Calculate the ionization constant, Ka, of phenol (HC6H5O), a weak acid, if a 0.25 M solution of it has a pH of 5.24.
1.3E-10
To calculate the ionization constant (Ka) of phenol (HC6H5O), we need to first understand the relationship between pH and the concentration of the acid (HC6H5O). The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration, [H+].
Given that the pH of the solution is 5.24, we can determine the hydrogen ion concentration [H+] using the equation:
[H+] = 10^(-pH)
[H+] = 10^(-5.24)
[H+] = 5.6 x 10^(-6) M
Phenol (HC6H5O) is a weak acid, which dissociates in water to produce hydrogen ions (H+) and phenolate ions (C6H5O-):
HC6H5O ⇌ H+ + C6H5O-
The ionization constant (Ka) is defined as the ratio of the concentration of the products (H+ and C6H5O-) to the concentration of the reactant (HC6H5O).
Ka = [H+][C6H5O-] / [HC6H5O]
We are given that the concentration of HC6H5O is 0.25 M. However, phenol is a weak acid, and it will not completely dissociate, so the concentration of HC6H5O will be the same as the initial concentration, 0.25 M.
Ka = [H+][C6H5O-] / [0.25]
Since phenol is a weak acid, we can assume that the concentration of H+ is equal to the concentration of C6H5O-. Therefore, we can replace [C6H5O-] with [H+].
Ka = [H+][H+] / [0.25]
Ka = [H+]^2 / [0.25]
Now, substitute the value of [H+] which we calculated to be 5.6 x 10^(-6) M:
Ka = (5.6 x 10^(-6))^2 / [0.25]
Ka = 3.14 x 10^(-11)
Therefore, the ionization constant (Ka) of phenol (HC6H5O) is approximately 3.14 x 10^(-11).