In: Chemistry
A certain weak acid, HA, has a Ka value of 6.7 X 10^-7.
Part A
Calculate the percent ionization of HA in a 0.10 M solution. Express your answer to two significant figures and include the appropriate units.
Part B
Calculate the percent ionization of HA in a 0.010 M solution.
Express your answer to two significant figures, and include the appropriate units.
Concepts and reason
This problem is based on the concept of ions and ionic equilibrium. A given substance on dissolution insolvent, which is generally water, yields either an ionic solution or a molecular solution. In an ionic solution substance splits up into ions, whereas in the latter, it is present.
Fundamentals
The extent of dissociation of a substance can be expressed in terms of the degree of dissociation, which is, by definition, equal to the fraction of total substance present in the form of ions.
(a) Consider a weak acid of type HA. It will dissociate as follows:
If \(\alpha\) is the degree of dissociation at a given concentration \(C\) of \(\mathrm{HA}\), then the concentrations in solution are as follows:
\([\mathrm{HA}]=C(1-\alpha)\)
\(\left[\mathrm{A}^{-}(a q)\right]=C \alpha\)
And, \(\left[\mathrm{H}^{+}(a q)\right]=C \alpha\)
since we know \(K_{a}\) is the dissociation constant and can be written as follows:
\(K_{a}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)
Substitute \(C(1-\alpha)\) for \([\mathrm{HA}], C \alpha\) for \(\left[\mathrm{A}^{-}\right]\) and \(\left[\mathrm{H}^{+}\right]\)
\(K_{a}=\frac{(C \alpha)(C \alpha)}{C(1-\alpha)}\)
\(=\frac{C^{2} \alpha^{2}}{C(1-\alpha)}\)
since \(\alpha\) is usually very small and thus negligible in comparison to unity. \((1-\alpha) \approx 1\)
Thus,
\(K_{a}=\frac{C \alpha^{2}}{1}\)
\(\alpha=\sqrt{\frac{K a}{C}}\)
Now substitute \(6.7 \times 10^{-7}\) for \(K_{a}\) and \(0.10 \mathrm{M}\) for \(C\) in the equation for \(\alpha\).
\(\alpha=\sqrt{\frac{6.7 \times 10^{-7}}{0.10}}\)
\(=0.00026\)
Percentage ionization is thus calculated as follows:
\(\%=\frac{(0.00026) 100}{0.10}\)
\(=0.26 \%\)
Here, \(\%\) is percentage ionization.
The value of the equilibrium constant is characteristic of a given weak electrolyte and depends only on the temperature. It is independent of the individual concentrations. Suppose a strong electrolyte is added to a weak electrolyte; even then, the same expression, as shown above, fits well. The effect of a strong electrolyte is to suppress the extent of dissociation of the weak electrolyte.
(b) The dissociation constant is written as follows:
$$ K_{a}=\frac{C \alpha^{2}}{1} $$
\(\alpha=\sqrt{\frac{K a}{C}}\)
Now substitute \(6.7 \times 10^{-7}\) for \(K_{a}\) and \(0.010 \mathrm{M}\) for \(C\) in the equation for \(\alpha\).
$$ \alpha=\sqrt{\frac{6.7 \times 10^{-7}}{0.010}} $$
\(=0.000081\)
Percentage ionization is thus calculated as:
\(\%=\frac{(0.000081) 100}{0.010}\)
\(=0.81 \%\)
Here, \(\%\) is percentage ionization.
The value of the equilibrium constant is characteristic of a given weak electrolyte and depends only on the temperature. It is independent of the individual concentrations. Suppose a strong electrolyte is added to a weak electrolyte; even then, the same expression, as shown above, fits well. The effect of a strong electrolyte is to suppress the extent of dissociation of the weak electrolyte.
Part a
The percent ionization is \(0.26 \%\).
Part b
The percent ionization is \(0.81 \%\).