# Dissociation Fraction

- Page ID
- 35933

Learning Objectives

- 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 (1-*f*)*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&}\\

(1-f)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):

\[K = \dfrac{(f\cancel{C})(fC)}{(1-f)\cancel{C}} = \dfrac{f^2 C}{1 - f} \label{2}\]

### 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* = 1.0e-7, 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 1.0e-3; 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 1e-7 to 100 in steps of 10 folds as given previously.

## Contributors and Attributions

Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)