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You weigh out 500mg of benzoic acid and dissolve it in 3mL of diethyl ether. This is then extracted with 3 mL of water. The ether layer is removed, dried with sodium sulfate 2 and the ether evaporated yielding 400mg of benzoic acid. What is the Kp value (partition coefficient) for the ether/water system

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ajeigbeibraheem

You weigh out 500mg of benzoic acid and dissolve it in 3mL of diethyl ether. This is then extracted with 3 mL of water. The Kp value (partition coefficient) for the ether/water system is 5

The Kp value (partition coefficient) is the ratio of unionized substance in the organic phase in relation to the aqueous phase at equilibrium.

From the information given:

  • The mass of the benzoic acid in 3mL of diethyl ether is = 500 mg
  • The volume of diethyl ether = 3 mL

Prior to contact between the layers, there are:

  • 0 mg of benzoic acid in 3 mL of water
  • 500 mg of benzoic acid in 3 mL of diethyl ether

After contact, i.e. at equilibrium

  • 400 mg of benzoic acid in 3 mL of ether
  • (500 - 400) = 100 mg of benzoic acid in 3 mL of water

Now, using the Nernst equation to determine the Kp value of the ether/water system, we have:

[tex]\mathbf{K_p = \dfrac{solubility \ of \ solute \ in \ solvent \ 1}{solubility \ of \ solute \ in \ solvent \ 2} }[/tex]

[tex]\mathbf{K_p = \dfrac{\dfrac{500}{3}}{\dfrac{100}{3}} }[/tex]

[tex]\mathbf{K_p = \dfrac{500}{3} \times {\dfrac{3}{100}} }[/tex]

Kp = 5

Therefore, we can conclude that the Kp value (partition coefficient) for the ether/water system is 5.

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