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X. Acid/Base Chemistry
Three definitions:
1) Arrhenius: Acid – increases [H+]; Base – increases [OH–]
2) Brønsted-Lowry: Acid – donates a proton; Base – accepts a proton
3) Lewis: Acid – electron-pair acceptor; Base – electron-pair donor
Brønsted-Lowry includes conjugates, e.g., HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq); Ka CH3COOH(aq) + H2O(l) CH3COO–(aq) + H3O+(aq); Ka NH3(aq) + H2O(l) NH4
+(aq) + OH–(aq); Kb a) Water behaves as both an acid and a base
H2O(l) + H2O(l) H3O+(aq) + OH–(aq) weak acid weak base conjugate acid conjugate base
OR H2O(l) H+(aq) + OH–(aq)
• water is a very weak acid and a very weak base • autoprotolysis or autoionization
2
H OHw H OH
H O
a aK a a H OH
a+ −
+ −+ − = = =
• value for Kw depends on T, at 298 K Kw = 10–14
T (K) 273 298 313 373 Kw 0.12×10–14 1.0×10–14 2.9×10–14 5.4×10–13
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b) pH • because of the wide range of concentrations of H+ commonly observed in
biological and chemical processes, a logarithmic scale is useful pH = - log aH+
• cannot measure γ for H+ in isolation, but we can estimate using the Debye-Hückel equation. For a solution of 0.050 M HCl
𝑙𝑙𝑙𝑙𝑙𝑙𝛾𝛾𝐻𝐻+ = −0.509𝑧𝑧2√𝐼𝐼 = −0.509(1)2√0.050 = −0.114
γ𝐻𝐻+ = 0.77
𝑝𝑝𝑝𝑝 = − log( ) =
• for dilute solutions at low ionic strength (i.e., [H+] ≤ 0.1 M and I ≤ 0.1 M), can usually approximate with
pH = –log[H+], i.e., 0.050 M HCl has a pH of 1.30
• also, pOH = –log[OH–] pH + pOH = 14
neutral solutions @ 25°C: [H+] = [OH–] = √Kw = √1.0×10-14 = 10–7 M and pH = 7.00
acidic solutions @ 25°C: [H+] > [OH–] pH
basic solutions @ 25°C: [H+] < [OH–] pH If we evaluate the pH of boiling water, Kw(100°C) = 5.4×10-13
[H+] = √Kw = 7.35×10-7 and pH = 6.13 is “neutral”. c) Dissociation of Acids and Bases
• strong acids and bases are assumed to completely dissociate in water and calculation of species concentrations is straight forward
• weak acids and bases partially dissociate in water and calculations of species concentrations is a little more complicated
• for a weak acid HA
HA(aq) + H2O(l) A-(aq) + H3O+(aq) ( )( )( )( )
3
2
H O Aa
HA H O
a aK
a a
+ −
=
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• the concentration of water in most solutions is close to that of pure water (i.e., 1 L contains 55.5 mol) and so we can consider it in its standard state (i.e., pure liquid) and approximate with
2H Oa = 1
( )( )[ ]
H Aa
HA HA
a a H AK
a HAγ γ
γ+ −
+ −+ − = =
• since HA is an uncharged species, we can set γHA ≈1 for dilute solutions and
substitute γ±2 for γ+γ-
[ ]
2
a
H AK
HAγ+ −± =
• if the acid is sufficiently weak, the concentration of the ions are low (≤ 0.050 M)
and we can ignore γ±2 and write
[ ]a
H AK
HA
+ − =
• the magnitude of Ka indicates the degree of dissociation of the acid and thus the acid strength (Refer to Table 8.1 – Ka & pKa values for common weak acids at 298 K) • small Ka, ( little / much ) dissociation, ( stronger / weaker ) acid
• large Ka, ( little / much ) dissociation, ( stronger / weaker ) acid
• can also use the percent dissociation to indicate strength
[ ]0100%eq
Hpercent dissociation
HA
+ = ×
• treatment of bases is similar, for NH3 the result is
Rxn: NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
4
3b
NH OHK
NH
+ − =
(Refer to Table 8.2 – Kb & pKb values for common weak bases at 298 K)
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Example: Calculate the concentration of the undissociated acid, the H+ ions and the CN– ions of a 0.050 M HCN solution at 25°C.
