Chemical Equilibria: Solutions.
Dynamic chemical equilibria
Provided the temperature is held constant,
an equilibrium constant retains its value even though the individual activities
may change.
If one substance is added to a mixture at equilibrium, the other
substances adjust their abundances to restore the value of K .
Proton transfer equilibria
Proton (H+ ), hydronium ion (H3 O+ )
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Brønsted–Lowry Theory
Acid: proton donor
HA(aq) + H2 O(l)
aH O+ aA−
K = 3
aHA aH2 O
H3 O+ (aq) + A− (aq)
A− = conjugate base of the acid HA
Base: proton acceptor
B(aq) + H2 O(l)
a +a −
K = BH OH
aB aH2 O
BH+ (aq) + OH− (aq)
BH+ = conjugate acid of the base B
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Brønsted–Lowry Theory
Autoprotolysis equilibrium (or autoionization)
2H2 O(l)
H3 O+ (aq) + OH− (aq)
aH O+ aOH−
K = 3
aH2 O
pH = − log aH3 O+
In elementry work, aH3 O+ = [H3 O+ ]
aH3 O+ =
[H3 O+ ]
with c
c
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= 1 mol/dm3 and γ = 1
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Protonation and Deprotonation
Dilute solution regarding the water present as being a nearly pure liquid
aH2 O = 1
Acidity constant, Ka
Ka =
aH3 O+ aA−
([H3 O+ ]/c )([A− ]/c )
[H3 O+ ][A− ]
=
=
aHA
[HA]/c
[HA]
pKa = − log Ka
∆r G
RT
> 0, corresponding to a deprotonation equilibrium lying strongly in
pKa ∝ ∆r G
If ∆r G
⇐ ln K = −
favour of reactants (the original acid molecules), then pKa > 0 too.
Weak acids and strong acids
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Protonation and Deprotonation
Basicity constant, Kb
Kb =
aBH+ aOH−
[BH+ ][OH− ]
=
aB
[B]
pKb = − log Kb
Strong base and weak base
Autoprotolysis constant for water, Kw
Kw = aH3 O+ aOH−
pKw = − log Kw
At 25 ◦ C, Kw = 1.0 × 10−14 , pKw = 14.00
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Protonation and Deprotonation
Relation between the acidity constant of the conjugate acid BH+
and the basicity constant of a base B
H3 O+ +B) and (B+H2 O BH+ +OH− )
aH O+ aB
a +a −
Ka Kb = 3
× BH OH = aH3 O+ aOH− = Kw
aBH+
aB
(BH+ +H2 O
As the strength of a base decreases, the strength of its conjugate acid increases,
and vice versa.
log Ka Kb = log Ka + log Kb = log Kw
pKa + pKb = pKw
log aH3 O+ aOH− = log aH3 O+ + log aOH− = log Kw
pH + pOH = pKw
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Protonation and Deprotonation
Acidity and basicity constants at 298.15 K
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Protonation and Deprotonation
Fraction deprotonated, fdeprotonated
equilibrium molar concentration of conjugate base
initial concentration of acid
−
[A ]equilibrium
=
[HA]as prepared
fdeprotonated =
Fraction protonated, fprotonated
equilibrium molar concentration of conjugate acid
initial concentration of base
+
[BH ]equilibrium
=
[B]as prepared
fprotonated =
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Protonation and Deprotonation
Example 8.1 Extent of deprotonation of a weak acid (0.15 M CH3 COOH(aq))
CH3 COOH+H2 O→H3 O+ +CH3 CO−
2 , pH and fdeprotonated =?
