47 trivial given even a 50% addend!

  7. Titration Curves (0.1 M acetic) • While [HA]/[A–] or [B]/[BH+] not near zero, buffering makes pH near pK •  pH changes slowly near ½ completion. • Near endpoint, those ratios vanish making [H+] very sensitive to titrant. •  pH changes very rapidly near endpoint!

  8. Strong/Strong Titration Curve

  9. Acid-Base Indicators • Indicators: molecules whose acid-base conjugates have distinct colors. • Color change occurs as acid/base ratio nears 1, i.e., as pHpKa (of indicator!) • Extreme sensitivity of pH to titrant volume near endpoint makes use of indicators quantitative. • Match pKindicator to pH at equivalence.

  10. pH at Equivalence • Sample is gone, replaced by conjugate at original number of moles. • [conjugate]0 = [sample]0 (V0 / Vtotal)  F • [conjugate]equilibrium = F – x (back rxn with water) • Kconjugate = x2 / (F – x) or x  [FKc]½ • pHequivalence = px or 14 – px = 8.72 (acetic) • pKindicator pHequivalenceis [Ind–]/[Ind]1.

  11. pH in the Buffer Region • Ka = [H+] [A–] / [HA] = [H+] [S]/[HA] • log Ka = log[H+] + log( [S]/[HA] ) • pKa = pH – log( [S]/[HA] ) • pH = pKa + log( [S]/[HA] ) neither S nor HA=0 • Henderson-Hasselbalch Equation • pOH = pKb + log( [BH +]/[B] ) • Concentration ratios = mole ratios!

  12. Solubility Product • AxBy(s)  x Ay+(aq) + y Bx–(aq) • Q = [Ay+]x [Bx–]yfor arbitrary concentrations • K =[Ay+]eqx [Bx–]eqyfor saturation conc. • Q < K implies no solid • Q = K implies saturated solution • Q > K super saturation difficult to achieve! Spontaneously precipitates.

  13. Calculating Solubility Product • Make a saturated solution. • Remove it from its precipitate. • Evaporate to dryness and weigh solid. • Convert to moles n of solid in original V. • If AxBy then [Ay+]=x(n/V) ; [Bx–]=y(n/V) • Ksp = (xn/V)x (yn/v)y • x and y have enormous influence

  14. Solubility and pH • If dissolved ions are conjugates of weak acid, say, both Ksp and Kb must be satisfied. • Ksp fixes [A–] at equilibrium value, and Kb establishes [OH–] and [HA], for example. • If Ka–1 and [H+] can lower [A–] below the solubility limit, acid can dissolve the solid. (Assuming solid is limiting reactant.)

  15. Dissolving Oxides • Ag2O + H2O  2 Ag+ + 2 OH– (410–16) • 2 H+ + 2 OH–  2 H2O (10+14)2 • Ag2O + 2H+ 2Ag+ + H2O (410+12) • Equilibrium lies far to right for modest acid. • Cu2O + H2O  2 Cu+ + 2 OH– (410–30) •  Cu2O + 2H+ 2Cu+ + H2O (410–2) • Only concentrated acids will suffice.

  16. Complex Equilibria • Empty or unfilled metal d-orbitals are targets for lone pair electrons in dative or coordinate-covalent bonding. • Square planar or octahedral (and beyond) geometries of ligands (e– pair donors) bind to metal atoms to make complexes. • Ligands can be neutral (H2O, NH3, CO … ) or charged (Cl–, CN–, S2O32– … ).

  17. Complex Equilibrium Constant • Exchange of ligands (labile) is governed by equilibrium constants. • Solid solubilities are thus influenced by ligand availability. • H2O always available (aq), but it’s not the strongest ligand. Serial replacement of H2O by other ligands leads to a sequence of equilibrium constants.

  18.  vs. K • Polyprotic acid constants proceed proton by proton: • HSO4–(aq)  H+(aq) + SO42–(aq) Ka2=10–2 • Ligand addition constants, , are cumulative instead: • Ag+(aq) + 2 I–(aq)  AgI2–(aq) 2=1011 • really Ag(H2O)4+ + 2 I– Ag(H2O)2I2– + 2 H2O

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