Dissociation Fraction
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.
A chemical equilibrium involving dissociation can be represented by the following reaction.
\(\ce{AB \rightleftharpoons A + B}, \hspace{20px} K = \ce{\dfrac{[A][B]}{[AB]}}\)
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 \(\ce{AB}\) before any dissociation takes place, and f is the fraction of dissociated molecules, the concentration of \(\ce{AB}\) is (1f)C when the system has reached equilibrium. The concentration of \(\ce{A}\) and \(\ce{B}\) 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:
\[\begin{array}{cccccl}
\ce{AB &\rightleftharpoons &A &+ &B&}\\
(1f)C &&fC &&fC &\leftarrow \ce{Concentration}
\end{array}\]
and the equilibrium constant, \(K = \ce{\dfrac{[A][B]}{[AB]}}\) can be written as (using the concentration below the formula):
\[K = \dfrac{(f\cancel{C})(fC)}{(1f)\cancel{C}} = \dfrac{f^2 C}{1  f} \label{2}\]
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, \(f \approx 1\), and for concentrated solutions, \(f \approx 0\). In solving Equation \(\ref{2}\) for f, we obtain:
\[f = \dfrac{  K + \sqrt{K^2 + 4 K C}}{2 C}\]
\(f\) is the fraction of molecules that have dissociated and 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. Table \(\PageIndex{1}\) illustrate the variation with a table below for a moderate value of \(K = 1.0 \times 10^{5}\).
C  \(1\times 10^{7}\)  \(1\times 10^{6}\)  \(1\times 10^{5}\)  \(1\times 10^{4}\)  \(1\times 10^{3}\)  \(1\times 10^{2}\)  0.1  1.0  10  100 

f  0.99  0.92  0.62  0.27  0.095  0.031  \(1\times 10^{2}\)  \(3\times 10^{3}\)  \(1\times 10^{3}\)  \(3\times 10^{4}\) 
There is little change in \(f\) when \(C\) decreases from \(1.0 \times 10^{7}\) to \(1.0 \times 10^{6}\), but the changes are rather somewhat 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 = 1.0e7, and we will never encounter solution as concentrated as 100 M either.
Questions
 What is the fraction of dissociation for a strong acid?
 What is the fraction of dissociation for a compound that does not dissociate?
 A 0.1 M solution of an acid \(\ce{HB}\) has half of its molecules dissociated. Calculate the acidity constant Ka.
 The equilibrium constant for a weak base \(\ce{B}\) is 0.05; what is the fraction of dissociation if the concentration is 0.10 M?
 The equilibrium constant for a weak base \(\ce{B}\) is 1.0e3; what is the fraction of dissociation if the concentration is 0.10 M?
Solutions

Answer: 1
Consider...
A strong acid is almost completely dissociated. 
Answer: 0
Consider...
Since there is no dissociation, the fraction is zero. This is a redundant question. 
Answer: 0.05
Consider...\(\begin{array}{cccccl}
\ce{HB &\rightleftharpoons &H+ &+ &B &}\\
0.05 &&0.05 &&0.05\: \ce M &\leftarrow \textrm{Concentrations at equilibrium}
\end{array}\)K = ?

Answer: 0.5
Consider...
See the previous question. 
Answer: 0.095
Consider...
Make a table to see the variation of f when the concentration changes from 1e7 to 100 in steps of 10 folds as given previously.
Contributors
Chung (Peter) Chieh (Chemistry, University of Waterloo)