Species:
Initial concentration (mol/dm3 )
3
Change to equilibrium (mol/dm )
3
Equilibrium concentration (mol/dm )
CH3 COOH
H3 O+
CH3 CO−
2
0.15
0
0
−x
+x
+x
0.15 − x
x
x
[H3 O+ ][CH3 CO−
x ×x
2 ]
Ka =
=
= 1.8 × 10−5
[CH3 COOH]
0.15 − x
p
x = 0.15 × 1.8 × 10−5 = 1.6 × 10−3
⇐ x 0.15
pH = − log(1.6 × 10−3 ) = 2.80
fdeprotonated =
[CH3 CO−
1.6 × 10−3
2 ]equilibrium
=
= 0.011
[CH3 COOH]as prepared
0.15
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Protonation and Deprotonation
Example 8.2 Estimating the pH of a dilute solution of a weak acid
1.5×10−4 M CH3 COOH(aq)
CH3 COOH
H 3 O+
CH3 CO−
2
−4
0
0
−x
+x
+x
1.5 × 10−4 − x
x
x
Species:
3
1.5 × 10
Initial concentration (mol/dm )
3
Change to equilibrium (mol/dm )
Equilibrium concentration (mol/dm3 )
Ka =
[H3 O+ ][CH3 CO−
x ×x
2 ]
=
= 1.8 × 10−5
[CH3 COOH]
1.5 × 10−4 − x
x 2 + (1.8 × 10−5 )x − (1.5 × 10−4 )(1.8 × 10−5 ) = 0
p
−(1.8 × 10−5 ) ± (−1.8 × 10−5 )2 + 4(1.5 × 10−4 )(1.8 × 10−5 )
x=
2
−5
−5
x = 4.4 × 10 or − 6.2 × 10
pH = − log(4.4 × 10−5 ) = 4.4
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Protonation and Deprotonation
Example 8.3 Extent of protonation of a weak base (0.010 M quinoline(aq))
Q+H2 O→QH+ +OH− , pKa = 4.88, pH and fprotonated =?
Species:
Q
OH−
QH+
Initial concentration (mol/dm3 )
0.010
0
0
Change to equilibrium (mol/dm3 )
−x
+x
+x
0.010 − x
x
x
Equilibrium concentration (mol/dm3 )
[OH− ][QH+ ]
x ×x
Kb =
=
= 10−pKb = 10−(14.00−pKa ) = 7.6 × 10−10
[Q]
0.010 − x
p
x = 0.010 × 7.6 × 10−10 = 2.8 × 10−6
⇐ x 0.010
pOH = − log(2.8 × 10−6 ) = 5.55
pH = 14.00 − 5.55 = 8.45
+
fprotonated =
[QH ]equilibrium
2.8 × 10−6
=
= 2.8 × 10−4
[Q]as prepared
0.010
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Polyprotic Acids
Molecular compound that can donate more than one proton
H2 SO4 donating up to two protons
H3 PO4 donating up to three protons
Best considered to be a molecular species that can give rise to a series of
Brønsted acids as it donates a succession of protons
Sulfuric acid: H2 SO4 itself, HSO−
4
2−
Phosphoric acid: H3 PO4 , H2 PO−
4 , HPO4
For a species H2 A
H2 A(aq) + H2 O(l)
H3 O+ (aq) + HA− (aq)
HA− (aq) + H2 O(l)
H3 O+ (aq) + A2− (aq)
aH3 O+ aHA−
aH2 A
aH3 O+ aA2−
Ka2 =
aHA−
Ka1 =
HA− : conjugate base of H2 A and acid
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Polyprotic Acids
Successive acidity constants of polyprotic acids at 298.15 K
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Polyprotic Acids
Example 8.5 Fractional composition of a solution of a diprotic acid
2−
Oxalic acid, H2 C2 O4 solution in equilibrium with HC2 O−
4 and C2 O4
H2 C2 O4 (aq) + H2 O(l)
Ka1 =
[H3 O+ ][HC2 O−
4 ]
[H2 C2 O4 ]
HC2 O−
4 (aq) + H2 O(l)
Ka2 =
H3 O+ (aq) + HC2 O−
4 (aq)
H3 O+ (aq) + C2 O2−
4 (aq)
[H3 O+ ][C2 O2−
4 ]
[HC2 O−
]
4
2−
Initial concentration O = [H2 C2 O4 ]as prepared = [H2 C2 O4 ] + [HC2 O−
4 ] + [C2 O4 ]
[HC2 O−
4 ] =
Ka1 [H2 C2 O4 ]
[H3 O+ ]
[C2 O2−
4 ] =
Ka2 [HC2 O−
Ka1 Ka2 [H2 C2 O4 ]
4 ]
=
[H3 O+ ]
[H3 O+ ]2
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Polyprotic Acids
Example 8.5 continued
Ka1 Ka2 [H2 C2 O4 ]
Ka1 [H2 C2 O4 ]
+
O = [H2 C2 O4 ] +
+
[H3 O ]
[H3 O+ ]2
Ka1
Ka1 Ka2
[H2 C2 O4 ]
= 1+
+
[H3 O+ ]
[H3 O+ ]2
n
o
1
=
[H3 O+ ]2 + Ka1 [H3 O+ ] + Ka1 Ka2 [H2 C2 O4 ]
+ 2
[H3 O ]
f (H2 C2 O4 ) =
[H2 C2 O4 ]
[H3 O+ ]2
=
+ 2
O
[H3 O ] + Ka1 [H3 O+ ] + Ka1 Ka2
f (HC2 O−
4 ) =
[HC2 O−
[H3 O+ ]Ka1
4 ]
=
+ 2
O
[H3 O ] + Ka1 [H3 O+ ] + Ka1 Ka2
f (C2 O2−
4 ) =
[C2 O2−
Ka1 Ka2
4 ]
=
O
[H3 O+ ]2 + Ka1 [H3 O+ ] + Ka1 Ka2
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Polyprotic Acids
Example 8.5 continued
H2 C2 O4 is dominant for pH<pKa1 .
