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Electrochemistry
November 25, 2015
Subject of electrochemistry:
dissociation (solutions of electrolytes, melts of salts)
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pH of strong acids and bases
Example. Calculate pH of aqueous HCl of concentration 0.01 mol dm−3.
100 % dissociation to the 1st degree:
cH+ = cHCl ⇒ pH = − log aH+ ≈ − log cH+ = 2
conductivity
Example. Calculate pH of aqueous H2SO4 of concentration 0.001 mol dm−3.
It dissociates completely to the 1st degree, partially to the 2nd degree, but
since here c K2 = 1.3×10−2, we can assume also 100 % dissociation.
phenomena at interfaces s/l (electrolysis, galvanic cells)
cH+ = 2cH2SO4 ⇒ pH = − log aH+ ≈ − log(2cH+ ) = 2.7
more accurate 2.75 (partial dissociation + Debye–Hückel)
Example. Calculate pH of aqueous NaOH of concentration 0.01 mol dm−3
at 25 ◦C.
cOH− = 0.01 mol dm−3, cH+ = 10−14/0.01 = 10−12 mol dm−3, pH = 12
Example. Calculate pH of aqueous Ca(OH)2 of concentration 0.001 mol
dm−3 at 25 ◦C.
cOH− = 0.002 mol dm−3, pOH = 2.7, pH = 14 − 2.7 = 11.3
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Solutions of electrolytes
strong electrolyte: (almost) completely ionizes (dissociates)
no uncharged molecules in the solution
H2SO4, KOH, Ca(OH)2, NaCl, BaSO4, . . .
not necessarilly to max. degree:
H2SO4
100 %
→
HA → H+ + A−
Determine pH of a weak acid from known concentration.
(analytic) concentration: c0
acidity constant: Ka
partially
H+ + HSO4− → 2 H+ + SO42−
weak electrolyte: contains neutral (not dissociated) molecules
organic acids and bases, NH3, H2O, . . .
Assumptions:
cOH− cH+
standard states: solvent (water): •; in dilute awater = 1
ions: [c] (ai = γici/cst)
γi = 1 (approximation of infinite dilution)
dissociation constant = equilibrium constant of the dissociation reaction
CH3 COOH
NH3 + H2O
(COOH)2
pH
pH = − log aH+
γH+ =1
=
→
→
→
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Dissociation of a weak acid I
CH3 COO− + H+
NH4+ + OH−
HOOC-COO− + H+
compound
0
eq.
HA
c0
c0 − x
A−
0
x
H+
0
x
Balance:
Correctly
c +
cH+
− log Hst = − log
≡ − log[H+]
c
mol dm−3
x2
(c0 −x)cst
Equation:
[H+][A−]
x2
=
= Ka
[HA]
c0 − x
= Ka. We will neglect cst (concentrations should be in-
serted in mol dm−3).
[H+] ≈ [H3O+]
log = log10, p = −log
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Dissociation of water
Dissociation of water:
Solution:
Ionic product (autoionization constant) of water Kw :
a +a −
c +c −
.
≡ [H+][OH−] = 1.00×10−14 (25 ◦C)
Kw = H OH ≈ H stOH
aH2O
(c )2
cH+ = x =
s
Summary of approximations:
Equivalently (at 25 ◦C):
pH + pOH = pKw = 14
x2
[H+][A−]
=
= Ka
[HA]
c0 − x
Equation:
H2O → H+ + OH−
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Dissociation of a weak acid II
Ka 2
Ka c0Ka p
+ Kac0 −
≈
K a c0
2
2
the acid is stronger than water (Ka Kw )
more accurate 13.997
conc. c0 is high enough (c0 Ka), then most of the acid is not ionized
depends on temperature: pKw (100 ◦C) = 12.29
conc. is not too high so that we can use the infinite dilution approximations (γi = 1)
heavy water: pKw (25 ◦C) = 14.87
(isotopic effect: D is more strongly bound)
Alternate form:
pH = 1
2 (pKa + pcA )
Degree of dissociation:
α≈
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Dissociation of water and the acidity constant
How much of deprotonated form can we find in a solution of given
pH (e.g., in a buffer)?
Ka = Kd
bases: dissociation constant of the conjugated acid
NH3 + H+
(Ka)
(Kw )
H2 O
→
NH3 + H2O
→
NH4+ + OH−
×(−1)
×(+1)
Kd = Kw /Ka
compound
0
eq.
HA
c0
c0 − x
A−
0
x
H+
cH+
cH+
x = cA− =
Example. Acidity constant of ammonium is pKa = 9.25 (at 25 ◦C). Calculate the dissociation constant of ammonium hydroxide (water solution if
NH3).
