The octanol-water partition coefficient (K_ow) for 2,4,6-trichlorophenol (C_6H_3

Question

The octanol-water partition coefficient (K_ow) for 2,4,6-trichlorophenol (C_6H_3CI_3O) is 4897.79. Determine the equilibrium concentration (in mol/L) of 2,4,6-trichlorophenol in octanol, if the equilibrium concentration of 2,4,6-trichlorophenol in water is 7.60 Times 10^-1. Report your answer to three significant figures in scientific notation. The octanol-water partition coefficient (K_ow) for phenol (C_6H_6O) is 31.62. At equilibrium, the concentration of phenol (C_6H_6O) was found to be 7.95 Times 10^-1 M in octanol. Determine the mass (in g) of phenol in 77.3 mL of water. Report your answer to three significant figures in scientific notation.

Explanation / Answer

Kow= [substance]octanol/[substance]water

1) Kow=4897.79= [2,4,6-trichlorophenol]octanol/7.60x10-1

[2,4,6-trichlorophenol]octanol= 3722.3mol/L = 3.72 x 103 M

2) Kow= 31.62= 7.95x10-1/[phenol]water --->[phenol]water= 0.025M ----> mol= 0.025M x 77.3x10-3L= 1.94x10-3mol

mass= 1.94x10-3mol x 94g/mol= 0.182g= 1.82x10-1g

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