Calculate the [H3O+] of the following polyprotic acid solution: 0.145 M H3C6H5O3?
To calculate the concentration of H3O+ in a polyprotic acid solution, we need to consider the ionization steps of the acid and the corresponding equilibrium constants.
H3C6H5O3 is a polyprotic acid, meaning it can donate more than one proton (H+). In this case, it has three ionization steps:
H3C6H5O3 ⇌ H+ + HC6H5O3- (Equation 1)
HC6H5O3- ⇌ H+ + C6H5O3^2- (Equation 2)
C6H5O3^2- ⇌ H+ + C6H5O3^3- (Equation 3)
The equilibrium constant for each equation is given by the acid dissociation constant (Ka).
Given that H3C6H5O3 is a weak acid, we assume that its ionization is incomplete, so we can use the initial concentration of the acid as the concentration of the acid before ionization.
To find the concentration of H+ from each ionization step, we can use the expression:
[H+] = √(Ka * [H3C6H5O3]) for Equation 1
[H+] = √(Ka * [HC6H5O3-]) for Equation 2
[H+] = √(Ka * [C6H5O3^2-]) for Equation 3
First, we need to determine the values of Ka for each equation. If the values of Ka are not provided, we can look them up in reference sources or use approximate values based on acid strengths.
Let's assume the following approximate Ka values for each equation:
Ka1 = 1.0 x 10^-5
Ka2 = 1.0 x 10^-8
Ka3 = 1.0 x 10^-10
Now, we can calculate the concentration of H+ from each equation separately.
[H+]1 = √(Ka1 * [H3C6H5O3]) = √((1.0 x 10^-5) * (0.145 M))
[H+]2 = √(Ka2 * [HC6H5O3-])
[H+]3 = √(Ka3 * [C6H5O3^2-])
Since [H+]2 and [H+]3 depend on the concentration of the previous species, we need first to solve Equation 1 before proceeding to Equation 2 and so on.
After solving all three equations and obtaining the concentrations of H+ from each ionization step, we can sum them up to get the total concentration of H+ ([H3O+]) in the solution:
[H3O+] = [H+]1 + [H+]2 + [H+]3
By plugging in the calculated values, you can find the final concentration of H3O+.