Aqueous Equilibria. II. Acids and Bases.


1. Arrhenius:

An acid produces H+ and a base produces OH-

Acid: HCl H+ + Cl-

Base: NaOH Na+ + OH-

2. Lowry and Brønsted:

An acid is a proton donor, a base is a proton acceptor.

HCl + H2O H3O+ + Cl-

acid base

3. Lewis:

An acid is an electron pair acceptor, a base is an electron pair donor in the formation of a covalent bond between the two.

Note that Brønsted and Lewis bases are often the same.

The Brønsted acid always involves the proton accepting the electron pair. This is only one possibility for a Lewis acid.

Conjugate Acid and Bases.

In a conjugate pair, the acid has the proton, the base does not.

Some species have both a conjugate acid and a conjugate base:

conjugate acid conjugate base
H3O+ H2O OH-
NH4+ NH3 NH2-
H2SO4 HSO4- SO42-
H2CO3 HCO3- CO32-


1 Using the Brønsted approach, pick out the acid-base conjugate pairs:

HBr + H2O H3O+ + Br-


NH4+ + H2O H3O+ + NH3

S2- + H2O HS- + OH-

H2O + H2O H3O+ + OH-

2. Give the conjugate acid and conjugate base for the following species:

conjugate acid conjugate base

The Equilibrium Constants, Ka, and Kb.

Ka is the equilibrium constant for an acid reacting with water as the base:

HA + H2O H3O+ + A-

Ka = {a(H3O+) * a(A-) } / {a(HA) * a(H2O)}

note that a(H2O) = 1 in a dilute solution.

Kb is the equilibrium constant for a base reacting with water as the acid:

B + H2O BH+ + OH-

Kb = {a(BH+) * a(OH-) } / {a(B) * a(H2O)}

and again, a(H2O) = 1.

Strong and Weak Electrolytes.

An electrolyte is a substance whose water solution conducts electricity due to the formation of ions.

A strong electrolyte produces a strong current due to high dissociation producing many ions.

A weak electrolyte produces a weak current due to low dissociation producing few ions.

Acids and bases which are highly dissociated and so produce many ions on dissolving in water are strong electrolytes and are termed strong acids or bases.

Acids and bases which are weakly dissociated produce only a few ions are weak electrolytes and are termed weak acids or bases.

In terms of the equilibrium constants, Ka, and Kb, an arbitrary cut off at 1 x 10-2 is often used to separate strong from weak.


Do not confuse strong and weak with dilute and concentrated.


The importance of the a(H3O+) in living systems makes a measure of this necessary. Because the dilute solutions involved have an a(H3O+) which varies from 1.0 to 1.0 x 10-14, the notion of pH was introduced.

pH = - log {a(H3O+)}

Note that higher values of pH indicate lower values of a(H3O+)

The use of "p" as an operature meaning "take - log of" is extended to:

pOH = - log {a(OH-)}

pKa = - log Ka

pKb = - log Kb

pKsp = - log Ksp

Average pH values for some well known Solutions:
pH Solution
0 1 M HCl
1 to 3 gastric juice
ca. 2 limes
2.3 lemons
2 to 4 soft drinks
2.6 to 4.4 orange juice
ca 3 vinegar, wine
ca. 3.5 dill pickles
4.5 to 8 urine
ca. 6.1 sour milk
ca. 6.6 milk
6.5 to 7.5 saliva
7.3 to 7.5 blood serum
ca. 8 egg white
ca. 8 sea water
ca. 10 detergent solution
ca. 10 milk of magnesia
ca. 11 bar soap
11.5 to 12 household ammonia
12 washing soda
14 1 M NaOH


Water is one of a few substances that can react with itself as both acid and base:

2 H2O H3O+ + OH-


Kw = { a(H3O+) * a(OH-) } / a(H2O)2

and again a(H2O) = 1.

At 25ºC, Kw = 1 x 10-14. (Experimental measurement)

So, in pure water at 25ºC:

1 x 10-14 = a(H3O+) * a(OH-)


a(H3O+) = a(OH-) = 1 x 10-7.

Taking logs of the equation for Kw:

- log Kw = - log{a(H3O+)} - log {a(OH-)}

pKw = pH + pOH

14 = pH + pOH (at 25ºC)

In pure water at 25ºC, pH = pOH = 7.

Because of the fixed relationship between pH and pOH there is only need for one scale, the pH scale.

Limits of the pH scale.

Note that the relationship:

pH = - log {a(H3O+)}

can be measured experimentally only for ideal (read dilute) solutions when a(H3O+) = {[H3O+] / 1 mol/L}.

When either the [H3O+] or the [OH-] becomes greater than 1 M the activities can no longer be measured and so neither can pH or pOH.

pH Calculations. I. Acids.

An acid in water donates a proton to the water:

HA + H2O H3O+ + A-

and the [H3O+] increases. Consequently the pH goes down from 7 (the pH of pure water.)

Calculation of the [H3O+] (acid solutions):

1. For a strong acid, assume 100% reaction to H3O+.

Example: calculate the pH of a 0.0100 M solution of HBr, a strong acid.

Assume that all HBr reacts to gibe H3O+, the [H3O+] is equal the initial [HBr] = 0.0100 M.

So pH = - log a(H3O+) = - log 0.0100 = 2.0

In this problem the [H3O+] from the water itself is ignored.

If the acid concentration is very low, this may not be possible:

Example: calculate the pH of a 1 x 10-8 M solution of HBr.

