Question

Sodium benzoate (NaC₇H₅O₂) is sometimes used as a preservative in foods. Benzoic acid is the conjugate acid of the benzoate ion, which has a K_a value of 6.3 x 10^-5.  

1. Calculate the K_b value for the benzoate ion. (1 mark)
2. Use the K_b value for the benzoate ion from part a, calculate the pH of a 0.215 M solution of the sodium benzoate.  (Ignore x method, 4 marks)

Answer 1

Kb is calculated:

Kb = Kw / Ka = 10 ^ -14 / 6.3E-5 = 1.60E-10

You have the reaction with water:

C7H5O2- + H2O = HC7H5O2 + OH-

We have the expression of Kb:

Kb = [HC7H5O2] * [OH-] / [C7H5O2-]

1.60E-10 = X ^ 2 / 0.215

Clears:

X = [OH-] = 5.90E-6 M

The pOH and pH are calculated:

pOH = - log [OH-] = - log 5.90E-6 = 5.23

pH = 14 - pOH = 14 - 5.23 = 8.77

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