1. Benzoic acid, C6H5COOH, dissociates in water. C6H5COOH (s) ---> C6H5COO-(aq) + H+(aq) ka= 6.46x10^-5. A 25.0ml sample of an aqueous soln of pure benz. acid it titrated usin standardized 0.150M NaOH
a.) after addition of 15.0ml of the 0.150M NaOH, the pH of the soln= 4.37 calculate the following:
* [H+] in soln
* [OH-] in soln
* the # of moles of NaOH added
* The # of moles of C6H5COO- (aq) in soln.
* The # of moles of C6H5COOH in soln.
b.) State whether the soln at the equivalence point of the titration is acidic, basic, or neutral. Explain your reasoning
In a different titration, 0.7529g sample of a mixture of solid C6H5COOH and solid NaCl is dissolved in water and titrated with 0.150M NaOH. The equivalence point is reached at 24.78mL of the base soln is added
c.) calculate each of the following:
* The mass, in grams of benzoic acid in the solid sample.
* The mass percentage of benzoic acid in the solid sample.
2007-04-09
16:43:46
·
2 answers
·
asked by
nickesha t
2