ACID-BASE CHEMISTRY

STRENGTH OF AN ACID OR BASE

• Strength: The tendency to donate or accept a proton, i.e., how readily does the substance donate or accept a proton?

• Weak acid has weak proton-donating tendency; a strong acid has a strong proton-donating tendency. Similarly for bases,

• Cannot define strength in absolute sense. Strength depends on both the acid and base involved in an acid-base reaction.

• Strength measured relative to some reference, in our case, the solvent water.

THE HYDRONIUM ION• The proton does not actually exist in aqueous

solution as a bare H+ ion. • The proton exists as the hydronium ion (H3O+).• Consider the acid-base reaction:

HCO3- + H2O H3O+ + CO3

2-

Here water acts as a base, producing the hydronium ion as its conjugate acid. For simplicity, we often just write this reaction as:

HCO3- H+ + CO3

2-

STRENGTH MEASURED QUANTITATIVELY BY THE IONIZATION CONSTANT

HA0 + H2O H3O+ + A-

orHA0 H+ + A-

][]][[

0HAAHK A

The larger KA, the stronger the acid; the smaller KA, the weaker the acid

DEFINITION OF pKA AND pH

pKA = - log KA

Thus, the larger pKA, the weaker the acid; the smaller pKA, the stronger the acid.

Similarly,pH = - log [H+]

pOH = - log [OH-]

H2CO30 + H2O H3O+ + HCO3

-

NH4+ + H2O H3O+ + NH3

0

CH3COOH0 + H2O H3O+ + CH3COO-

H2O + H2O H3O+ + OH-

The stronger an acid, the weaker the conjugate base, and vice versa:

• The conjugate bases of weak acids are strong, and the conjugate bases of strong acids are weak.

CONJUGATE ACIDS-BASESAcids (blue) on left, conjugate bases (green) on right

• The strength of an acid is expressed by the value of the equilibrium constant for its dissociation reaction.

• Consider: HCO3- H+ + CO3

-

• The dissociation constant for this reaction at 25°C is:

• This can also be expressed as pK a = 10.3• The larger the pK a, the weaker the acid. Bicarbonate

is considered to be a very weak acid.

STRENGTH OF ACIDS AND BASES- I

10-10.3 = a[H+] a[CO3]2- / a[HCO3-]

HCO3- + H2O H3O+ + CO3

2-

CO32- is a stronger base than H2O

pKB of H2O = 14

pKB of CO32- = 3.7

At high pH:

HCO3- + OH- H2O + CO3

2-

(OH)- is a stronger base than CO32-

pKB of (OH)- = 0

STRENGTH OF ACIDS AND BASES- II

Explanation:For the reaction, HCO3- + H2O H3O+ + CO3

2-

KA = 10-10.3, so pKA = 10.3

KB = KW/KA, so pKB = pKW- pKA= 14 – 10.3 = 3.7

For the reaction, HCO3- + OH- H2O + CO3

2-,

KB for OH- is based on the reaction:

OH- + H2O OH- + H2O

K for the reaction is clearly = 1Thus, pKB = 0

STRENGTH OF ACIDS AND BASES- IV

HNO30 + H2O H3O+ + NO3

-

HNO3 is a stronger acid than H3O+

pKa of HNO3 is strongly negative

pKa of H3O+ = 0

All acids with pKa smaller than 0, are completely dissociated in water

• Most acid-base reactions in aqueous solutions are very fast (almost instantaneous); thermodynamic equilibrium is attained and thermodynamic principles yield correct answers.

Kinetics

SELF-IONIZATION OF WATER AND NEUTRAL pH

H2O H+ + OH-

Neutrality is defined by the condition: [H+] = [OH-]

Kw = [H+]2

log Kw = 2 log [H+]-log Kw = -2 log [H+]

14 = 2 pHpHneutral = 7

14

2

10][

]][[

OHOHHKw At 25oC and

1 bar

AMPHOTERIC SUBSTANCE

• Now consider the acid-base reaction:NH3 + H2O NH4

+ + OH-

In this case, water acts as an acid, with OH- its conjugate base. Substances that can act as either acids or bases are called amphoteric.

• Bicarbonate (HCO3-) is also an amphoteric

substance:Acid: HCO3

- + H2O H3O+ + CO32-

Base: HCO3- + H3O+ H2O + H2CO3

0

DISSOCIATION CONSTANTS OF SELECTED WEAK ACIDS AT

25°CAcid pK1 pK2 pK3Acetic (CH3COOH) 4.75 ----- -----

Boric (H3BO3) 9.2 ----- -----Carbonic (H2CO3) 6.35 10.33 -----

Phosphoric (H3PO4) 2.1 7.0 12.2Hydrosulfuric (H2S) 7.0 13.0 -----

Silicic (H4SiO4) 9.9 11.7 9.9Hydrofluoric (HF) 3.2 ----- -----Arsenic (H3AsO4

0) 2.2 7.0 11.5Sulfurous (H2SO3

0) 1.7 6.9 -----

EXAMPLES OF ACIDS AND BASES PRESENT IN NATURAL WATERS

• Most important base: HCO3-

• Other bases: B(OH)4-, PO4

3-, NH30, AsO4

3-, SO4

2-, CO32-, etc.

