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 + Band the equilibrium constant, K = [A][B]/[AB] can be written as (using the concentration below the formula):
(1-f)C fC fC - - - Concentration
f^{2} C K = -------- 1 - f
In solving the above equation for f, we obtain:
- K + (K^{2} + 4 K C)^{1/2} f = -------------------- 2 CNote 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.
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.
Answer 1
Consider...
A strong acid is almost completely dissociated.
Answer 0
Consider...
Since there is no dissociation, the fraction is zero.
This is redundant question.
Answer 0.05
Consider...
HB = H^{+} + B^{-} 0.05 0.05 0.05 M <-- Concentrations at equilibrium.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.