9
Some remarks on the theory of the rocking extraction.' BY Uno Boklund. (From the medico-chemical Institute of Lund.) (With I graph In the test.) On extraction of organic acids by the rocking extraction method2 it will be proved that, with a sufficient degree of exactitude, the extrac- tion process can he put into the formula 1 a t a--5 1; = -log - the validity of which has been tried for various organic acids. The following discussion shows quite theoretically that this formula can be applied: If the rocking velocity, the inclination of the apparatus and the liquid volumes are constant, the velocity of the extraction is determined for the most essential part by the solubility of the molecules in the ex- tracting solvent. Now as in each rocking period the extracting solvent comes approximately into contact with equal volumes of acid, thus the number of molecules, x,, which in each period is removed by the extracting solvent will be determined by the actual molary concen- tration of the acid, or in other words: the extraction velocity in eac.h period will be proportional to the concentration. This signifies that the less the proportion is to the total extraction time, T, and the number of rocking periods, n, the better the diff.-equatioa d z = k (a -3) d t will be realized on the simultaneous presumption thet too concentrated solutions are not used. The ideal realization comes about on condition Der Redaktion am 29. Juni 1927 zugegangen. E. Widmark: Dies. Archiv. 1926. Bd. XLVIII. p. 61.

Some remarks on the theory of the rocking extraction

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Page 1: Some remarks on the theory of the rocking extraction

Some remarks on the theory of the rocking extraction.' BY

Uno Boklund.

(From the medico-chemical Institute of Lund.) (With I graph In the test.)

On extraction of organic acids by the rocking extraction method2 it will be proved that, with a sufficient degree of exactitude, the extrac- tion process can he put into the formula

1 a t a--5 1; = -log -

the validity of which has been tried for various organic acids. The following discussion shows quite theoretically that this formula can be applied:

If the rocking velocity, the inclination of the apparatus and the liquid volumes are constant, the velocity of the extraction is determined for the most essential part by the solubility of the molecules in the ex- tracting solvent. Now as in each rocking period the extracting solvent comes approximately into contact with equal volumes of acid, thus the number of molecules, x,, which in each period is removed by the extracting solvent will be determined by the actual molary concen- tration of the acid, or in other words: the extraction velocity in eac.h period will be proportional to the concentration.

This signifies that the less the proportion is to the total extraction time, T, and the number of rocking periods, n, the better the diff.-equatioa d z = k (a -3) d t will be realized on the simultaneous presumption thet too concentrated solutions are not used. The ideal realization comes about on condition

Der Redaktion am 29. Juni 1927 zugegangen. E. Widmark: Dies. Archiv. 1926. Bd. XLVIII. p. 61.

Page 2: Some remarks on the theory of the rocking extraction

SOME REMARKS ON THE THEORY OF THE ROCKING EXTRACTION. 177

The extraction time for a partly dieeoeiated acid. Now we are going to regard the state of an extraction, where the

acid is partly dissociated. A survey of the length of the extraction time for such states of dissociation has a great practical importance, as the conditions for an extraction of labil substances on a comparatively low and sparing hydrogen ion concentration are hereby illustrated. In the calculation we presume that the degree of dissociation is u, the initial concentration of the acid a and further that u by aid of buffer is kept constant. At the beginning of the extraction then there are “(1-a) acid molecules and a-u ions. After the time tz niolecules are supposed to be extracted away. These molecules then must be taken partly from the already existing undissociated molecules, but partly even from the ions (by regeneration of molecules) and this in such proportions that the relation between the remaining dissociated molecules and the total number of inolecules preserves its-by the buffer-fixed value u. If the total number of removed molecules is zx, are supposed to be taken from the molecules and xcj indirectly from the ions.

The condition for u being a constant will be then a - n -

U xj - a-x

ise: x j = u x

or a x + x m = x whence x, = x(1-a).

but x,+ x,= x

Put into the equation of velocity this will be: ax = IC[U(I-~)-Z, ,J at d x = k[a(I-a)-~(l-a)]d t

k = - Ill-. or when integrated

1 a t ( 1 - a ) a - m

A comparison between the equations (I) and (2) shows, that the dissociation setting in, the extraction velocity constant is reduced from

Thus the extraction velocity is directly proportional to the amount of acid niolecules.

If (1--cc) is replaced by the hydrogen ion concentration [HI and the dissociation constant of the acid K , we will get for the extraction constant at partial dissociation (kcws):

k to (1-u)k.

Skandinav. Archiv. LIU. 12

Page 3: Some remarks on the theory of the rocking extraction

UNO ROKLUND

Jidiss e [HI + K, [HI

178

or

In the same way the equation (2) can be reduced to the following expression

(3). a kt = (1 + l O p " - ~ s ) log a--2

which consequently describes the extraction period even in its depen- dence of the dim.-const. of the acid and on the reaction of the medium.

The table below shows graphically the 'relation between kdiss, pH and pS.

Fig. 1.