Rxn: I C E
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d) Relationship between Ka and Kb • for a weak acid HA: HA(aq) + H2O(l) A–(aq) + H3O+(aq)
[ ]a
A HK
HA
− + =
• and for its conjugate base: A–(aq) + H2O(l) HA(aq) + OH–(aq)
[ ]b
HA OHK
A
−
−
=
• If we were to add the two chemical equations together, i.e.,
HA(aq) + H2O(l) A–(aq) + H3O+(aq); Ka
A–(aq) + H2O(l) HA(aq) + OH–(aq); Kb
H2O(l) + H2O(l) H3O+(aq) + OH–(aq); K = ?
[ ][ ]
a b
w
A H HA OHK K
HA A
H OH
K
− + −
−
+ −
= ×
= =
e) Salt Hydrolysis
• ions that result from the dissociation of salts can undergo hydrolysis resulting in acid or basic solutions
• common examples are salts of weak acids or bases, e.g., sodium acetate forms a basic solution; ammonium chloride forms an acidic solution
• small, highly charged cations such as Be2+, Al3+ and Bi4+ can also hydrolyze forming acidic solutions
e.g., AlCl3(s) Al3+(aq) + 3Cl–(aq)
The Al3+ has a hydration sphere of six water molecules and the dense charge on the cation polarizes the O-H bonds resulting in loss of a proton to the bulk water.
Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+(aq) + H3O+(aq); Ka = 1.4×10–5
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f) Polyprotic acids • more complicated than monoprotic • we will only consider biologically important carbonic acid and phosphoric acid
i) Carbonic acid
• carbon dioxide on dissolving in water reacts to form carbonic acid, a diprotic acid, i.e.,
CO2(aq) + H2O(l) H2CO3(aq)
• the K for this reaction is small (0.00258) but experimentally we can’t distinguish between CO2(aq) and H2CO3(aq), so we generally treat them as the single species H2CO3(aq)
• the first dissociation constant is H2CO3(aq) H+(aq) + HCO3
–(aq)
[ ]1
3 7
2 3
4.2 10a
H HCOK
H CO
+ −−
= = ×
• the conjugate base from the 1st dissociation becomes the acid for the 2nd dissociation, i.e.,
HCO3–(aq) H+(aq) + CO3
2–(aq)
2
23 11
3
4.8 10a
H COK
HCO
+ −−
−
= = ×
• using these equations, we can calculate the concentration of the carbonate species at any pH (refer to Example 8.2)
• since 1aK >>
2aK we can make some generalizations
(Fig 8.1)
• in a solution of carbonic acid, most of the H+ is generated by the 1st dissociation, and the 2nd dissociation is negligible – mathematically, this results in the concentration of the conjugate base from the 2nd dissociation is numerically equal to
2aK
• no more than two of the carbonate species will be present in significant concentrations at any one pH
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ii) Phosphoric Acid • an important triprotic acid, i.e.,
H3PO4(aq) H+(aq) + H2PO4–(aq)
1aK
H2PO4(aq)- H+(aq) + HPO42–(aq)
2aK
HPO42-(aq) H+(aq) + PO4
3–(aq) 3aK
• we can use the same procedure used to calculate the concentrations of all the species used above to calculate all four phosphate species
(Fig 8.2)
g) Buffers • are solutions containing a weak acid and its conjugate base or a weak base and its
conjugate acid at similar concentrations. • resist changes in pH on addition of small amounts of acid or base.
• pH of bodily fluids varies greatly depending on location (blood plasma pH 7.4, gastric juices pH 1.2) and maintaining these pH’s is essential
• enzymes will only work properly in a small range of pH values • pH balance also maintains the balance of osmotic pressure
• equation generally used to determine the pH of a buffer is the Henderson-Hasselbalch equation
[ ][ ]
log
aconjugate base
pH pKconjugate acid
= +
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• the pH of the buffer depends on the pKa of the weak acid, and the relative amounts of the two conjugates present
• consider a buffer made from acetic acid and sodium acetate • the acid partially dissociates producing acetate and hydronium ions
CH3COOH(aq) + H2O(l) CH3COO–(aq) + H3O+(aq) • the base partially hydrolyzes producing acetic acid
CH3COO–(aq) + H2O(l) CH3COOH(aq) + OH–(aq) • however, the presence of the acetate shifts the acid reaction to the left, and
the presence of the acetic acid shifts the base reaction to the left – essentially, no reaction occurs and we can consider the concentrations at equilibrium to be the same as the initial concentrations.