0.01
[A−]
Ka
=
[HA] [H+]
c0Ka
cH+ + Ka
CH3COOH
0.01 mol dm-3
pKa = 4.76
Degree of dissociation:
α=
1
cH+ /Ka + 1
We have assumed the infinite dilution
approximations
[CH3COO--]
[CH3COOH]
c/mol dm-3
CH3 COOH → CH3 COO− + H+
H+ + OH−
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HA → H+ + A−
acids: dissociation constant of the acid
→
Ka
c0
Dissociation of a weak acid III
Acidity constant (ionization constant)
Ka = equilibrium constant of deprotonisation
NH4+
s
0
2
3
4
5
6
pH
α=1
2 for pH=pKa
7
8
1.78×10−5
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Calculate pH of rainwater at 25 ◦C and 1 bar.
Henryho constant of dissolution of CO2: Kh = 0.033 mol dm−3 bar−1
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Simultanneous equilibria: more ionized states
Aminoacids, ionization degree according to pH. E.g., HIS, LYS:
Acidity constant of CO2: pKa1 = 6.37. This is total for reactions
AH32+ → AH2+ + H+
(Ka1)
AH → A− + H+
(Ka3)
AH2+ → AH + H+
CO2 + H2O → H2CO3 → H+ + HCO3−
CO2 in the air (2014): y = 400 ppm (18th cent.: y = 280 ppm)
H+ easily detaches from AH32+, hardly from AH
⇒ Ka1 > Ka2 > Ka3 or pKa1 < pKa2 < pKa3.
CO32− can be neglected
OH− can be neglected
Balance:
cCO2 = Khyp = 0.033 mol dm−3 bar−1 × 0.000400 × 1 bar
[A−] + [AH] + [AH2+] + [AH32+] = c0
= 1.297×10−5 mol dm−3
Equation:
?
Balance: cH+ = cHCO− (Not cCO2 = cCO2,0 − cH+ , as if the initial concen3
tration of CO2 were known.)
Equation:
[H+] =
p
[H+][HCO3−]
= Ka1
[CO2]
[AH2+][H+]
= Ka1
[AH32+]
[AH][H+]
= Ka2
[AH2+]
− log 1.297×10−5
.
1
pH = 1
2 (pKa1 + p[CO2 ]) = 2 (6.37 + 4.88) = 5.62
Ka1[CO2],
the real pH is usually lower (HNO3, H2SO4)
[A−][H+]
= Ka3
[AH]
(18th cent.: pH = 5.70)
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pH of a weak base
Example. Calculate pH of aqueous solution of ethylamine of concentration
c0 = 0.01 mol dm−3 at 25 ◦C. The acidity constant of ethylammonium is
Ka = 1.6×10−11.
[cd ../maple; xmaple lysine.mws] 14/25
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Lysine
NH2-(CH2)4-CH(NH2)-COOH, pKa1 = 2.15, pKa2 = 9.16, pKa3 = 10.67
1
1×10−14
Kw
=
= 0.000 625
Dissociation constant od ethylamine: Kd =
Ka
1.6×10−11
compound
0
eq.
C2H5NH2
c0
c0 − x
C2H5NH3+
0
x
OH−
0
x
s
x = [OH−] =
or:
ci/mmol.dm-3
Assumptions: [H+] [OH−], γ = 1. And in the same way as for weak acids:
Balance:
[C2H5NH3+][OH−]
x2
=
= Kd
[C2H5NH2]
c0 − x
Equil.:
LysLys
LysH+
LysH2+
0.5
Kd 2
K c0Kd p
≈
+ Kdc0 − d
Kdc0
2
2
0
pH = 1
2 (pKa + pKw − pc0 )
0
1
2
3
4
5
6
0
eq.
MA
(Ka)
A− + H2O
compound
balance
[A−] + [HA] = c0
[H+] − ([A−] + [OH−]) = 0
HA
c0
c0 − x
A
A−
0
x
charge
H+
0
x+y
OH−
0
y
Equation
[H+][A−]
= Ka
[HA]
Equation
4 equations, 4 unknowns:
[H+], [HA], [A−], [OH−]
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[H+] =
√
[H+] = Kac0
(approximate formula)
c0
c0 − x
OH−
0
x
HA
0
x
H+
0
Kw /x
c0 K w
+
Ka
x (hydrolysis)
conditions
K
Ka =
w (c − x)
[H+][A−]
0
≈ x
[HA]
x
for [OH−] [H+]
r
Kw 2 Kw
c0Kw
−
≈
2Ka
2Ka
Ka
pH ≈ 1
2 (pKw + pKa − pc0 )
⇒
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Little soluble salts of strong electrolytes
γi = 1 assumed for ions
BaSO4 → Ba2+ + SO42− Ks =
6
Mg(OH)2 → Mg2+ + 2 OH−
As2S3 → 2 As3+ + 3 S2−
pH 5
but: S2− + H2O → HS− + OH− . . .
