Calculate the pH of an aqueous solution at 25 degress C that is .41 M in phenol (C6H5OH). Ka for phenol=1.3 x 10^-10.

Same type problem

To calculate the pH of the aqueous solution of phenol, we need to determine the concentration of H3O+ ions in the solution. This can be done using the equilibrium constant expression, also known as the acid dissociation constant (Ka).

The chemical equation for the dissociation of phenol (C6H5OH) in water is:

C6H5OH (aq) ⇌ C6H5O- (aq) + H+ (aq)

Given that the concentration of phenol is 0.41 M, we can assume that the concentration of C6H5O- is negligible compared to phenol's initial concentration. This is because phenol is a weak acid, and only a small fraction of it will dissociate into H+ ions and C6H5O- ions in water.

Using the Ka expression for phenol:

Ka = [C6H5O-][H+] / [C6H5OH]

We can rearrange the equation to solve for [H+]:

[H+] = (Ka * [C6H5OH]) / [C6H5O-]

Substituting the given values:

Ka = 1.3 x 10^-10
[C6H5OH] = 0.41 M

[H+] = (1.3 x 10^-10 * 0.41) / [C6H5O-]

Now, we need to determine the concentration of C6H5O-. This can be done by assuming that phenol fully dissociates, and using the principle of conservation of mass:

[C6H5OH] = [C6H5O-] + [H+]

Since [C6H5O-] is small compared to [C6H5OH], we can approximate the equation as:

[C6H5OH] ≈ [H+]

Now we can solve for [H+]:

[H+] = (1.3 x 10^-10 * 0.41) / [C6H5O-]
[H+] = (1.3 x 10^-10 * 0.41) / 0.41

[H+] ≈ 1.3 x 10^-10

Now that we know the concentration of H+ ions, we can calculate the pH. The pH is defined as the negative logarithm (base 10) of the concentration of H+ ions:

pH = -log[H+]
pH = -log(1.3 x 10^-10)

Calculating this value gives pH ≈ 9.89.

Therefore, the pH of the aqueous solution of phenol (C6H5OH) is approximately 9.89 at 25 degrees Celsius.