Calculate the relative concentration of phenol (ka=1.05x10^10) and phenolate ion in a solution with 0.05M total phenol (pH=7.5)
I believe you've made a typo with Ka.
Call phenol just HP.
HP ==> H^+ + P^-
Ka = (H^+)(P^-)/(HP)
Convert pH to (H^+) and substitute that along with Ka value and solve for the ratio of HP/P
To calculate the relative concentration of phenol and phenolate ion in a solution, we first need to determine the equilibrium constant (Ka) and then use the pH of the solution to calculate the concentrations.
Step 1: Convert the pH to the concentration of H+ ions
In this case, the pH is given as 7.5. The concentration of H+ ions can be calculated using the formula: [H+] = 10^(-pH).
Therefore, [H+] = 10^(-7.5).
Step 2: Use the equilibrium expression for the ionization of phenol to find the concentration of the phenolate ion (C6H5O-):
Ka = [C6H5O-][H+] / [C6H6]
Given that the total phenol concentration is 0.05 M, we can assume that initially, the concentration of phenol and phenolate ion are equal.
So, [C6H5O-] = [H+] = 10^(-7.5) (From Step 1)
Step 3: Calculate the concentration of phenol (C6H6):
Using the equation [C6H6] = total phenol concentration - [C6H5O-]
[C6H6] = 0.05 M - 10^(-7.5)
Now, let's calculate the relative concentration of phenol and phenolate ion:
[C6H5O-] / [C6H6] = (10^(-7.5)) / (0.05 M - 10^(-7.5))
Please note that the concentration of phenol is calculated by subtracting the concentration of phenolate ion from the total phenol concentration.
Using this equation, you can now substitute the values and calculate the relative concentration of phenol and phenolate ion in the solution.