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

The Kb of the following reaction = 6.3x10^-5

C6H5COO- + H2O -------------------> C6H5COOH + OH-

0.1 - 0 0

0.1 -x - x x

The equilibrium constant Kb = [C6H5COOH] [OH-]/[C6H5COO-]

= x² / (0.1-x)

= 6.3x10^-5

Assuming x to be very samll compared to 0.1, we can take the expression as

x2/0.1 = 6.3x10^-5 and solving for x

We get x = 2.5 x10^-2

^thus [C6H5COO-] = 0.1- 2.5 x10^-2

= 0.075M

and [C6H5COOH] ={OH-] = 2.5x10^-2 M

d) delta G = -2.303 x RT x log K

   = -2.303 x 8.314 x298 x log 6.3x10^-5

   = 23964.56J

   = + 23.96 kJ