Transcript Zumdahl’s Chapter 15 - University of Texas at Dallas

Zumdahl’s Chapter 15
Applications of
Aqueous Equilibria
Chapter Contents

Acid-Base Equilibria
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Common Ion Effect
Buffers
Titration Curve
Indicators
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Solubility
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Solubility Product
Common Ion Effect
pH and Solubility
Complex Equilibria
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Complexes and
Solubilities
Acid-Base Titrations

Le Châtlier: restoration of equilibrium
replaces species lost. QK
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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.
Common Ion in Acid-Base

Le Châtlier: restoration of equilibrium
consumes addends. QK
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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
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Kb = [BH+][OH–]/[B]
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Ka = [H+][A–]/[HA]
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If [B]=[BH+], then [OH–] = Kb
If [HA]=[A–], then [H+] = Ka
Furthermore, in either case, excess H+
or OH– finds abundance of its reactant!
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Associated robust pH, a buffer hallmark.
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)
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Kb = (0.1–0.05+x) (0+x) / (0.1+0.05–x)
x = [OH–]new  (3 Kb)½ or pOHnew = 4.27
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5% rule OK due to starting point of full reaction!
pOH = 0.47 trivial given even a 50% addend!
Titration Curves
While [HA]/[A–] or [B]/[BH+]
not near zero, buffering makes
pH near pK
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 pH changes slowly near ½
completion.
Near endpoint, those ratios
vanish making [H+] very
sensitive to titrant.
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 pH changes very rapidly near
endpoint!
Titration: Weak Acid by Strong Base
14
12
10
8
pH
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(0.1 M acetic)
6
4
2
0
0
0.05
0.1
0.15
Volume of Base
0.2
0.25
Strong/Strong Titration Curve
V base
0
50
90
95
99
99.9
100
V total
100
150
190
195
199
199.9
200
[H+]
1M
.5/1.5
.1/1.9
.05/1.95
.01/1.99
.001/1.999
0/2
pH
0
0.48
1.28
1.59
2.30
3.30
7.00
Acid-Base Indicators
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Indicators: molecules whose acid-base
conjugates have distinct colors.
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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.
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Match pKindicator to pH at equivalence.
pH at Equivalence
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Sample is gone, replaced by conjugate
at original number of moles.
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[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  pHequivalence is [Ind–]/[Ind]1.
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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
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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–]y for arbitrary concentrations

K = [Ay+]eqx [Bx–]eqy for 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
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x and y have enormous influence
Solubility and pH
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If dissolved ions are conjugates of weak
acid, say, both Ksp and Kb must be
satisfied.
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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
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)
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Equilibrium lies far to right for modest acid.
Cu2O + H2O  2 Cu+ + 2 OH– (410–30)
  Cu2O + 2H+  2Cu+ + H2O (410–2)
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Only concentrated acids will suffice.
Complex Equilibria
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Empty or unfilled metal d-orbitals are
targets for lone pair electrons in dative
or coordinate-covalent bonding.
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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
Exchange of ligands (labile) is governed
by equilibrium constants.
 Solid solubilities are thus influenced by
ligand availability.
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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.
 vs. K
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Polyprotic acid constants proceed
proton by proton:
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HSO4–(aq)  H+(aq) + SO42–(aq) Ka2=10–2
Ligand addition constants, , are
cumulative instead:
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Ag+(aq) + 2 I–(aq)  AgI2–(aq)
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2=1011
really Ag(H2O)4+ + 2 I–  Ag(H2O)2I2– + 2 H2O