If the [H3O+] from water is ignored, this will give a pH of 8, which is basic!

2 H2O H3O+ + OH-
I 1 x 10-8 0
C x x
E 1 x 10-8 + x x

Kw = 1 x 10-14 = (1 x 10-8 + x)x

x = 9.5 x 10-8

[H3O+] = 1.05 x 10-7

pH = 6.98

Example: calculate the pH of 0.0100 M hyperchlorous acid, HOCl, pKa = 7.2

Step A: Analysis

HOCl + H2O H3O+ + ClO-
I (mol/L) 0.0100 0 0
C -x x x
E 0.0100 - x x x

Ka = {a(H3O+) * a(ClO-)} / a(HOCl)

6.3 x 10-8 = x * x /(0.0100-x)

unknown: pH, calculated from x.

Step B: Brainstorm (trivial)

Step C: Calculate

If it is suspected that x is very small compared with the value to which it is added or subtracted (K is very small), the an approximate solution can be quickly obtained by ignoring the x in such sums or differences.


6.3 x 10-8 = x * x /(0.0100)

so an approximate value for x is found to be 2.53 x 10-5.

Validate this approximation by checking that x < 5% of the value to which it is added or subtracted:

{x / 0.0100}*100% = 2.53 x 10-5*100% / 0.01

= 0.25%

so the approximation is valid.

x = 2.53 x 10-5

pH = 4.6

What to do if the Approximation is not Valid.


Calculate the pH of a 0.0100 M solution of HF. pKa = 3.17

Step A: Analysis.

HF + H2O H3O+ + F-
I 0.0100 0 0
C (-1)x (+1)x (+1)x
E 0.0100 - x > x

Ka = a(H3O+) * a(F-) / a(HF) = 6.76 x 10-4

= x * x / (0.0100 - x) = 6.76 x 10-4

Use the approximation x << 0.01

6.76 x 10-4 = x * x / 0.0100

from which x1 = 0.00260.

Test the approximation: 0.0026 * 100% / 0.01 = 26%.

The approximation is not valid.

Use the method of successive approximations:

Substitute into the quadratic expression the approximate value obtained for x, and resolve to obtain a second approximation.

Repeat this to get a third approximation, and then continue until two successive approximations are the same.

Second approximation:

6.76 x 10-4 = x * x / (0.0100 - 0.00260)

from which x2 = 0.00224

Third approximation (you must do this one!)

6.76 x 10-4 = x * x / (0.0100 - 0.00224)

from which x3 = 0.00229

Check successive approximations from now on:

x2 and x3 not same, so find x4:

Fourth approximation:

6.76 x 10-4 = x * x / (0.0100 - 0.00229)

from which x4 = 0.00228.

x4 is (to < ½%) the same as x3.

Thus we take x = 0.00228 as the correct result.

From which: [H3O+] = 0.00228 M = a(H3O+)

and pH = 2.64.

Note: you could also solve the quadratic equation.

Question: what acids are strong, what acids are weak?


Strong acids are:

HCl, HBr, HI, HNO3, H2SO4, HClO4

Memorize this list.

Weak acids are:

every other acid.

If the acid is not on the above list it is weak.

Note about calculations in general.

When a relationship can be developed between several variables, knowledge of all but one can allow the calculation of that one.

In the case of acid/base equilibria:

HA + H2O H3O+ + A-
I (mol/L) a 0 0
C (-1)x x x
E (a - x) x x

Ka = x * x /(a - x)

This is a relationship in three variables, given any two the other can be calculated.

Problems can then:

give Ka and starting acid concentration, calculate pH (x)

give Ka and pH, calculate starting acid concentration.

give pH and starting acid concentration, calculate Ka.

This is a reminder and, of course, it applies to any relationship we have met or will meet in this course.

Calculation of H3O+: base solutions.

1. Strong bases.

This term base here is a hang over from the Arrhenius definition of a base (a substance that produces OH-.) A Brønsted base is a proton acceptor, strong bases in fact just dissociate in water:

NaOH(s) Na+ + OH-

We will assume that strong bases are 100% dissociated, so that the [OH-] from them is equal to the original [MOH].

Calculate the pH of 0.01 M Mg(OH)2.

Answer: [OH-] = 0.02,

pOH = 1.7

pH = 12.3 (pH + pOH = 14)

CARE: The calculated concentration is the [OH-], and - log of this is the pOH.

2. Very dilute solutions of strong bases.

Use same technique as for very dilute acid solutions.

3. Weak bases.

Weak bases do react with water to give OH-:

B + H2O BH+ + OH-

To find the [OH-] and hence the pH, requires the use of an equilibrium calculation.

Example: calculate the pH of a 0.0100 M solution of ammonia.

Step A: Analysis.

NH3 + H2O NH4+ + OH-
I (mol/L) 0.0100 0 0
C (-1)x x x
E 0.0100 - x x x

Kb = {a(NH4+) * a(OH-)} / a(NH3)

1.8 x 10-5 = x * x / (0.0100 - x)

Unknown, pH, obtainable from x.

Step B: Brainstorm (trivial)

Step C: Calculate

Try the approximation

1.8 x 10-5 = x * x / 0.0100

x = 4.24 x 10-4

Check the approximation: 4.24 x 10-4 * 100% / 0.01 = 4.24%

the approximation is valid.

x = [OH-] = 4.24 x 10-4 M

pOH = 3.4 (don't forget this!!!)

pH = 10.6