A student was performing an extraction procedure of o-Xylene from water into octanol. The total mass of o-Xylene was 0.10 gram (molecular weight 106.2 g/mole). The volume of water was 990mL and volume of octanol 10 mL. There is a head space of 200 mL above the solution and it is sealed. Temperature was 25°C. The Henry’s Law constant for o-Xylene is 0.2Matm-1 and logKOW is 3.12. Calculate the concentrations of o-Xylene in the air above the solution in ?g/m3, concentration in water and octanol in mg/L.

## Calculate concentrations in air, water and octanol

# 3 Answers

^{-1}) * 0.1 (g) / 106.2 (g/mole) * 1/1 (L) , since total volume of liquid is 1L.

^{-1}) * 0.2 (L) * 1/298 (K) / 0.082 (L*atm mol

^{-1}K

^{-1})

^{-4}g (in 200cc). Its concentration in g m

^{-3}is

^{-4}(g) * 1/200 (cm

^{3}) * 10

^{6}(cm

^{3}/ m

^{3}), or 0.8185 g m

^{-3}.

_{OW}= 3.12. Volume of water and xylene are 990 mL and 10mL respectively.

^{-1}.

_{OW}= 3.12 , hence Conc in octanol / Conc in water is 10

^{3.12}= 1318.257.

^{-1}, then

^{-1}, and the concentration of xylene in octanol is 100.9335 mg L

^{-1}. These values were obtained by multiplying (1-x) and x respectively with 101.01. The units of concentration did not make a difference, as we were using ratios of concentration in the partition coefficient equation.

Did the person who wrote this problem even try solving it? It's actually not solvable with the information given. When I tried it, I actually got more xylene than is actually present, when I added the three amounts together.

Partly, this is because you were given the wrong Henry's constant. A quick Google search told me that the Henry's constant for o-xylene in water is 0.2 M/atm. This is irrelevant, however, since octanol is the top layer.

Another problem is that Henry's Law relates the partial pressure of gas to the solubility of that gas in a liquid. Solubility is different from actual concentration. Thus, you can't use Henry's Law for this problem. You should be using vapor pressure calculations, instead.

If we were ignoring the air concentration, the first answer would have almost the right procedure for solving the octanol and water concentrations, but there's an error. First, there's only 0.1 g of xylene present. Second, the oil-water partition coefficient uses molar concentrations. Thus, the numerator is the concentration of xylene in octanol (we'll call this co), while the demoninator is the concentration of xylene in water (cw).

The octanol concentration is moles of xylene (no) per liter of octanol (0.01 L), and moles of xylene is mass of xylene (mo) divided by its molar mass (106.2 g/mol). Thus, co = mo/0.01L/(106.2) g/mol = mo/1.062 M.

We can do something similar for water concentration, and get cw = mw/95.6 M.

Now, going back to the earlier o-w partition coefficient,

1318 = co/cw = (mo/1.062) / (mw/95.6)

simplifying, we get mo/mw = 14.64.

Remembering that the total mass of xylene is 0.1 g,

mo + mw = 0.1, or

mw = 0.1 - mo,

we can substitute to get

14.64 = mo / (0.1-mo)

Finally,

mo = 0.0936 g

mw = 0.0064 g

Or, if we're correctly following the sig fig rules,

mo = 0.09 g

mw = 0.01 g

Helllo Brook,

I expect that "logKOW" means the log of the octanol-water partition coefficient. Taking the inverse log of 3.12, we get 1318.256739. If before the extraction we have 0.10g of o-xylene in water, then post extraction we have x grams of o-xylene in octanol, and 1-x is the grams of o-xylene left in the water. Set up an equation as follows:

1318.256739 = [x/10. mL] / [(1-x)/990. mL] = (x/10)(990/1-x) = (990x/10-10x)

Solving for x we get x=0.930 g o-xylene, and 1-x=0.0699 g water

Convert your grams to mg: 930 mg o-xylene and 69.9 mg water

The volume is 990. mL + 10. mL = 1000. mL = 1.00 L

I don't understand what "?g/m3" represents.

Randall

## Comments

remove that '?' and you get grams per cubic meter, which would be the concentration of xylene in the air.

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