4
2
-1
-
aBa2+ aSO2−
4
aBaSO4
= [Ba2+][SO42−]
Ks = [Mg2+][OH−]2
Ks = [As3+]2[S2−]3
As3+ + 3 OH− → H3AsO3 . . .
Example. How much mg Mg(OH)2 dissolves in 1 L of pure water?
Data: Ks = 2.6×10−11, M(Mg(OH)2) = 58.3 g mol−1
3
In a realistic region of concentrations the simplified formula is sufficient
A−
(100 %)
AH + OH−
activities of salts (s) are asalt = 1
7
+ Kac0 − K2a
c0
M+ + A−
Solubility product = equilibrium constant of dissociation.
8
Ka
2
eq.
c0
→
→
√
where the last approximation holds for c0 KKw & x Kw , i.e. c0 Ka
CH3COOH, pKa = 4.76
2
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a
Very dilute solutions
r
0
M+
x=
2 equations, 2 unknowns: x, y
“exact”
compound
Solution:
s
[H+][OH−] = Kw
(x + y)x
= Ka
c0 − x
(x + y)y = Kw
9 10 11 12 13 14
Salt of weak acid + strong base
(Kw )
“Implicit” balance
“Explicit” balance
8
E.g., M=Na, A=CH3COO.
Dissociation of water has to be taken into account.
HA → H+ + A−
H2O → H+ + OH−
7
pH
pH=11.34, approx.: 11.40; α = 0.22
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Simultaneous equilibria: very dilute solutions
compound
(Ka2)
0
1
2
3
4
5
6
p[C2H5COOH]
7
8
9
10
Balance: [OH−] = 2[Mg2+] = 2c
Equation: Ks = [Mg2+][OH−]2 = c(2c)2 = 4c3
Solution: c = [Mg(OH)2] = (Ks/4)1/3 = 0.0001866 mol dm−3,
cw = cMMg(OH)2 = 11 mg dm−3
Often problems with hydrolysis, complexation, . . .
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Little soluble salts: more ions
Calcium oxalate: Ks(CaC2O4) = 3.9×10−9.
250
a) how much dissolves in pure water?
b) how much dissolves in blood?
([Ca2+] = 2.4 mmol dm−3)
11
200
10
Ks = [Ca2+][C2O42−]
Solution:
√
a) [C2O42−] = [Ca2+] = Ks
.
−3
= 62 µmol dm
hydrolysis of C2O42− unimportant (62.49 vs. 62.45 µmol dm−3)
.
b) [C2O42−] = Ks/[Ca2+] = 1.6 µ mol dm−3
150
100
8
oxalis
50
Solubility decreses in a presence of one of the ions (CaCl2, Na2C2O4)
7
18th century
now
0
-4 -3 -2 -1 0 1 2 3 4 5 6
But: upon adding H2C2O4 the solubility slightly increases, because at
lower pH C2O42− is protonated
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HCO3−
H2 O
→
→
→
→
log(yCO2/ppm)
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Buffers
Buffer = solution able to keep pH (almost) constant if acid or base is added.
Karsts, seawater,a . . .
CaCO3(s)
18th century
now
6
-4 -3 -2 -1 0 1 2 3 4 5 6
log(yCO2/ppm)
Other ions (e.g., NaCl) slightly increase solubility (more later . . . )
Case study: system CaCO3(s,aq.) + CO2(g,aq)
9
pH
cw(Ca)/g.m-3
CaC2O4 → Ca2+ + C2O42−
CO2 + H2O
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Case study: system CaCO3(s,aq.) + CO2(g,aq)
Ca2+ + CO32−
H2CO3 → H+ + HCO3−
H+ + CO32−
H+ + OH−
Typical example: solution of a weak acid HA ([HA] = cacid) + its salt MA
(of strong base, [MA] = [M+] = cbase), e.g., CH3COOH + CH3COONa
pKs = 8.35 (calcite)
pKa1 = 6.37
acid
pKa2 = 10.25
pKw = 14
compound
0
M+
cbase
cbase
A−
cbase
cbase + x
Henry constant of CO2 in water : Kh = 0.033 mol dm−3 bar−1
Partial ressure of pCO2 is known (atmosphere)
Balance
CO2 given by pCO2
⇒ no balance of HxCO3
base
eq.