H2 C2 O4 and HC2 O−
4 have the same concentration at pH=pKa1 .
2−
HC2 O−
4 is dominant for pH>pKa1 , until C2 O4 becomes dominant.
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Polyprotic Acids
Example 8.6 Fractional composition of a solution of a triprotic acid
2−
3−
Phosphoric acid, H3 PO4 solution in equilibrium with H2 PO−
4 , HPO4 , PO4
H3 PO4 (aq) + H2 O(l)
Ka1 =
[H3 O+ ][H2 PO−
4 ]
[H3 PO4 ]
H2 PO−
4 (aq) + H2 O(l)
Ka2 =
H3 O+ (aq) + HPO2−
4 (aq)
[H3 O+ ][HPO2−
4 ]
[H2 PO−
]
4
HPO2−
4 (aq) + H2 O(l)
Ka3 =
H3 O+ (aq) + H2 PO−
4 (aq)
H3 O+ (aq) + PO3−
4 (aq)
[H3 O+ ][PO3−
4 ]
2−
[HPO4 ]
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Polyprotic Acids
Example 8.6 continued
P = [H3 PO4 ]as prepared
2−
3−
= [H3 PO4 ] + [H2 PO−
4 ] + [HPO4 ] + [PO4 ]
H = [H3 O+ ]3 + Ka1 [H3 O+ ]2 + Ka1 Ka2 [H3 O+ ] + Ka1 Ka2 Ka3
[H3 O+ ]3
H
Ka1 [H3 O+ ]2
−
f (H2 PO4 ) =
H
Ka1 Ka2 [H3 O+ ]
2−
f (HPO4 ) =
H
Ka1 Ka2 Ka3
3−
f (PO4 ) =
H
f (H3 PO4 ) =
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Polyprotic Acids
For each conjugate acid–base pair with acidity constant Ka
The acid form is dominant for pH<pKa .
The conjugate pair have equal concentrations at pH=pKa .
The base form is dominant for pH>pKa .
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Amphiprotic Systems
Amphiprotic species:
molecule or ion that can both accept and donate protons.
2−
HCO−
3 as an acid (to form CO3 )
as a base (to form H2 CO3 )
Is the solution of NaHCO3 acidic on account of the acid character of HCO−
3 ,
or is it basic on account of the anion’s basic character?
pH of amphiprotic salt solution
pH =
1
(pKa1 + pKa2 )
2
provided the molar concentraton of the salt is high in the sense that
Kw
S
S
and
Ka1
c
Ka2
c
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Amphiprotic Systems
For the salt MHA with initial concentration S
Species:
H2 A
HA−
A2−
H 3 O+
0
S
0
0
+x
−(x + y)
+y
+(y − x)
x
S−x −y
y
y −x
3
Initial concentration (mol/dm )
3
Change to equilibrium (mol/dm )
Equilibrium concentration (mol/dm3 )
Ka1 =
[H3 O+ ][HA− ]
(y − x)(S − x − y )
=
[H2 A]
x
Ka2 =
[H3 O+ ][A2− ]
(y − x)y
=
−
S−x −y
[HA ]
(y − x)2 y
y
= [H3 O+ ]2 × = [H3 O+ ]2
x
x
xKa1 = (y − x)(S − x − y ) = Sy − y 2 − Sx + x 2 + xy
Ka1 Ka2 =
very small xKa1 , x 2 , y 2 , xy
0 ≈ Sy − Sx
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∴x ≈y
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Amphiprotic Systems
Brief illustration 8.3 The pH of a solution of an amphiprotic species
Sodium hydrogencarbonate, NaHCO3
pH =
1
1
(pKa1 + pKa2 ) = (6.37 + 10.25) = 8.31
2
2
Beacause
Kw
= 2 × 10−4 and Ka1 = 4.3 × 10−7 , reliable provided
Ka2
[NaHCO3 ]as prepared
2 × 10−4 ([NaHCO3 ]as prepared 0.2 mmol/dm3 ).
