To find out the Ka of ethanoic acid, chloroethanoic acid and dichloroethanoic acid.

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Experiment 17        21-4-99

Aim:

To find out the Ka of ethanoic acid, chloroethanoic acid and dichloroethanoic acid.

Procedure:

  1. The pH meter is calibrated, using a buffer solution of accurately known pH.
  2. 20.0cm3 of 0.10M ethanoic acid was pipetted into a conical flask.
  3. 0.10M sodium hydroxide solution was titrated using phenolphthalein as indicator, until the solution was just turned pink.
  4. A further 20.0cm3 of the same ethanoic acid solution was added to the flask and was mixed thoroughly.
  5. The pH of the resulting solution was determined.

Result:

Volume of NaOH used is recorded below:

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pH of chloroethanoic acid and dichloroethanoic acid is given by the teacher.

Calculation:

Ka =

  1. It is reasonable to make the following assumptions:

[HX] total acid concentration

[X-] total salt concentration

because CH3COOH is a weak acid which is only slightly ionized and thus contributed only a little bit to total salt concentration. Therefore [HX] tends to total acid concentration, and [X-] is mostly contributed by the salt resulting from neutralization.

Using these assumptions, we have:

[H+]  Ka 

  1. If the concentrations of the acid and salt in the mixture are equal, then:

[H+]  Ka × 1

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