Question

Question:

Consider a .045 M solution of benzoic acid (C₆H₅CO₂H) whose K_a value is 6.5 x 10^-5.

A). write the acid-base reaction equation for this acid with water:

B). write the equilibrium expression (K_a) for this reaction:

C). Calculate the concentration of benzoic acid, benzoate ion (the conjugate base), and hydronium ion for this solution.

D). Calculate the pH of this solution.

E). Calculate the concentration of hydroxide ions in this solution.

Answer 1

A) C6H5COOH(l) + H2O(l) <----> H3O+(aq) + C6H5COO-(aq)

B) Ka = [C6H5COO-][H3O+]/[C6H5COOH]

C) Ka = 6.5*10^-5

(6.5*10^-5 ) = X*X/(0.045-X)

X = 0.001678 M

so that,

[C6H5COO-] = X = 0.001678 M

[H3O+] = X = 0.001678 M

[C6H5COOH] = 0.045-0.001678 = 0.0433 M

D) pH = -log[H3O+]

      = -log(0.001678)

      = 2.77

e) [OH-][H3O+] = KW = 1*10^-14

[OH-]= KW/[H3O+]

        = (1*10^-14)/(1.678*10^-3)

        = 5.96*10^-12 M