c
Potassium dihydrogenphosphate, KH2 PO4
pH =
1
1
(pKa2 + pKa3 ) = (7.21 + 12.67) = 9.94
2
2
Beacause
Kw
= 0.05 and Ka2 = 6.2 × 10−8 , reliable provided
Ka3
[KH2 PO4 ]as prepared
0.05 ([KH2 PO4 ]as prepared 0.05 mol/dm3 ).
c
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Salts in Water
Ions when a salt is added to water
⇒ acids or bases
⇒ affecting the pH of the solution
Ammonium chloride (NH4 Cl) to water
−
Providing both an weak acid (NH+
4 ) and a very weak base (Cl )
Acidic solution
Sodium acetate (NaCH3 CO2 ) to water
Neutral ion (Na+ ) and a base (CH3 CO−
2 )
Basic solution
To estimate the pH of the solution
Same way as for the addition of a ‘conventional’ acid or base
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Salts in Water
Brief Illustration 8.4 The pH of 0.010 M NH4 Cl(aq) at 25◦ C
+
NH+
4 +H2 O→H3 O +NH3
Species:
3
Initial concentration (mol/dm )
3
Change to equilibrium (mol/dm )
Equilibrium concentration (mol/dm3 )
NH+
4
H 3 O+
NH3
0.01
0
0
−x
+x
+x
0.01 − x
x
x
[H3 O+ ][NH3 ]
x ×x
=
= 5.6 × 10−10
+
0.01
−x
[NH4 ]
p
x = 0.01 × 5.6 × 10−10 = 2.37 × 10−6
⇐ x 0.15
Ka =
pH = − log(2.37 × 10−6 ) = 5.63
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Acid–Base Titrations
Stoichiometric point
Stage at which a stoichiometrically equivalent amount of of acid has been
added to a given amount of base
Equivalence point
Analyte, titrant, pH curve
Plot of the pH of the analyte against the volume of titrant
Titration of a strong acid with a strong base
Titration of a weak acid with a strong base
Titration of a weak base with a strong acid
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Acid–Base Titrations
Titration of a strong acid with a strong base
HCl(aq)+NaOH(aq)→ NaCl(aq)+H2 O(l)
The ions present at the
stoichiometric point (the Na+ ions
from the strong base and the Cl−
ions from the strong acid) barely
affect the pH, so the pH is that of
almost pure water, namely pH=7.
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Acid–Base Titrations
Titration of a weak acid with a strong base
CH3 COOH(aq)+NaOH(aq)→ NaCH3 CO2 (aq)+H2 O(l)
At the stoichiometric point
Na+ ions and CH3 CO−
2 ions
pH>7 due to the Brønsted base
CH3 CO−
2
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Acid–Base Titrations
Titrating 25 cm3 of 0.1 M CH3 COOH(aq) with 0.2 M NaOH(aq) at 25 ◦ C
pH= 2.9 at start of a titration
Addition of titrant ⇒ (some acid → conjugate base)
CH3 COOH(aq)+OH− (aq)→H2 O(l)+CH3 CO−
2 (aq)
Supposing enough titrant to produce a concentration [base] of the conjugate
base and simultaneously reduce the concentration of acid to [acid]
The acid and its conjugate base remain at equilibrium.
