The ionization constant, dissociation constant or acidity constant, is a property that reflects the tendency of a substance to release hydrogen ions; that is, it is directly related to the strength of an acid. The higher the value of the dissociation constant (Ka), the greater the release of hydrogen ions by the acid..
When it comes to water, for example, its ionization is known as 'autoprotolysis' or 'autoionization'. Here, a water molecule yields an H+ to another, producing the H ions3OR+ and OH-, as seen in the image below.
The dissociation of an acid from an aqueous solution can be outlined as follows:
HA + HtwoOR <=> H3OR+ + TO-
Where HA represents the acid that is ionized, H3OR+ to the hydronium ion, and A- its conjugate base. If the Ka is high, more of the HA will dissociate and there will therefore be a higher concentration of the hydronium ion. This increase in acidity can be determined by observing a change in the pH of the solution, whose value is below 7.
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The double arrows in the upper chemical equation indicate that a balance is established between reactants and product. As all equilibrium has a constant, the same happens with the ionization of an acid and is expressed as follows:
K = [H3OR+][TO-] / [HA] [HtwoOR]
Thermodynamically, the constant Ka is defined in terms of activities, not concentrations. However, in dilute aqueous solutions the activity of water is around 1, and the activities of the hydronium ion, the conjugate base and the undissociated acid are close to their molar concentrations..
For these reasons, the use of the dissociation constant (ka) was introduced which does not include the water concentration. This allows the weak acid dissociation to be schematized in a simpler way, and the dissociation constant (Ka) is expressed in the same form..
HA <=> H+ + TO-
Ka = [H+][TO-] / [HA]
The dissociation constant (Ka) is a form of expression of an equilibrium constant.
The concentrations of the undissociated acid, the conjugate base, and the hydronium or hydrogen ion remain constant once the equilibrium condition is reached. On the other hand, the concentration of the conjugate base and that of the hydronium ion are exactly the same.
Its values are given in powers of 10 with negative exponents, so a simpler and more manageable form of expression of Ka was introduced, which they called pKa.
pKa = - log Ka
PKa is commonly called the acid dissociation constant. The pKa value is a clear indication of the strength of an acid.
Those acids that have a pKa value less or more negative than -1.74 (pKa of the hydronium ion) are considered as strong acids. While acids that have a pKa greater than -1.74, they are considered non-strong acids..
An equation that is extremely useful in analytical calculations is deduced from the Ka expression..
Ka = [H+][TO-] / [HA]
Taking logarithms,
log Ka = log H+ + log A- - log HA
And solving for log H+:
-log H = - log Ka + log A- - log HA
Then using the definitions of pH and pKa, and regrouping terms:
pH = pKa + log (A- / HA)
This is the famous Henderson-Hasselbalch equation.
The Henderson-Hasselbach equation is used to estimate the pH of buffers, as well as how the relative concentrations of conjugate base and acid influence pH..
When the concentration of the conjugate base is equal to the concentration of the acid, the relationship between the concentrations of both terms is equal to 1; and therefore, its logarithm is equal to 0.
As a consequence, pH = pKa, this being very important, since in this situation the buffering efficiency is maximum..
The pH zone where there is the maximum buffering capacity is usually taken, that where the pH = pka ± 1 pH unit.
The dilute solution of a weak acid has the following concentrations in equilibrium: undissociated acid = 0.065 M and concentration of the conjugate base = 9 · 10-4 M. Calculate the Ka and pKa of the acid.
The concentration of the hydrogen ion or the hydronium ion is equal to the concentration of the conjugate base, since they come from the ionization of the same acid.
Substituting in the equation:
Ka = [H+][TO-] / HA
Substituting in the equation for their respective values:
Ka = (910-4 M) (910-4 M) / 6510-3 M
= 1,246 10-5
And then calculating its pKa
pKa = - log Ka
= - log 1,246 10-5
= 4,904
A weak acid with a concentration of 0.03 M, has a dissociation constant (Ka) = 1.5 · 10-4. Calculate: a) pH of the aqueous solution; b) the degree of ionization of the acid.
At equilibrium, the acid concentration is equal to (0.03 M - x), where x is the amount of acid that dissociates. Therefore, the concentration of hydrogen or hydronium ion is x, as is the concentration of the conjugate base.
Ka = [H+][TO-] / [HA] = 1.5 · 10-6
[H+] = [A-] = x
Y [HA] = 0.03 M-x. The small value of Ka indicates that the acid probably dissociated very little, so (0.03 M - x) is approximately equal to 0.03 M.
Substituting in Ka:
1.5 10-6 = xtwo / 3 10-two
xtwo = 4.5 10-8 Mtwo
x = 2.12 x 10-4 M
And since x = [H+]
pH = - log [H+]
= - log [2.12 x 10-4]
pH = 3.67
And finally, regarding the degree of ionization: it can be calculated using the following expression:
[H+] or [A-] / HA] x 100%
(2.12 10-4 / 3 10-two) x 100%
0.71%
I calculate Ka from the percentage of ionization of an acid, knowing that it ionizes by 4.8% from an initial concentration of 1.5 · 10-3 M.
To calculate the amount of acid that is ionized, its 4.8% is determined.
Ionized amount = 1.5 · 10-3 M (4.8 / 100)
= 7.2 x 10-5 M
This amount of ionized acid is equal to the concentration of the conjugate base and the concentration of the hydronium or hydrogen ion at equilibrium..
The acid concentration at equilibrium = initial acid concentration - the amount of the ionized acid.
[HA] = 1.5 · 10-3 M - 7.2 10-5 M
= 1,428 x 10-3 M
And then solving with the same equations
Ka = [H+][TO-] / [HA]
Ka = (7.2 · 10-5 M x 7.2 10-5 M) / 1,428 10-3 M
= 3.63 x 10-6
pKa = - log Ka
= - log 3.63 x 10-6
= 5.44
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