H+
0
x
HA
cacid
cacid − x
Ka =
[H+][A−] x(cbase + x) xcbase
=
≈
[HA]
cacid − x
cacid
Approximate solution—Henderson–Hasselbalch:
c
x ≈ Ka acid
cbase
CaCO3(s) is in surplus ⇒ no balance of Ca
charge balance:
c
[H+] = Ka acid
cbase
Assumptions and generalization:
[OH−], [H+] cacid, cbase; cacid
2 [Ca2+] + [H+] − 2 [CO32−] − [HCO3−] − [OH−] = 0
[cd ../maple; xmaple vapenec+co2.mws] 19/25
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roughly
≈
cbase, cacid, cbase Ka; γi = 1
holds true also for a mixture of a weak base B (B + H2O → BH+ + OH−)
and its salt BHX (of strong acid), cacid = [BHX] = [X−], cbase = [B]
credit: www.explorecrete.com
Case study: system CaCO3(s,aq.) + CO2(g,aq)
c
pH = pKa + log10 base
cacid
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Buffer capacity
Let us add a small amount dc of a strong base MOH ⇒ dc of HA is
neutralized to MA (=“base”).
Add strong acid = remove strong base
[CO2] = yCO2 pKh
(given)
[Ca2+][CO32−] = Ks
[H+][HCO3−]
= Ka1
[CO2]
(2)
[H+][CO32−]
= Ka2
[HCO3−]
(3)
[H+][OH−] = Kw
(4)
+ charge balance:
2 [Ca2+] + [H+] − 2 [CO32−] − [HCO3−] − [OH−] = 0
cacid → cacid − dc
cbase → cbase + dc
(1)
(5)
⇒ 5 equations, 5 unknowns:
[Ca2+], [CO32−], [H+], [HCO32−], [OH−].
Buffer capacity = β =
dc
dc
= − ln 10 [H+]
d(pH)
d[H+]
“the amount of a strong base needed to increase pH by 1”
The same assumptions as Henderson–Hasselbalch:
c
[H+](c) = Ka acid
cbase
c
− dc
c
1
1
[H+](c + dc) = Ka acid
= Ka acid 1 − dc
+
cbase + dc
cbase
cacid cbase
d[H+]
c
1
1
c
+ cbase +
= −Ka acid
+
= − acid
[H ]
dc
cbase cacid cbase
cacidcbase
β = ln 10
Solubility of limestone in rainwater
(recalculated to Ca2+)
19.7 mg dm−3 today (400 ppm CO2 (2015)), pH=8.23
17.6 mg dm−3 before industrial revolution (280 ppm CO2), pH=8.33
For comparison:
4.8 mg dm−3 in pure water (no CO2, pH=9.9)
√
2.7 mg dm−3 hydrolysis neglected (cst Ks, pH=7)
Contents of Ca2+ and pH (more ions present. . . ):
Blood: 100 mg dm−3, pH = 7.35–7.45
Sea: 400 mg dm−3, pH = 7.5–8.4
cacidcbase
cacid + cbase
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Maximum capacity
β = ln 10
cacidcbase
cacid + cbase
Given A (cA,total = cacid + cbase), β reaches maximum at cacid = cbase
Maximum buffer capacity is reached for an equimolar mixture
Example: acetate buffer, cacid = cbase = 0.1 mol dm−3, pH = pKa = 4.76
7
0.12
0.1
6
β/mol dm-3
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pH
Case study: system CaCO3(s,aq.) + CO2(g,aq)
.
ln 10 = 2.3026
Equation:
5
0.08
0.06
0.04
4
0.02
3
-0.1
-0.05
0
∆cbase/mol
0.05
0.1
dm-3
graphs: exact solution with γ = 1
0
-0.1
-0.05
0
∆cbase/mol dm-3
0.05
0.1
Bicarbonate (hydrogen carbonate) buffer
Main part of the blood buffer system
Henry constant of CO2 in water at body temperature:
Kh = 0.025 mol dm−3 bar−1
Acidity constant of CO2 at body temperature: pKa1 = 6.1 for
CO2 + H2O → H2CO3 → H+ + HCO3−
Contents of hydrogen carbonates (mostly NaHCO3):
[HCO3−] = 24 mmol dm−3 = cbase
pH = 7.35 to 7.45
c
[H+] = Ka1 acid
cbase
cacid =
[H+]cbase
,
Ka1
c
[H+]cbase
pCO2 = acid =
Kh
Ka1Kh
⇒ pCO2 = 5.4 to 4.3 kPa (∼ 5 obj.% in alveolar air)
Outside range ⇒ respiratory acidosis/alkalosis
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