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ikadub [295]
3 years ago
6

A 25.00 mL sample of vinegar was titrated with 39.27 mL of 0.4293 M NaOH. Calculate the concentration of acetic acid in the vine

gar sample.
HC2H302 + NaOH NaC2H302 + H2O
Chemistry
1 answer:
QveST [7]3 years ago
4 0

Answer:

0.6743 M

Explanation:

HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O

First we <u>calculate how many NaOH moles reacted</u>, using the <em>definition of molarity</em>:

  • Molarity = moles / volume
  • moles = Molarity * volume
  • 0.4293 M * 39.27 mL = 16.86 mmol NaOH

<em>One NaOH moles reacts with one acetic acid mole</em>, so <u>the vinegar sample contains 16.86 mmoles of acetic acid as well</u>.

Finally we <u>calculate the concentration (molarity) of acetic acid</u>:

  • 16.86 mmol HC₂H₃O₂ / 25.00 mL = 0.6743 M
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Question is incomplete, complete question is as follows :

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[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

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As we know that the rate depednant step is the slowest step of the reaction, rate law is :

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But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

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