• on addition of a small amount of strong acid to this buffer, the acetate would react with it producing acetic acid
• on addition of a small amount of strong base, acetic acid would react with it producing acetate
• the base form of the H-H equation is
[ ][ ]
log
b
conjugate acidpOH pK
conjugate base= +
i) Effect of Ionic Strength and Temperature on Buffers
• a more rigorous treatment requires using activities • for a monoprotic acid HA, assuming the activity coefficient for HA is 1:
[ ]log 0.509
ApH pKa I
HA
− = + −
• the addition of the last term can have significant effects on pH calculations, especially in buffers where there is a significant ionic strength
• most Ka values are measured at 25°C, but most biological processes occur at 30 – 40°C
• Ka can be calculated at the higher temperature using the van’t Hoff equation and ΔH° and then the pH can be calculate using the H-H equation
Buffer Capacity - the amount of acid or base which must be added to a buffer to produce a change in pH of one unit
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ii) Preparing buffers • weak acids buffer best within one pH unit of their pKa • to prepare a buffer of a specific pH
• choose an appropriate acid/base pair (pKa ≈ pH) • use the H-H equation to calculate the ratio of base to acid • mix the acid and base together in this ratio • this is one of three ways to prepare a buffer:
• mix conjugates together in approximately equal amounts e.g., 1 CH3COOH + 1 CH3COONa • partially neutralize a weak acid by mixing weak acid and strong base in
approximately a 2:1 ratio e.g., 2 CH3COOH + 1 NaOH • partially neutralize a weak base by mixing weak base and strong acid in
approximately a 2:1 ratio e.g., 2 CH3COONa + 1HCl h) Acid-Base Titrations
• titration is an important analytical technique for determining concentrations of solutions
• acid/base titrations involving adding base from a buret to a solution of acid (or vice versa) until the equivalence point is reached
• the equivalence point can be detected by a number of methods, including using a pH-meter and acid-base indicators
• for titration of a strong acid with a strong base • the pH before the equivalence point is determined by the unreacted acid and is
usually strongly acidic • the pH at the equivalence point is _____ • the pH after the equivalence point is determined by the excess base added and
is usually strongly basic
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• for titration of a weak acid with a strong base • the pH before the equivalence point is determined by the unreacted acid and is
usually weakly acidic • the pH at the equivalence point is weakly ____________ due to the presence
of the conjugate base • the pH after the equivalence point is determined by the excess base added and
is usually strongly basic • for titration of a weak base with a strong acid
• the pH before the equivalence point is determined by the unreacted base and is usually weakly basic
• the pH at the equivalence point is weakly ____________ due to the presence of the conjugate acid
• the pH after the equivalence point is determined by the excess acid added and is usually strongly acidic
Fig 8.4
i) Acid-Base Indicators
• weak acids and bases where the two conjugates are different colours. • colour changes when pH of sol’n is within one pH unit of the pKa of the indicator
pKIn ± 1 • choose an indicator whose pKIn is close to the pH at the equivalence point • the point at which the indicator changes colour is the end point of the titration –
ideally your end point should be at your equivalence point • Refer to Table 8.5 for some common acid-base indicators
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j) Amino Acids • behave as both acids and bases – ampholytes • in solution exist as zwitterions +NH3CHRCOO–
• behaves as a base when titrated with hydrochloric acid • behaves as an acid when titrated with sodium hydroxide
Fig 8.5 - titration of glycine
• pH at 1st half-equivalence point is pKa1 (or pKa′)
• pH at 2nd half-equivalence point is pKa2 (or pKa″)
• pH at first equivalence point where zwitterion is predominate species is the average of pKa1 and pKa2
k) Isoelectric Point (pI) • pH at which zwitterion does not move in an electric field – net charge on molecule
is zero • determine which structure has a net zero charge, and average pK’s on either side
to determine pI e.g., Aspartic acid
2
3
1 2.09 3.86
9.82
+ + -3 2 3 2
+ - - - -3 2 2 2
: 1 : 0
: 2: 1
NH CH(CH COOH)COOH NH CH(CH COOH)COO
NH CH(CH COO )COO NH CH(CH COO )COO
pK pKa a
pKa
A B
DC
= =
=
+
−−
• B is electrically neutral and the isoelectric point is
2.09 3.86 2.982
pI += =
• pI values for amino acids are listed in Table 8.6
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k) Titration of proteins • proteins can be titrated to determine the number of dissociable protons
(Fig 8.7)
• many equivalence points and much care must be taken for accurate assignments • can match pK values of these protons to those for amino acids to determine which
amino acid they came from • comparing these results to amino acid analysis of the protein can reveal
information on the structure of the protein • any dissociable protons not accounted for in the titration must be on groups
within the interior of the protein where they are not available for titration • these protons can be titrated if the protein is denatured.
l) Maintaining the pH of Blood Reading assignment – p. 293 - 297
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