• Most important acid: CO2(aq) or H2CO30

• Other acids: H4SiO40, NH4

+, B(OH)30,

H2SO40, CH3COOH0 (acetic), H2C2O4

0

(oxalic), etc.Note: B(OH)3 looks more like an acid written as H3BO3

NUMERICAL EQUILIBRIUM CALCULATIONS

Monoprotic acidWhat are the pH and the concentrations of

all aqueous species in a 5 x 10-4 M solution of aqueous boric acid (B(OH)3)?

Steps to solution1) Write down all species likely to be present

in solution: H+, OH-, B(OH)30, B(OH)4

-.

Note on boric acid

Boric acid might conventionally be written as H3BO3, which would imply that it would be capable of donating three protons. However, it is a very weak acid and effectively is capable of donating only one proton. In fact, it is better conceived as a Lewis acid; i.e. a hydroxyl acceptor rather than a proton donor. The formula of boric acid is then better written as B(OH)3 . Hence, its dissociation is a hydrolysis reaction: H2O + B(OH)3 = B(OH)4

- + H+

2) Write the reactions and find the equilibrium constants relating concentrations of all species:

H2O H+ + OH-

10

203

4 107]][)([])(][[

xOHOHB

OHBHK A

14

2

10][

]][[

OHOHHKw (I)

(II)

B(OH)30 + H2O B(OH)4

- + H+

3) Write down all mass balance relationships:

5 x 10-4 M = B

= [B(OH)4-] + [B(OH)3

0] (III)

4) Write down a single charge-balance (electroneutrality) expressions:

[H+] = [B(OH)4-] + [OH-] (IV)

5) Solve n equations in n unknowns.

EXACT NUMERICAL SOLUTION

Eliminate [OH-] in (I) and (IV)

[H+][OH-] = Kw

[OH-] = Kw/[H+]

[H+] = [B(OH)4-] + Kw/[H+]

[H+] - [B(OH)4-] = Kw/[H+]

][])([][ 4

H

OHBHKw (V)

Solve (III) for [B(OH)30]

[B(OH)30] = B - [B(OH)4

-]

AKOHBBOHBH

])([])(][[

4

4

[H+][B(OH)4-] = KA(B - [B(OH)4

-]) (VI)

Now solve (V) for [B(OH)4-]

- [B(OH)4-] = Kw/[H+] - [H+]

[B(OH)4-] = [H+] - Kw/[H+]

Substitute this into (VI)

[H+]([H+] - Kw/[H+]) = KA(B - [H+] + Kw/[H+])

[H+]2 - Kw = KAB - KA[H+] + KAKw/[H+]

[H+]3 - Kw[H+] = KAB[H+] - KA[H+]2 + KAKw

[H+]3 + KA[H+]2 - (KAB + Kw)[H+] - KAKw = 0

[H+]3 + (7x10-10)[H+]2 - (3.6x10-13)[H+] - (7x10-24) = 0

We can solve this by trial and error, computer or

graphical methods. From trial and error we obtain

[H+] = 6.1x10-7 M or pH = 6.21

[OH-] = Kw/[H+][OH-] = 10-14/10-6.21

[OH-] = 10-7.79 M

[B(OH)4-] = [H+] - Kw/[H+]

[B(OH)4-] = 6.1x10-7 - 1.62x10-8

[B(OH)4-] = 5.94x10-7 M

[B(OH)30] = B - [B(OH)4

-]

[B(OH)30] = 5x10-4 - 5.94x10-7 M = 4.99x10-4 M

APPROXIMATE SOLUTION

• Look for terms in additive equations that are negligibly small (multiplicative terms, even if very small, cannot be neglected.

• Because we are dealing with an acid, we can assume that [H+] >> [OH-] so that the mass balance becomes:

[H+] = [B(OH)4-]

and then

[B(OH)30] = B - [H+]

10

203

4 107]][)([])(][[

xOHOHB

OHBHK A

][][ 2

HBHK A

(ii)

[H+]2 = KA·B-KA[H+]

[H+]2 + KA[H+] - KA·B = 0

This is a quadratic equation of the form:ax2 + bx + c = 0

and can be solved using the quadratic equation

aacbbx

242

In our case this becomes:

24

][2 BKKK

H AAA

Only the positive root has any physical meaning.[H+] = 5.92 x 10-7

We could have made this problem even simpler. Because boric as is a quite weak acid (i.e., veryKA value, very little of it will be ionized, thus

[B(OH)30] >> [B(OH)4

-]

B [B(OH)30] = 5 x 10-4 M

10

203

4 107]][)([])(][[

xOHOHB

OHBHK A

1003

2

107])([

][

xOHBHK A

104

2

107105

][

xxHK A

[H+]2 = 3.5 x 10-15

[H+] = 5.92 x 10-7 M

It is wise to check your assumptions by back substitutinginto original equations. If the error is 5%, the approxi-mation is probably justified because KA values are at least

this uncertain!