CH3 COOH(aq)+H2 O(l) H3 O+ (aq)+CH3 CO−
2 (aq)
aH3 O+ aCH3 CO−
aH3 O+ [base]
Ka [acid]
aH3 O+ ≈
aCH3 COOH
[acid]
[base]
[acid]
log aH3 O+ ≈ log Ka + log
[base]
[acid]
pH ≈ pKa − log
Henderson–Hasselbalch equation
[base]
Ka =
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2
≈
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Acid–Base Titrations
Example 8.7 Estimating the pH at an intermediate stage in a titration
0.10 M CH3 COOH(aq) 25.00 cm3 with 0.20 M NaOH(aq) 5.00 cm3
[acid]
[base]
[acid]
nacid /V
nacid
=
=
[base]
nbase /V
nbase
pH ≈ pKa − log
nOH− = (5.00 × 10−3 dm3 ) × (0.20 mol/dm3 ) = 1.00 × 10−3 mol
⇒ Converting 1.00 mmol CH3 COOH to the base CH3 CO−
2
nCH3 COOHinitial = (25.00 × 10−3 dm3 ) × (0.10 mol/dm3 ) = 2.50 × 10−3 mol
pH ≈ 4.75 − log
김진곤 (Pusan National University)
1.50 × 10−3 mol
= 4.6
1.00 × 10−3 mol
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Acid–Base Titrations
pH halfway to the stoichiometric point
When enough base has been added to neutralize half the acid,
the concentrations of acid and base are equal
pH ≈ pKa
In titrating 25 cm3 of 0.1 M CH3 COOH(aq) with 0.2 M NaOH(aq) at 25 ◦ C
pH ≈ 4.75
At the stochicometric point
Enough base has been added to convert all the acid to its base.
The solution consists primarily of CH3 CO−
2 ions.
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Acid–Base Titrations
Brief Illustration 8.5 The pH at the stoichiometric point
in titrating 25 cm3 of 0.1 M CH3 COOH(aq) with 0.2 M NaOH(aq)
−
CH3 CO2 + H2 O → OH
−
[CH3 CO2 ] =
nCH CO−
3
2
Vsolution
−
+ CH3 COOH
=
2.50 × 10−3 mol
37.5 × 10−3 dm3
3
Vsolution = Vanalyte + Vtitrant = 25 cm +
2.50 × 10−3 mol
0.2 mol/dm3
Species:
Initial concentration (mol/dm3 )
Change to equilibrium (mol/dm3 )
Equilibrium concentration (mol/dm3 )
Kb =
x =
[OH− ][CH3 COOH]
p
[CH3 CO−
2 ]
=
= 6.67 × 10
3
mol/dm
−3
= 37.5 × 10
CH3 CO−
2
6.67×10−2
−x
6.67×10−2 − x
OH−
0
+x
x
dm
3
CH3 COOH
0
+x
x
x ×x
−10
= 5.6 × 10
6.67 × 10−2 − x
6.67 × 10−2 × 5.6 × 10−10 = 6.11 × 10
pH = 14 − pOH = 14 + log(6.11 × 10
김진곤 (Pusan National University)
−2
−6
−6
) = 8.79
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Acid–Base Titrations
The pH curve for the titration of a weak base with a strong acid
The stoichiometric point occurs at pH<7.
The final pH of the solution approaches that of the titrant.
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Buffer Action
Ability of a solution to oppose
changes in pH when small amounts
of strong acids and bases are added
Slow variation of the pH when the
concentrations of the conjugate acid
and base are nearly equal, when
pH≈pKa
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Buffer Action
Acid buffer solution
One that stabilizes the solution at a pH below 7
Typically, a weak acid (such as CH3 COOH) + a salt that supplies its
conjugate base (such as CH3 COONa)
The abundant supply of A− ions (from the salt) can remove any H3 O+ ions
brought by additional acid.
The abundant supply of HA molecules can provide H3 O+ ions to react with
any base that is added.
Base buffer solution
One that stabilizes the solution at a pH above 7
Typically, a weak base (such as NH3 ) + a salt that supplies its conjugate
acid (such as NH4 Cl)
The weak base B can accept protons when an acid is added and its
conjugate acid BH+ can supply protons if a base is added.
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Buffer Action
Brief Illustration 8.6 The pH of a buffer solution
formed from equal amounts of KH2 PO4 (aq) and K2 HPO4 (aq)
H2 PO−
4 (aq) + H2 O(l)
H3 O+ (aq) + HPO2−
4 (aq)
⇒ needed pKa of the acid form H2 PO−
4
= pKa2 of phosphoric acid H3 PO4 = 7.21 from Table 8.2
Close to pH = 7
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Buffer Action
Example 8.8 Illustrating the effect of a buffer
When a drop (0.20 cm3 ) of 1.0 mol/dm3 HCl(aq) to 25 cm3 of pure water
⇒ 0.0080 mol/dm3 increase of hydronium ion concentration, pH 7.0 → 2.1
In an accetate buffer solution
that is 0.040 mol/dm3 NaCH3 CO2 (aq) and 0.080 mol/dm3 CH3 COOH(aq), ?
