Find the pH of the mixture 0.095 M in HF and 0.225 M in HC6H5O

Answer 1

To find the pH of the mixture, we first need to determine the concentration of H+ ions in the solution.

HF is a weak acid and partially dissociates in water as follows:

HF + H2O ⇌ H3O+ + F-

The equilibrium constant expression for this reaction is:

Ka = [H3O+][F-]/[HF]

where Ka is the acid dissociation constant of HF.

Similarly, HC6H5O (also known as phenol or C6H5OH) is a weak acid that partially dissociates in water as follows:

HC6H5O + H2O ⇌ H3O+ + C6H5O-

The equilibrium constant expression for this reaction is:

Ka = [H3O+][C6H5O-]/[HC6H5O]

where Ka is the acid dissociation constant of HC6H5O.

We can use the concentrations of HF and HC6H5O and their respective Ka values to calculate the concentrations of H3O+ and F- or C6H5O- ions in the solution. Then, we can use the following equation to calculate the pH:

pH = -log[H3O+]

To simplify the calculation, we can assume that the contribution of the autoionization of water to the concentration of H3O+ is negligible compared to the contributions from HF and HC6H5O.

Using the Ka values of HF and HC6H5O, we can calculate the equilibrium concentrations of H3O+ and F- or C6H5O- ions:

For HF:
Ka = [H3O+][F-]/[HF]
[H3O+] = sqrt(Ka[HF]) = sqrt(6.6 x 10^-4 x 0.095) = 0.031 M
[F-] = [HF] - [H3O+] = 0.095 - 0.031 = 0.064 M

For HC6H5O:
Ka = [H3O+][C6H5O-]/[HC6H5O]
[H3O+] = sqrt(Ka[HC6H5O]) = sqrt(1.3 x 10^-10 x 0.225) = 1.03 x 10^-6 M
[C6H5O-] = [HC6H5O] - [H3O+] = 0.225 - 1.03 x 10^-6 = 0.225 M (assuming that [H3O+] is negligible compared to [HC6H5O])

Now we can calculate the total concentration of H3O+ in the solution:

[H3O+]total = [H3O+] from HF + [H3O+] from HC6H5O
[H3O+]total = 0.031 + 1.03 x 10^-6 = 0.031 M

Finally, we can calculate the pH of the solution:

pH = -log[H3O+]total = -log(0.031) = 1.51

Therefore, the pH of the mixture 0.095 M in HF and 0.225 M in HC6H5O is 1.51.