AcidBase: Fraction of Dissociation

Skills to develop

Define the fraction of dissociation of a weak electrolyte

Calculate the fraction of dissociation of a weak acid or base

Sketch the fraction of dissociation as a function of concentration

Fraction of Dissociation

A chemical equilibrium involving dissociation can be represented by the following reaction. AB = A + B, - - - K = [A][B]/[AB]

The concept equilibrium has been discussed in Mass action law and further discussed in Weak Acids and Bases, K_a, K_b, and K_w, and Exact pH Calculations.

Further, if C is the initial concentration of AB before any dissociation takes place, and f is the fraction of dissociated molecules, the concentration of AB is (1-f)C when the system has reached equilibrium. The concentration of A and B will each then be f C. Note that C is also the total concentration.

For convenience, we can write the concentration below the formula as:

      AB    =   A  +  B

    (1-f)C     fC    fC  - - -  Concentration

and the equilibrium constant, K = [A][B]/[AB] can be written as (using the concentration below the formula):

           f² C
    K = --------
          1 - f

Variation of f as a Function of C

How does f vary as a function of C? Common sense tells us that f has a value between 0 and 1 (0 < f < 1). For dilute solutions, f => 1, and for concentrate solutions, f => 0.

In solving the above equation for f, we obtain:

        - K + (K² + 4 K C)^1/2
   f = --------------------
               2 C

Note that f is the fraction of molecules that have dissociated. It is also called the degree of ionization. When converted to percentage, the term percent ionization is used. Even with the given formulation, it is still difficult to see how f varies as C changes. In the DOS version CACT, there is a program associated with this module for you to investigate the relationship between f and C. The program plots f as a function of C for a specified value of K. We now, illustrate the variation with a table below for a moderate value of K = 1.0e-5.

Variation of fraction of dissociation f as a function of concentration C when K = 1.0e-5.
C 1e-7 1e-6 1e-5 1e-4 1e-3 0.01 0.1 1.0 10 100
f 0.99 0.92 0.62 0.27 .095 .031 1e-2 3e-3 1e-3 3e-4

**Variation of fraction of dissociation f as a function of concentration C when K = 1.0e-5.**
C	1e-7	1e-6	1e-5	1e-4	1e-3	0.01	0.1	1.0	10	100
f	0.99	0.92	0.62	0.27	.095	.031	1e-2	3e-3	1e-3	3e-4

There is little change in f when C decrease from 1.0e-7 to 1.0e-6, but the changes are rather some what regular for every 10 fold decrease in concentration. Please plot f against a log scale of C to see the shape of the variation as your activity. Normally, we will not encounter solution as dilute as C = 1.0e-7, and we will never encounter solution as concentrate as 100 M either.

Confidence Building Questions

What is the fraction of dissociation for a strong acid?
Answer 1
Consider...
A strong acid is almost completely dissociated.
What is the fraction of dissociation for a compound that does not dissociate.
Answer 0
Consider...
Since there is no dissociation, the fraction is zero. This is redundant question.
A 0.1 M solution of an acid HB has half of its molecules dissociated. Calculate the acidity constant K_a.
Answer 0.05
Consider...
```
 HB  =  H⁺  +  B^-
0.05   0.05   0.05 M  <-- Concentrations at equilibrium.
```
K = ?
The equilibrium constant for a weak base B is 0.05, what is the fraction of dissociation if the concentration is 0.10 M?
Answer 0.5
Consider...
See the previous question.
The equilibrium constant for a weak base B is 1.0e-3, what is the fraction of dissociation if the concentration is 0.10 M?
Answer 0.095
Consider...
Make a table to see the variation of f when the concentration changes from 1e-7 to 100 in steps of 10 folds as given previously.