On the abscissa p H is put as a function of pS. So the point pS means that p H and pS have the same numerical value, pS + 1 means that p H is one unit greater than pS and so on. It is evident that at pS-2 the extraction velocity begins to diminish. At pS it is only half of the primary velocity and at pS+ 2 as a matter of fact no extraction can come about. This signifies that at pS the extraction time has to be redoubled and at pS + 1 it has to be increased somewhat more than tenfold.

Consequently the extraction at partial dissociation ought not to be done at a hyarogm ion concentration less than the nuinerical value of the disso- ciation. constunt, so that the extraction time shall not be too long.

So the acetic acid the pS of which is 4-75 can be extracted at p H 2.75 with the same velocity as in a strong acid solution. If p H is 4-75 the extraction time has to be redoubled and if p H is 5-75 it has to be in- creased tenfold.

Page 4: Some remarks on the theory of the rocking extraction

SOJlE REMARKS ON TIIE THEORY OF THE ROCKING EXTRACTIOX. 179

Fractionated extraction.

Another problem which can be solved by aid of the rocking method is the extractive separation of acids with different pS. We are only looking at the case of two acids, S and A!&, with the exponents of disso- ciation pS and pS,. Evidently the question is to find the p H , at which the extraction of one acid comes about as soon as possible but that of the other as slowly as possible, which signifies that one acid ougth to be undissociated while the other ought t o be dissociated as completely as possible. As a matter of fact this is the case with regard to one acid at

p H = pS-2

and with regard to the other at p H = p S , + 2.

As these equations are valid for the same p H we will get by addition PS + P S I p H = 2 (4)

The extractive separation of two acids then ought to come about in n medium the hydrogen ion exponent of zchich is the arithmetical mediuwb of the dissociation exponents of the acids.

The following simple experiment beautifully shows such a separation. Into the dimitting vessel we pour down 20 cc. of phosphate mixture

whith the reaction p H = 6.2. After that we put in some drops of the two indicators phenolphtalein and bromphenolblue, representing two acids with the exponents of dissociation 8.4 and 4.0. On extraction against alkali and with ether as extracting medium we will find that the alkali will be red and on acidifying colourless. The phenolphtalein which a t p H = 6 .2 exists as molecules can consequently pass trough the ether, while the stronger acid bromphenolblue remains as a salt in the dimitting solution.

The greater the difference is between these exponents the sooner we will reach a complete separation, because the ideal state will thus be the better realized, which is as mentioned above that one acid is un- dissociated while the other is completely dissociated. At the same extraction velocity constants this will be the case if pS-pS, = 4. Such favourable conditions however are only exceptionally to hand and therefore it will be interesting to make out how the extraction time changes with varying difference between the exponents. By this we examine how the halving times (i. e. the time for extraction of 50% of the acid) for the two acids are in relation to each other.

12*

Page 5: Some remarks on the theory of the rocking extraction

180 Uivo BOKLUSD:

By emplognieiit of the equation (3) we will get for one of tllc acids

(5)

(6)

a 1;t = (1 4- l O a ~ - y o , a ~

h! t = ( 1 + l o P ~ ~ - ~ ~ l ) l o ~ - - * and f9r t,he other

b 1 1 b - Y

But according to the equation (4) pIiI = PS +PSI

2 whence

and

PS - PSI pH -pps = -

pH--pS , = + 2

PS -PSI.

So (5) and (6) will be converted into

b

(7)

If then we put in z = u and y = b t and t, will mean the required halving times and w e get

p s - ps,

PS - PSI

-- 1 d = ( 1 + 1 0 ) lo@

1; t = (1 + lo++ log 2 1 1

whence by division

This is illustrated by the table below

~

p S - p S , i 2.2 I 2.4 I 2.6 I 2.8 1 3.0 1 3.2 I 3.4 I 3.6 I 3.8 I 4.0 I

If the numerical difference between the dissociation exponents of the acids is for instance 1-4 the table shows that the proportion t,kJtk will be 5.012. Then in order to calculate from this the proportion

Page 6: Some remarks on the theory of the rocking extraction

SOME REMARKS O N THE THEORY OF THE ROCKING EXTRACTION. 181

between the halving times we will have to put in the nuiiierical value of k1:k which is presumed to be known. Suppose that this value = 1. Then t,:t will be 6.012, i. e. the acid A', is extracted five times slower than the acid S. If the proportion k,:k ha8d bee11 = 1/2, t,: 1 had been = 2 x 5.012 = 10.024 consequently the extraction of 8, would have heen done ten times as slowly as the extraction of 8.

A greater practical interest is however offered by the knowledge of the quantity of each acid, which a t certain time has been extracted over to the receiving vessel.

