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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 l.jpg

Zumdahl’s Chapter 15

Applications of

Aqueous Equilibria


Chapter contents l.jpg

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


Acid base titrations l.jpg
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.


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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 !


Buffer solutions l.jpg
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.


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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!


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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!



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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.


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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.


Ph in the buffer region l.jpg
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!


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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.


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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


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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.)


Dissolving oxides l.jpg
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.


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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– … ).


Complex equilibrium constant l.jpg
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.


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 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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