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 Na

Question

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 24.78 mL of the base solution is added. Calculate each of the following.

a. The mass, in grams, of benzoic acid in the solid sample
b. The mass percentage of benzoic acid in the solid sample

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Cherry 4 years 2021-07-30T20:55:42+00:00 1 Answers 56 views 0

Answers ( )

  1. MichaelMet
    0
    2021-07-30T20:57:16+00:00
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    Answer:

    a. 0.4539g benozic acid

    b. 60.29% of benzoic acid in the solid sample

    Explanation:

    The benzoic acid reacts with NaOH as follows:

    C6H5COOH + NaOH → C6H5COONa + H2O

    Where 1 mole of the acid reacts per mole of NaOH

    The NaCl doesn’t react with NaOH

    To solve this question we must find the moles of NaOH added = Moles of benzoic acid. With the moles of the acid and its molar mass (C6H5COOH = 122.12g/mol) we can find the mass of the acid and its mass percentage:

    a. Moles NaOH = Moles Benzoic acid:

    24.78mL = 0.02478L * (0.150mol / L) = 0.003717 moles Benozic acid

    Mass benzoic acid:

    0.003717 moles Benozic acid * (122.12g / mol) = 0.4539g benozic acid

    b. Mass percentage is:

    0.4539g / 0.7529g * 100 = 60.29% of benzoic acid in the solid sample

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