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3) 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.

3) 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

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