An unknown weak monprotic acid with the general formula HA has a Ka=5.5 x 10^-7. what is the value of the equilibrium constant that describes the following reaction?
A- +H2O = HA +OH-
Keq = (HA)(OH^&-)/(A^-)(H2O)
To find the value of the equilibrium constant (Kw) that describes the given reaction, we need to understand the relationship between acid dissociation constant (Ka) and the ion product of water (Kw).
The ion product of water is defined as the equilibrium constant for the autoionization of water:
2H2O ⇌ H3O+ + OH-
At 25°C, the value of Kw is approximately 1.0 x 10^-14. This value represents the product of the concentrations of H3O+ and OH- ions in water at equilibrium.
Now, let's look at the provided reaction:
A- + H2O ⇌ HA + OH-
In this reaction, A- is the conjugate base of the weak acid HA. When HA reacts with water, it donates a proton (H+) to form H3O+. Simultaneously, OH- is produced as A- accepts a proton from water.
Here's the key point: the concentration of H3O+ in the reaction is equal to the concentration of HA, and the concentration of OH- is equal to the concentration of A-.
Therefore, we can write the equilibrium concentrations as:
[H3O+] = [HA]
[OH-] = [A-]
Now, if we compare these concentrations to the ion product of water (Kw), we get:
Kw = [H3O+][OH-] = [HA][A-]
Since [HA][A-] is the product of the concentrations of the acid and its conjugate base, it represents the equilibrium constant for the given reaction.
Finally, substituting the given Ka value (5.5 x 10^-7) for [HA][A-], we have:
Kw = [HA][A-] = 5.5 x 10^-7
Thus, the value of the equilibrium constant (Kw) that describes the reaction A- + H2O ⇌ HA + OH- is 5.5 x 10^-7.