Calculate [H3O^-] in a 0.15M solution of benzoic acid, HC7H5O2(aq) having a Ka of 6.4 x 10^-5
I got.
6.4x10^-5=(x)(x)/0.15
6.4x10^-5= x^2/0.15
6.4x10^-5 x 0.15 = x^2
square root of 9.6x10^-6 = square root of x^2
x= 0.003098
The pH solution is -log [0.003098]
= 2.50
b.) Calculate the percent ionization of the HC7H5O.
%ion= (H^+)/0.15x100???? not sure of this?
a is ok.
b is ok if you used
%ion = [(H^+)/0.15]*100 = about 2%
To calculate [H3O^-] in a 0.15M solution of benzoic acid (HC7H5O2), we can use the expression for the ionization of the acid:
Ka = [H3O^+][C7H5O2^-] / [HC7H5O2]
Given that the Ka value for benzoic acid is 6.4x10^-5, we can assume that the ionization of benzoic acid is small, so we can ignore the [C7H5O2^-] concentration compared to the initial concentration of HC7H5O2.
Therefore, we have:
Ka = [H3O^+][C7H5O2^-] / [HC7H5O2]
6.4x10^-5 = [H3O^+][C7H5O2^-] / 0.15M
Since [H3O^+] is equal to [C7H5O2^-], we can substitute [H3O^+] with x:
6.4x10^-5 = x^2 / 0.15
Rearranging the equation gives us:
x^2 = 6.4x10^-5 * 0.15
x^2 = 9.6x10^-6
x = √(9.6x10^-6)
x ≈ 0.0031
Therefore, [H3O^-] is approximately 0.0031M.
For part b, to calculate the percent ionization of HC7H5O2, we can use the equation:
% ionization = ([H3O^+]/[HC7H5O2]) * 100
Substituting the values:
% ionization = (0.0031M / 0.15M) * 100
% ionization = 2.0667%
Therefore, the percent ionization of HC7H5O2 is approximately 2.07%.
To calculate the concentration of H3O+ in a solution of benzoic acid, you need to consider the dissociation of benzoic acid (HC7H5O2) in water.
The equation for the dissociation of benzoic acid is:
HC7H5O2(aq) ⇌ H+(aq) + C7H5O2^-(aq)
The Ka value represents the equilibrium constant for this dissociation reaction. In this case, the Ka value is given as 6.4 x 10^-5.
To calculate [H3O+], you can use the equation for the Ka expression:
Ka = [H+][C7H5O2^-] / [HC7H5O2]
Since [C7H5O2^-] and [HC7H5O2] are both initially equal to 0.15 M (as given in the problem), you can substitute these concentrations into the equation:
6.4 x 10^-5 = [H+][0.15] / 0.15
Simplifying the equation, you get:
[H+] = 6.4 x 10^-5 M
Therefore, the concentration of H3O+ in the 0.15 M solution of benzoic acid is 6.4 x 10^-5 M.
Now, moving on to part b), to calculate the percent ionization of benzoic acid, you need to determine the concentration of ionized benzoic acid ([C7H5O2^-]) relative to the initial concentration of benzoic acid ([HC7H5O2]).
You can use the equation:
% ionization = ([C7H5O2^-] / [HC7H5O2]) x 100
Since [C7H5O2^-] can be considered equal to [H+], you can use the concentration of H3O+ you calculated earlier:
% ionization = (6.4 x 10^-5 / 0.15) x 100
Calculating this, you get:
% ionization = 0.043%
Therefore, the percent ionization of benzoic acid in this solution is 0.043%.