Question

The acid-dissociation constant for benzoic acid ( C 6 H 5 COOH) is 6.3× 10 −5 .

Part A Calculate the equilibrium concentration of H3O+ in the solution if the initial concentration of C6H5COOH is 5.7×10−2 M . Express your answer using two significant figures.

Part B Calculate the equilibrium concentration of C6H5COO− in the solution if the initial concentration of C6H5COOH is 5.7×10−2 M . Express your answer using two significant figures.

Part C Calculate the equilibrium concentration of C6H5COOH in the solution if the initial concentration of C6H5COOH is 5.7×10−2 M . Express your answer using two significant figures.

Answer 1

part A

acid-dissociation constant for benzoic acid(ka) = 6.3*10^-5

from ionic equilibrium

ka = Cx^2

c = initial concentration = 5.7*10^−2 M

(6.3*10^-5 ) = (5.7*10^−2)*x^2

x = degree of dissociation = 0.033

[H3O+] at equilibrium = cx

         = (5.7*10^−2)*0.033

        = 0.0019 M

part B

   C6H5COOH(aq) <-----> c6h5coo-(aq) + H3O+(aq)

[C6H5COO-] = [H3O+] at equilibrium = cx

= (5.7*10^−2)*0.033

= 0.0019 M

part C

       
[C6H5COOH] at equilibrium = C - cx

          C = initial concentration = 5.7*10^−2 M

cx = concentration of C6H5COOH reacted = 0.0019 M

[C6H5COOH] at equilibrium = (5.7*10^−2) - 0.0019

            = 0.055 M