Answer:
a) To find the equilibrium constant, we can use the following equation:
Kw = Ka x Kb
where Kw is the ion product constant of water (1.0 x 10^-14 at room temperature), Ka is the acid dissociation constant for benzoic acid, and Kb is the base dissociation constant for the conjugate base of benzoic acid (which is the sodium benzoate in this case).
Since pOH = 5.31, we can find the pH as follows:
pH + pOH = 14
pH = 14 - 5.31 = 8.69
Now, we can use the pH to find the [H+] concentration:
pH = -log[H+]
[H+] = 10^(-pH) = 1.28 x 10^(-9) M
Since sodium benzoate is a salt of a weak acid and a strong base, it will undergo hydrolysis in water:
C6H5COO- + H2O ⇌ C6H5COOH + OH-
The equilibrium constant for this reaction can be expressed as:
Kb = [C6H5COOH][OH-] / [C6H5COO-]
Since Kb and Kw are related, we can solve for Ka using the following equation:
Ka = Kw / Kb = 1.0 x 10^-14 / Kb
Substituting the given values, we get:
Kb = [OH-]^2 / [C6H5COO-] = (1.28 x 10^-9)^2 / 0.15 = 1.09 x 10^-17
Therefore,
Ka = 1.0 x 10^-14 / Kb = 9.17 x 10^-8
b) The Ka value for benzoic acid is 9.17 x 10^-8.
c) Since the pH of the saturated solution is 2.81, we can find the [H+] concentration as:
pH = -log[H+]
[H+] = 10^(-pH) = 1.26 x 10^(-3) M
We can assume that the dissociation of benzoic acid is negligible in this case, since the pH is well below its pKa (4.20). Therefore, the molar solubility (S) of benzoic acid can be expressed as:
S = [C6H5COOH] = Ksp / [H+]^2
where Ksp is the solubility product constant for benzoic acid. Since benzoic acid is a weak acid, we can assume that it dissociates very little in water, and therefore, the solubility product can be expressed as:
Ksp = [C6H5COOH]^2
Substituting the given values, we get:
S = Ksp / [H+]^2 = [C6H5COOH]^2 / [H+]^2 = 4.67 x 10^-4 M
Therefore, the molar solubility of benzoic acid in water is 4.67 x 10^-4 M.
Step-by-step explanation: