Titration of benzoic acid with NaOH

C6H5COOH ( benzoic acid ) = C6H5COO- (aq) + H+ (aq) Ka = 6.46 x 10^-5
Benzoic acid dissociates in water as shown in the equation above. A 25.0 ml sample of an aqueous solution of pure benzoic acid is titrated using standardized 0.150 M NaOH

A) After addition of 15.0ml of the 0.150 M NaOH , the pH of the resulting solution is 4.37. Calculate each of the following
I) [H+] in the solution

Ii) [OH-] in the solution

Iii) The number of moles of NaOH added

Iv) The number of moles of C6H5COO- (aq) in the solution

V) The number of moles of C6H5COOH in the solution

B) State whether the solution at the equivalence point of the titration is acidic , basic or neutral. Explain your reasoning
In a different titration , a 0.7529 g sample of a mixture of solid. C6H5COOH and solid NaCL is dissolved in water and titrated with 0.150 M NaOH. The equivalence point is reached when 24.78ml of the base solution is added.

C) calculate each of the following
(I) The mass, in grams, of benzoic acid in the sample
(Ii) The mass percentage of benzoic acid in solid sample

Solution Summary

It investigates the titration process of benzoic acid with NaOH. The solution is detailed and well presented.