Question

Sodium benzoate NaC6H5CO2, kb = 6.3×10^-5. is iften used as a preservative in food products if it is added water to bring the concentartaion to .10M. the benzoate ion, C6H5CO^-2 reacts with water according to the following equiliburium:
C6H5CO^-2 + H2O ➡ C6 H5 CO2 +OH^-1 Kb= 6.3×10^-5

a. write the equibulirum expression for the reactions.

b. complete an ICE table using the initial concentration of .10 M venzoate ion.

c. determined what equibulirum concentration of (C6H5CO2^-1), ( C6H5CO2H) and (OH^-1).

d. determined the Delta G for the reaction.

Answer 1

C6H5COO- + H2O --------------> C6 H5 CO2H + OH-      Kb= 6.3×10^-5

a)

equilibrium expression = [C6 H5 CO2H] [OH-] / [C6H5COO-]

b)

C6H5COO- + H2O --------------> C6 H5 CO2H + OH-

0.10                                                       0                0          -----------> I

-x                                                         + x              + x        ------------> C

0.10-x                                                   x                  x          ------------> E

c)

Kc = [C6 H5 CO2H] [OH-] / [C6H5COO-]

Kb = x^2 / 0.1 - x

6.3×10^-5 = x^2 / 0.1 - x

x = 2.48 x 10^-3

[[C6 H5 CO2H] = 2.48 x 10^-3 M

[OH-] = 2.48 x 10^-3 M

[C6 H5 CO2-] = 0.0975 M

d)

delta G = - RT ln K

            = - 8.314 x 10^-3 x 298 x ln (6.3 x 10^-5)

           = 23.96 kJ/mol