For the calculation of this thc equations (7) and (8) will be used, the assumption being made that 6 = n - c b and then we will get

a Is. t - p s - p S 1 r

log--- - -___ a--a

1 + l O 2

log- n.a = k1 fl--. n.a - y PS - PSI

1 + 1 0 + 2

According t o (9) the halving time for the acid S is

Then the time 10t stands for just about coniplete extraction i.e. after the time 10 tx will be = a (0,999a) and by making t, in the equation (13) equal to 101 we will get the quantity, y, which has been extracted over from the acid S,, when the acid S is totally extracted. The equa- tion runs:

or reduced:

In order to get a survey of how the extractions are going on at the varying value of the factor (pS-pS,) we presume that n = 1 and k, = k. Then we will get the following table, where y1 means the quantity of the acid S,, which has been extracted over, when the acid S was totally extracted, yz the quantity of S, which has passed over, the extraction having been repeated once more with the quantities, that we get as a final result in the receiving vessel at the preceding extraction, and y3 the result analogous to this after the third extraction.

Page 7: Some remarks on the theory of the rocking extraction

182

0.99 a 0.99 a j 0.97 a 0.94 a

UNO ROKLUND:

i 1

- .... - 0 0.2 0.4 0.6 0.8 1.0 1 .2 1 - 4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3 .2 2.4

0.89 a 0.82 a 0.75 a

0.98 a 0.97 a 0.94 a 0.88 a 0.79 a 0.67 a 0.56 a 0.45 a 0.35 a 0.25 a 0.18 a

0.97 a 0.96 a 0.91 a 0.83 a 0.70 a 0.55 a 0.42 a 0.30 a 0.21 a 0.13 a

1 O . U ~ a 0.35 a 0.12 a 11 0.04 a

u 08 a I 0.02 a 1 0.29 a

1

1 0.24 a 11 0-06 a 0.01 a 0.20 a 0.04 a 0.01 a

/ I 0.16 a 0.03 a 0.01 a

0-50 a 0.42 a

- - /I 0.13 a 11 0.02 a 0.00 a 3.6 i l 0.10 a 1 0.01 a 1 0.00 a 3.8 0.08 a 1 0.01 a 0.00 a 4.0 /j 0.07 a 1 0.01 a 0.00 a

If - the extraction finished - at most 10% of the acid 8, is allowed to be left in the receiving vessel we will have t o fix the practical limit of the applicability of the method at a numerical difference between the strength exponents of the acids of about 2.2. At this value - as the table shows - we are obliged to resort to three extractions for getting the acid S, in the receiving vessel to be 10% of the acid S. If the diffe- rence between the strength exponents is at least 2.6 two extractions will be sufficient and in case of this difference being 2 3.6 our condition will be fulfilled already at the first extraction.

It ougth to be noticed that of course the number of extractions can be increased and in this way a still more effective separation of the acids can be obtained than that of the table, where the calculations are based on three extractions. The trouble is inconsiderable for the apparatus having once been set in work it will manage by itself entirely.

This table however, as mentioned before, is applicable only when n = 1 and kl = k. Practically these factors are varying to an essential degree and thus displace the extraction possibility in a favourable or unfavourable direction. It is not possible to indicate in general when the extractive separation is practically realizable or not, but we have to examine the circumstances from one case to another.

Page 8: Some remarks on the theory of the rocking extraction

SOME REMARKS ON THE THEORY OF THE ROCKING EXTRACTIOS. 183

An example will illustrate the procedure: we have two acids, 5' and S,, with the strength exponents 5.3 and 4.1. By extractions, performed in a strong acid medium, k and k, have been settled to respec- tively 0.00300 and 0.00075. The concentration of 8, is for instance SOO/o of S (i. e. n -- 0.8). The calculations will be as follows:

The waction of ihe mediztw. According to the equation (4) we will get

whence p H = 4.7.

The extraction then will be performed in a medium the hydrogen ion exponent of which is 4.7.

The time for the extraction. R e will get the halving time for the acid S from the formula (9)

PS - PSt l i t= 1 + l o - 2 ) log2 (

1.2

t - - 0 * 00300

t = 125.5 min.

The time for total extraction is calculated from this to 1%5min.

The result of the first extraction. After this time then the acid S is extracted over completely. And

so we are going to examine how much of the acid S, has during the same time passed over.

According to the equation (14) we will get 1.2

0.8 a =-. 0.00075 1 0 1 - T . log 2 log 0.8 a - y 0*00300

y = 0-28a

i. e. when the acid S has been extracted over completely Z8°/0 of the acid S, also has been gathered in the receiving vessel, but as our con- dition was that a t most 10% of the acid s, should be allowed to be

Page 9: Some remarks on the theory of the rocking extraction

184 UNO BOKLUND: S O M E REMARKS ON THE THEOXY OF THE ROCKISG ETC.

found there we are obliged to resort to another extraction. In exactly the same way we will get the result of this second extraction. Thus y will be = 0.09. And so the condition is fulfilled and the result will be that there is found in the receiving vessel 100°/o S and 9% S, i. e. on the basis of a numerical difference beticeen the exponenfs of disso- ciation we have got a satisfactory separation of tro acids.

In the same way the separation can be continued still longer by a third extraction, and thus only 2% of S, will have passed over when S is totally extracted.