3.3. TASKS FOR SELF-STUDY

Sample problem № 1. An aqueous solution of succinic acid with a concentration of 5.98 × 10–2 mol/L is in equilibrium with its ethereal solution of a concentration of 1.1 × 10–2 mol/L at 25°С. What is the concentration of succinic acid in the aqueous solution, which is in equilibrium with the ethereal solution containing 8.6 ×·10–3 mol/L of this acid at 25°С?

Solution. Substance A is distributed between two immiscible solvents according to the partition law:

(KD )A = CA (org) =const,

СA (aq)

where (KD)A is the distribution coefficient of Substance A between two phases, and СА(org) and СА(aq) the equilibrium concentrations of this substance in the organic and aqueous phases in mol/L.

The distribution constant, which depends on temperature, and the nature of solvent and solute, but does not depend on the concentration in the area of dilute solutions, is found as follows:

Since the distribution coefficient does not depend on concentration, then

Сsucc.acid (aq) = Сsucc.acid (ether) = 4.67 10−2 mol/L.

KD

Answer: the concentration of succinic acid in water solution equals 4.67·10-2 mol/L.

Sample problem № 2. Iodine from 4 L of an aqueous solution with a concentration of 0.1 g/L is extracted with 100-mL portions of carbon disulphide at 25°С. What is the amount of iodine remained in the extracted solution after the first extraction procedure. How many extractions should be done to separate 99% of iodine? Take the distribution coefficient as equal to 590.

Solution. The amount of iodine in the initial aqueous solution is found by the formula,

m0=V·C=4·0.1 = 0.4 g.

Then remnants of non-extracted iodine remained after the extraction are

calculated by the formula,

mn = m0 (1 – β) = 4·10–3 g,

81

where β = 0.99, and stands for the degree of extraction according to the problem statement.

The weight of the extracted substance remained in the initial solution after single extraction is calculated by the following equation:

where V is the volume of a solution of the extracted substance, and VE is the volume of the used extracting agent.

The volume measurement units should be the same in both cases.

The remnants of iodine after the first extraction make m1 = 2.54·10–2 g, where (KD )I2 = 590 is the distribution coefficient of iodine between

carbon disulphide and water.

The remaining amount of iodine after multiple extractions is calculated by the following equation:

where VE/п is the volume of a portion of the extracting solvent (extracting agent), and п is the number of extractions.

The data from the problem statement are inserted in the above equation, as follows:

The value, n, is found after transforming the equation and taking its logarithm, in the following way:

0.01 = (4/63)n,

Answers:

1) The remaining amount of iodine after the single extraction makes m1 = 2.54·10–2 g.

2)The number of needed extractions is 2.

82

Sample problem № 3. Calculate the volume of air oxygen that will dissolve in 1 m3 of water at Т = 293 К and Р = 5·105 N/m2. The distribution coefficient of oxygen in water at 293К is 1.365·10–8 kmol/N·m.

Solution. The partial pressure of oxygen in air at a total pressure of 5·105 makes Po2 = 0.21·5·105 = 1.05·105 N/m2.

From the equation,

c = A P ,

where c is the concentration of gas in a fluid phase, A is the distribution constant, and Р is the partial pressure of gas, it can be found that

c = 1.365·10–8 · 1.05·105 = 1.432 · 10–3 kmol/m3.

Answer: The volume of oxygen dissolved in 1 m3 of water and normalized for standard conditions is 1.432·10–3 м3.

Sample problem № 4. A 0.02 M aqueous solution of picric acid is in equilibrium with a 0.07 M solution of picric acid in benzene. Calculate the distribution coefficient of picric acid between benzene and water, if its molecular weight is regular in a benzene solution, while in an aqueous solution it is dissociated partially. The degree of dissociation makes 0.9.

Solution. To determine the distribution constant of picric acid between benzene and water, one should divide the concentration of this acid in benzene by the concentration of its non-dissociated part in an aqueous solution, as follows:

K = 0.7 = 350. 0.02 (1−0.9)

Answer: The distribution coefficient of picric acid between benzene and water equals 350.

Sample problem № 5. Upon distribution of salicylic acid between benzene and water at Т=298, the following data has been obtained:

where C1 is the concentration of salicylic acid in the aqueous layer expressed in kmol/m3, C2 is its concentration in the benzene layer in kmol/m3. Determine graphically the n and K values.

83

Solution. Let us calculate the K values by the equation,

K = C1 . C2

К = 1.97; 1.33; 0.96; 0.86; 0.64; 0.53; 0.34; 0.27; 0.21.

Since the ratio, C1/C2, does not remain constant, it means a change in the molecular weight of a distributed substance is observed. Let us apply the partition law in its general form, as follows:

where M ′′ is the average molecular weight of a distributed substance in the first phase, and M ′ is the average molecular weight of a distributed substance in the second phase.

Taking a logarithm of both parts of the expression, one finds that

n lgC1 – lgC2=lgK

or

lgC2=n lgC1 – lgK.

The obtained equation is an equation of the straight line, the tangent of a slope angle of which with

respect to the axis of abscisses equals п, and the intercept on the axis of ordinates equals lgK.

The graph of the function, lgC2 = f(lgC1), is plotted on the basis of the above data we build (Fig. 6). The tangent of a slope angle of the line with respect to the axis of abscisses and the section, lgK (the line’s intercept on the axis of ordinates) are found from the graph, as follows:

tgα = 11..6600 =1.66 ; n=1.66;

84