1. Oct 15, 2007

### dingol

1. The problem statement, all variables and given/known data

The distribution coefficient, k = (conc. in CH2Cl2/conc in H2O), between dichloromethane and water for solute A is 7.3.

What weight of A would be removed from a solution of 10 g of A in 100 mL of water by a single extraction with 100 mL of dichloromethane?

What weight of A would be removed by three successive extractions with 33 mL portions of dichloromethane?

How much dichloromethane would be required to remove 98.5% of A in a single extraction?

2. Oct 15, 2007

### dingol

is this method right>>?>??

Basic numbers :

10 g of A total
100 ml of H2O, 100 ml of dichloromethan ( Iwilll abbreviate as dich.)
10 g of A in 100 ml of h2o by a single extraction w/ 100 mL of dich.?
if (x) g = the amount that stays in the water, 7.3(x)g of A must go into dich. phase.

Wt. of A in Dich. = 100 ml * 7.3(x)g/ml = 730 (x)g
Wt. of A in H2O = 100ml * (x) g/ml = 100(x)g

Total weight of A = 10 g (now partitioned between two layers)

add 100 (x) g + 730(x)g = 830 (x)g

830 (x)g = 10g
(x) = .0120481928

now, plugging x back into the equations:

wt. of A in Dich. = 730 (.0120481928)g = 8.7965 g

wt. of A in H20 = 100 (.0120481928)g = 1.205 g
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Pt. 2: what weight og A owuld be removed by 3 successive extractions with 33 ml portions of Dich.?

Everything is the same, except instead of one extraction there will be 3 extractions with 33 mL portion s of dich. instead of 100 mL.

Wt. of A in dich. = (33mL)(7.3)(x)g/mL = 240.9 (x)g
Wt. of A in H20 = 100mL (x)g/ml = 100 (x) g

Total weight = 10 g

after adding 100 x + 240.9 x = 340.9 x

340.9(x)g = 10 g

x= 0.0293341156

after substituting back into equations:

Wt. of A in dich. = (33mL)(7.3)(.0293341156)g/mL = 6.824697g
Wt. of A in H20 = 100mL (.0293341156)g/ml = 2.83341156 g

DIch. later is removed. and second round begins

Wt. of A in dich. = 240.9g
wt. of A in H20 = 100g

total wieght = 2.833 g

340.9 (x)g = 2.833g

(x) = .0083103549

Weight of A in Dich. = 240.9 (.0083103549) = 1.99947g
weight of A in H2O = 100 (.0083103549) = .86103549g

Now layer is removed and third round begins

weight of A in Dich. = 240.9
Wt. of A in water = 100

total weight = 340.9

there fore,

340.9 x = .83g

x= .00243g

wt. of A in dich = 240.9 * .00243g = .5855387g
wt. of A in H2o = 100 * .00243g = .243g

total % weight = 6.825 + 1.999+ .5855 = 9.04095 from 10 g

(9.04095 / 10) * (100%) = 94%

3. Oct 15, 2007

### dingol

continuation

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Pt 3:

How much dich. would be required to remove from 98.5% of A in a single extraction?

98.5% = 9.85g of total weight of A in dichloro

wt. of A in Dich = (x)(y) = .985g

x= x
y = orig wt. of dich.
z = x*y

z= 10g

wt. of A in Dich. = y * 7.3x = y7.3 x
wt. of A in water = 100mL * 7.3 x = 730x

to be continued....