A 0.25 mol/L solution of benzoic acid, KC7H5O2 and antiseptic also used as a food preservative, has a pH of 2.40. Calculate the Ka of benzoic acid at SATP
How does the benzoic acid ionization go?
C6H5COOH ==> C6H5COO^- + H^+
The ionization of benzoic acid (C7H6O2) in water can be represented by the following equation:
C7H6O2 + H2O ⇌ C7H5O2- + H3O+
In this equation, benzoic acid (C7H6O2) reacts with water (H2O) to form the benzoate ion (C7H5O2-) and the hydronium ion (H3O+). The Ka value represents the acid dissociation constant and can be calculated using the equilibrium concentrations of the ions.
To find the Ka value, we need the initial concentration of benzoic acid, the concentration of the benzoate ion at equilibrium, and the concentration of hydronium ion at equilibrium.
Given that the benzoic acid is in a 0.25 mol/L solution, we know that the initial concentration of benzoic acid ([C7H6O2]) is 0.25 mol/L.
Since benzoic acid is a weak acid, we assume that the concentration of benzoic acid that ionizes ([C7H6O2]) at equilibrium is negligible compared to the initial concentration. Therefore, we can assume that the change in concentration of benzoic acid ([C7H6O2]) is equal to -x (the moles that react).
As a result of the ionization, an equal molar amount of benzoate ion ([C7H5O2-]) and hydronium ion ([H3O+]) will be produced. Therefore, the change in concentration of benzoate ion ([C7H5O2-]) and hydronium ion ([H3O+]) will be +x.
At equilibrium, the concentration of benzoic acid ([C7H6O2]) is equal to the initial concentration of benzoic acid minus the change in concentration (-x), which is [C7H6O2] - x.
Similarly, the concentration of benzoate ion ([C7H5O2-]) at equilibrium will be equal to the change in concentration (+x), which is [C7H5O2-] = x.
The concentration of hydronium ion ([H3O+]) at equilibrium will also be equal to the change in concentration (+x), which is [H3O+] = x.
The pH of the solution is given as 2.40, which indicates the concentration of hydronium ion ([H3O+]). pH is defined as the negative logarithm (base 10) of the concentration of hydronium ion. Therefore, we can find the concentration of hydronium ion as 10^(-pH).
Now, we can set up the equilibrium expression and solve for Ka:
Ka = ([C7H5O2-] * [H3O+]) / [C7H6O2]
Substituting the previously determined concentrations:
Ka = (x * x) / (0.25 - x)
We need to remember that the value of x is very small compared to the initial concentration (0.25 mol/L). Therefore, we can assume that 0.25 - x ≈ 0.25.
By substituting this approximation, the equation becomes:
Ka = (x^2) / 0.25
Finally, we substitute the value of x as 10^(-pH) and the value of pH as 2.40 in the equation to calculate the Ka:
Ka = [(10^(-2.40))^2] / 0.25
Simplifying further, we get the Ka value of benzoic acid at SATP.