Volume of base required to neutralize 50 ml of benzoic acid solution after two e

Question

Volume of base required to neutralize 50 ml of benzoic acid solution after two extraction with 5.0 ml of CH_2Cl_2 each: Mole of benzoic acid neutralized: Grams of benzoic acid neutralized: Grams of benzoic acid originally present in 50 ml before extracting with CH_2 Cl_2 Grams of benzoic acid extracted in CH_2Cl_2 Percent of benzoic acid in extracted in CH_2Cl_2: Calculate amount of benzoic acid which would be extracted in two 5.0 ml extractions with CH_2Cl_2 based on your K_2 obtained in part 2:

Explanation / Answer

Part A.

I assume your base concentration = 0.0315 M and volume of base used = 11 ml

a. Volume of base used = 11 ml

b. moles of benzoic acid = moles of base = 0.0315 x 0.011 = 3.465 x 10^-4 mols

c. mass of benzoic acid = 3.465 x 10^-4 x 122.12 = 0.0423 g

d. original mass of benoic acid = 0.122 g

e. amount of benzoic acid extracted = 0.122 - 0.0423 = 0.0797 g

f. % of benzoic acid extracted = (0.797/0.122) x 100 = 65.31%

Part B.

Kd = 6.182

let x amount of benzoic acid is extracted in CH2Cl2 after first extraction

6.182 = (x/5)(0.122-x)/(volume of water phase)

Since you have not mentioned volume of water phase, you can add this value to calculate amount extracted after first extrcation.

Then for second used 0.122-x as the amount present in water phase and repeat the same procedure. Add the two masses and that should be your extracted mass of benzoic acid.

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