Question

20 mL of benzoic acid of unknown concentration is titrated with potassium hydroxide whose concentration is 0.095 M. Calculate the pH at: (a) 0 %; b) 25 %; (c) 100 %; (d) 150 % degree of titration. Up to the equivalence point 13.56 mL of KOH solution is consumed

Answer 1

(a) n(KOH) = 13.56 × 0.095 = 1.2882 mmol 

n(KOH) = n(benzoic acid) = 1.2882 mmol 

c(benzoic acid) = 1.2882/20 = 6.441 x 10^-2 M

K_a = 6.7 × 10^–5 

As we do not have the three orders of magnitude difference between the concentration and the acidic ionization constant, the more complicated formula has to be used for the calculation of the pH.

(b) n(KOH) = 0.25 × 13.56 × 0.095 = 0.32205 mmol 

n(benzoic acid)sum = 1.2882 mmol  

We get a basic buffer solution of: 

n(benzoic acid)rest = 0.96615 mmol 

n(potassium benzoate) = 0.32205 mmol

(c) n(potassium benzoate) = 1.2882 mmol 

 c(potassium benzoate)

The salt, potassium benzoate is a weak base, so first the basic ionization constant should be calculated. 

(d) n(KOH) = 1.50 × 13.56 × 0.095 = 1.9323 mmol 

n(benzoic acid)_sum = 1.2882 mmol 

n(potassium benzoate) = 1.2882 mmol 

n(KOH)excess = 1.9323 - 1.2882 = 0.6441 mmol 

As we have a weak base and a strong base in the system, it is the strong base, namely the KOH solution in excess that determines the pH of the solution.