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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