Question

Calculate the pH of a 0.50 M solution of sodium benzoate (NaC₆H₅COO) given that the K_a of benzoic acid (C₆H₅COOH) is 6.50 x 10^-5.

Answer 1

Ka of C6H5COOH= 6.5*10^-5
Kb of NaC6H5COO =10^-14/Ka
                                       =(10^-14)/ (6.5*10^-5)
                                        =1.54*10^-10

NaC6H5COO + H2O ----> CH3COOH + NaOH
0.5                                                      0                   0   (initial)
0.5-x                                                   x                   x   (at equilibrium)

Kb = [CH3COOH][NaOH] / [NaC6H5COO ]
1.54*10^-10 = x*x / (0.5-x)
Since Kb is very small, x will be small and can be ignored as compared to 0.5
1.54*10^-10 = x*x / 0.5
x=8.77*10^-6
so[NaoH] = x = 8.77*10^-6
pOH = -log [OH-]
   = -log (8.77*10^-6)
   =5.06
pH =14-pOH
    =14 - 5.06
    =8.94
Answer: 8.94