Question

1. The octanol-water partition coefficient (Kow) for 1,2-diphenylhydrazine (C12H12N2) is 21.88. Determine the equilibrium concentration (in mol/L) of 1,2-diphenylhydrazine in octanol, if the equilibrium concentration of 1,2-diphenylhydrazine in water is 6.23 x 10-3. Report your answer to three significant figures in scientific notation. Submit Answer Tries 0/13 2. The octanol-water partition coefficient (Kow) for trimethylamine (C3H9N) is 1.45. At equilibrium, the concentration of trimethylamine (C3H9N) was found to be 5.57 x 10-1 M in octanol. Determine the mass (in g) of trimethylamine in 90.5 mL of water. Report your answer to three significant figures in scientific notation. Submit Answer Tries 0/13

Answer 1

1. The octanol-water partition co-efficient for 1,2-diphenylhydrazine (abbreviate as DPH in this problem) is K_OW = 21.88.

Since K_OW is an equilibrium constant, it can be written as

K_OW = [DPH]_O/[DPH]_W

Given that K_OW = 21.88 and [DPH]_W = 6.23*10^-3 mol/L, plug in values in the expression and get

21.88 = [DPH]_O/(6.23*10^-3 mol/L)

======> [DPH]_O = 21.88*(6.23*10^-3 mol/L)

======> [DPH]_O = 0.1363 mol/L

======> [DPH]_O ≈ 0.136 mol/L (correct to 3 sig. figs) = 1.36*10^-1 mol/L

The equilibrium concentration of 1,2-diphenylhydrazine in octanol is 1.36*10^-1 mol/L (ans).

2. The octanol-water partition co-efficient for trimethylamine (abbreviate as TMA in this problem) is K_OW = 1.45.

Since K_OW is an equilibrium constant, it can be written as

K_OW = [TMA]_O/[TMA]_W

Given that K_OW = 1.45 and [TMA]_O = 5.57*10^-1 mol/L, plug in values in the expression and get

1.45 = (5.57*10^-1 mol/L)/[TMA]_W

======> [TMA]_W = (5.57*10^-1 mol/L)/(1.45)

======> [TMA]_W = 0.3841 mol/L

======> [TMA]_W ≈ 0.384 mol/L (correct to 3 sig. figs) = 3.84*10^-1 mol/L

The equilibrium concentration of trimethylamine in water is 3.84*10^-1 mol/L.

Volume of water taken = 90.5 mL = (90.5 mL)*(1 L)/(1000 mL)

= 0.0905 L.

Mol(s) TMA in water = (0.0905 L)*(3.84*10^-1 mol/L)

= 0.034752 mol.

The atomic masses are

C: 12.011 g/mol

H: 1.008 g/mol

N: 14.007 g/mol

Gram molar mass of trimethylamine, C₃H₉N = (3*12.011 + 9*1.008 + 1*14.007) g/mol

= 59.112 g/mol.

Mass of C₃H₉N in 90.5 mL water = (0.034752 mol)*(59.112 g/mol)

= 2.054 g

≈ 2.05 g (correct to 3 sig. figs)

The mass of trimethylamine in 90.5 mL water = 2.05 g (ans).