ACIDS & BASES - a partial treatment

(C) - Copyright, 2000 F.W. Boyle, Jr., Ph.D.

Acids are bases are special types of chemicals. They are separated from other materials because of their properties.

As a general rule, an acid is any substance which releases or provides H+ ions. This H+ ion then reacts with H2O to form the hydronium ion, H3O+, because of the charge on the ion and the polar nature of water. The hydronium ion, H3O+, is the reactive ion in solution. In order to save time, many chemists still refer only to the H+.

Complementary to acids are the compounds known as bases. Bases accept H+ ions from or release OH- ions to the solution.

The Bronsted-Lowry model of acids and bases states the following:

    1) an acid is a proton donor
    2) a base is a proton acceptor
    3) in an acid-base reaction, a proton is transferred from the 
       acid to the base.
The question then arises, "What is a proton donor?"

Consider that an acid releases a H+ ion into solution. Look at the H+ ion.

An H atom contains 1 proton in the nucleus and has a single electron orbiting that nucleus.

An H+ ion is an H atom that has lost an electron. An H atom that loses an electron is only a proton since there only was 1 proton and 1 electron at the start. Thus the H+ ion is a proton nd the terminology of Bronsted and Lowry uses this fact.

Acids and bases are classified according to the level to which they release H+ or OH-.

Strong acids release all their H+ and are said to dissociate completely. The following example uses the generic term, "A", to be the anion of the acid.

          HA <-->  H+ + A-

          Ka = [H+][A-]

With a strong acid there is no HA left so it does not appear in the equation for the constant.

Strong bases release all their OH- ions to solution and are also said to dissociate completely.

          MOH <-->  M+ + OH-

          Kb = [M][OH-]
M is a generic symbol for any metal.

As with the strong acid, a strong base has no MOH left so it is not included in the equation for the K.

Alternatively, the Bronsted-Lowry model shows the transfer of the H+ ion from one acid, HB, to another acid, HA.

          HB   +   A-    <-->    HA   +    B-
          |        |             |         |
  conjugate acid---|-------------|------conjugate base
               conjugate     conjugate
                 base          acid
In this version, the A- ion is known as the conjugate base of the acid, HA, because it accepts the H+ from HB. The B- ion is the conjugate base of acid, HB, and accepts the H+ from the HA in the reverse reaction.

The ionization constant for water

When an acid and a base are both introduced to the same solution, the acid releases its H+ ion and the base releases its OH- ion. The following reaction then occurs:

H+(aq) + OH-(aq) --> H2O(l)
Ionzation of water is the reverse of the above reaction.

H2O(l) <--> H+(aq) + OH-(aq)
Just as a constant is developed for other reactions, one is developed for the ionization of water.
             K = ---------
Now the [H2O] is constant in dilute solutions.

Note: Remember water's concentration is 55.55 M!!!!!!

So we modify the K above by leaving out the [H2O] since [H2O] is constant. Our new constant is called Kw and is shown below.

Kw = [H+]OH-] = 1.0x10-14
Also note that the -log of 1.0x10-14 is 14. It is the constant for this equilibrium that was used to create the pH scale. See my paper on pH for more info.

     ACID                          NEUTRAL                        BASE
pH   0                               7                              14
[H+] 1                             10-7                           10-14
     <-----  acidity increases              basicity increases  ----->

     Strong acids         Weak acids    Weak bases        Strong bases

We can develop both pH and pOH values for any system. Because the relationship between [H+] and [OH-] is constant, knowing one of the two concentrations determines the other.

pH = -log[H+] pOH = -log[OH-] pH = 14 - pOH

pOH = 14 - pH
If the pH or pOH is known, the [H+] and [OH- ] can be calculated.


A water sample is found to have a pH of 8.3. What is it's [H+]?

pH = -log[H+] = 8.3

Multiply both sides by -1.

log[H+] = -8.3

Take the antilog of both sides.

10log[H+] = 10-8.3

This is a mathematical identity.

10LOG(X) = X


[H+] = 10-8.3

And in more convenient numbers:

[H+] = 5.01 x 10-9 M = 0.00000000501 M


A water sample is found to have a pOH of 8.3. What is the pH of the sample?

Using the relationship: pH = 14 - pOH we find:

pH = 14 - 8.3 = 5.7

pH = -log[H+] = 5.7

Multiply both sides by -1.

log[H+] = -5.7

Take the antilog of both sides.

10log[H+] = 10-5.7

This is a mathematical identity.

10LOG(X) = X


[H+] = 10-5.7 M

And in more convenient numbers:

[H+] = 1.995 x 10-6 = 0.000001995 M

Please note that a solution with a pH of 5.7 has nearly 400 times more [H+] than a solution with a pH of 8.3.

Weak Acids and Bases
Contrary to strong acids and bases, weak acids and bases release only part of their H+ or OH- ions in to solution. One can still tritrate ALL the available H+ because as one H+ is used in the reaction another one will release from a molecule of the weak acid until all the available H+ ions are neutralized. A good indication that a weak acid of bsae is being titrated is that following the addition of the titrant, the indicator stays colored for some time but then loses its color as more ions are released. A weak base woks the same way in that the weak base only releases a portion of its total basicity to the solution at any given moment.

Recall that with strong acids the Equilibrium Constant, Ka, was:

K =[H+][A-]

Consider the dissociation of weak acid, HB, where B represents any anion.

HB <--> H+(aq) + B-(aq)

Because the dissociation is partial, some HB remains. When we form the EQ constant, we get:

         Ka = -------

In general, Ka for weak acids have values much less than 0.1.

The student should note from the text that any substance or ion that can cause the release of H+ ions to solution is considered an acid. If the K for the reaction is small (less than 0.01), the substance is considered a weak acid.

Example 3:

A solution of acetic acid is prepared. The acid concentration is 0.001 M. What is the pH of this solution?

The reaction:

HC2H3O2(aq) <--> H+ + C2H3O2-

The K for the reaction:

                        Ka = ------------

The method:

Set up a table for initial and final conditions. Look up the Ka for acetic acid.
Ka = 1.8x10-5

substance initial [ ] final [ ]
HC2H3O2 0.001 0.001 - x
H+ 0 x
C2H3O2- 0 x

Then substitute values into the Ka.

               Ka = ----------- = 1.8 x 10-5
For the first approximation, assume x is small compared to 0.001 so we have:

                -------  =  1.8 x 10-5


                x2 = 1.8 x 10-5 * 0.001

                x2 = 1.8 x 10-8

                  x = 1.34 x 10-4

Substitute this x in the denominator:

                 ---------------- = 1.8 x 10-5

And solve a second time:

                 ------------ = 1.8 x 10-5

                  x2 = 1.8 x 10-5 * 0.000866

                        x2 = 1.559 x 10-8

                        x = 1.25 x 10-4
The [H+] is 1.25 x 10-4 M.

The pH is:

pH = -log(1.25 x 10-4)

pH = 3.90

On to weak bases:

When considering weak bases, the same pattern of logic holds except that the substance accepts H+ ions from the solution.

Here's a weab base ionization to consider:

NH3(aq) + H2O <--> NH4+ + OH-
Write the Kb for the reaction.

        Kb = ----------

Note that the water is not included because its concentration is constant.

To determine the pH one could easily find the pOH and subtract that from 14 to get pH.

Ka to Kb
Ka is related to Kb by the Kw in the following way.

Ka*Kb = Kw

So if you know Ka or Kb, you can calculate the other.

        Ka = -------


        Kb = -------

On to the next chapter.