CH3 COOH(aq) + H2 O(l)
−
+
H3 O (aq) + CH3 CO2 (aq)
Initial pH of the buffer solution, pH = pKa − log
−
In buffer solution n(CH3 CO2 ) = (25 × 10
−3
n(CH3 COOH) = (25 × 10
+
n(H3 O ) = (0.20 × 10
−
n(CH3 CO2 )
−3
3
[acid]
0.080
= 4.75 − log
= 4.45
[base]
0.040
3
3
dm ) × (0.040 mol/dm ) = 1.0 mmol
−3
3
3
dm ) × (0.080 mol/dm ) = 2.0 mmol
3
dm ) × (1.0 mol/dm ) = 0.20 mmol from HCl(aq) drop
1.0 ⇒ 0.8 mmol
n(CH3 COOH)
2.0 ⇒ 2.2 mmol
2.2
Ignoring volume change pH = 4.75 − log 25 = 4.31
0.8
25
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Indicators
Rapid change of pH near the stoichiometric point of an acid–base titration
A water-soluble organic molecule
with acid (HIn) and conjugate base (In− ) forms that differ in colour.
Tow forms are in equilibrium in solution:
HIn(aq) + H2 O(l)
KIn =
H3 O+ (aq) + In− (aq)
aH3 O+ aIn−
aH O+ [In− ]
≈ 3
aHIn
[HIn]
[In− ]
KIn
≈
[HIn]
aH3 O+
log
[In− ]
≈ log KIn − log aH3 O+ = pH − pKIn
[HIn]
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Indicators
Indicator colour changes
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Indicators
For a strong acid–strong base titration, the stoichiometric point is
indicated accurately by an indicator that changes color at pH = 7
(such as bromothymol blue).
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Indicators
For a weak acid–strong base titration, an indicator with pKIn ≈ 7 (the
lower band, like bromothymol blue) would give a false indication of the
stoichiometric point.
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Indicators
Color change over a range of pH,
typically from pH ≈ pKIn − 1 to pH ≈ pKIn + 1
End point of the indicator
The pH halfway through a color change,
when pH ≈ pKIn and the two forms, HIn and In− , are in equal abundance
With a well-chosen indicator,
the end point coincides with the stoichiometric point of the titration.
Indicator with pKIn ≈ 7 for strong acid–strong base titrations
pKIn < 7 for strong acid–weak base titrations
pKIn > 7 for weak acid–strong base titrations
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Solubility Equilibria
Solid solute–saturated solution in heterogeneous equilibrium
Molar solubility of the solid, s
= molar concentration in saturated solution
Sparingly soluble
Ca(OH)2 (s)
Ks =
Ca2+ (aq) + 2OH− (aq)
2
aCa2+ aOH
−
2
= aCa2+ aOH
−
aCa(OH)2
Solubility constant
For an ionic compound of the form Ax By made up of Aa+ and Bb−
Ks = aAx a+ aBy b− = [Aa+ ]x [Bb− ]y = (xs)x (ys)y = x x y y sx+y
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Solubility Equilibria
Solubility constants at 298.15 K
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Solubility Equilibria
Brief Illustration 8.8 The solubility of a salt
Ca(OH)2 (s)
[Ca2+ ] = s
Ca2+ (aq) + 2OH− (aq)
[2OH− ] = 2s
Ks = s × (2s)2 = 4s3
13
1
Ks
5.5 × 10−6 3
s=
=
≈ 1 × 10−2
4
4
only approximate because ion–ion interactions have been ignored
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Common-Ion Effect
Effect on the solubility of a compound of the presence of another freely
soluble solute that provides an ion in common with the sparingly soluble
compound already present
Ex.) adding sodium chloride to a saturated solution of silver chloride
Common ion Cl−
Solubility of the original salt ↓
⇐ Le Chatelier’s principle
1
2
In pure water, s ≈ Ks
Adding Cl− ions (C mol/dm3 ), which greatly exceeds the concentration of
the same ion that stems from the presence of the silver chloride
0 [Ag+ ]
C
s
C
Ks = aAg+ aCl− ≈
=
c
c
c
c
2
K
×
c
s
s0 =
C
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