J. Solution Chem.
SUPPORTING INFORMATION
A thermodynamic method to study the interaction between NaOH and a
highly carboxylate polymeric particles
J. M. del Río1, Santillán R2., J.-P. E. Grolier3, M. Corea2
1.- Instituto Mexicano del Petróleo, Programa de Aseguramiento de la Producción, Eje
Central Lázaro Cárdenas 152, Col. San Bartolo Atepehuacan,Del. Gustavo A. Madero, C.P.
07730 México, D.F., México
2.- Instituto Politécnico Nacional, ESIQIE, UPALM, Edificio Z-6, Primer Piso, San Pedro
Zacatenco, Del. Gustavo A. Madero, México, D.F., México
3.- Institut de Chimie de Clermont-Ferrand UMR 6296. 24 avenue des Landais – BP 80026.
63171 Aubière Cedex. France.
Supporting information content:
1.- Contributions to the partial properties of interaction
2.- Simulation of the thermodynamic basis
A thermodynamic method to study the interaction between NaOH and a highly carboxylate polymeric
particles
1.- Contributions to the partial properties of interaction
Partial volume of polymeric particles can be broken down in the following terms [1]:
v 2;1  v 2;1/ atom  v 2;1/ void  v 2;1/ hyd
(1)
where v2;1/atom is the atomic contribution, v2;1/void is the free volume contribution and
v2;1/hyd is the hydration contribution. For the electrolyte the voidness contribution is not
taken into consideration[2]:
v3;1  v3;1/ atom  v3;1/ hyd
(2)
In in a similar way, the contributions of the complex solute can be written as:
vS;1  vS;1/ atom  vS;1/ cav  vS;1/ hyd
(3)
The contribution of the interactions to the partial volume of the complex solute was defined
as (see equation 14 of the paper):
vS;1  vS;1  (t s2 v 2;1  t s3 v3;1 )
(4)
Substituting equations (1)-(3) in (4) it is obtained that:
vS;1  vS;1/ void  vS;1/ hyd
(5)
vS;1/ void  vS;1/ void  t s2 v2;1/ void
(6)
vS;1/ hyd  vS;1/ hyd  (t s2 v 2;1/ hyd  t s3 v3;1/ hyd )
(7)
where:
and
In equation (5) the atomic contribution does not appear because it was supposed that:
vS;1/ atom  (t s2 v 2;1/ atom  t s3 v3;1/ atom )  0
(8)
substituting (5) in the following equation (equation in the equation (19) of the paper):
v 2;1,3  vS;1 
dvS;1
dt s3
 t s3
(9)
J. M. del Río, Santillán R., J.-P. E. Grolier, M. Corea
Journal of Solution Chemistry
it is obtained that:
v 2;1,3  v 2;1,3/ void  v 2;1,3/ hyd
(10)
where:
(11)
v 2;1,3/ void  vS;1/ void 
v 2;1,3/ hyd  vS;1/ hyd 
dvS;1/ void
dt s3
dvS;1/ void
dt s3
 t s3
 t s3
(12)
In a similar way it is obtained that:
v3;1,2  v3;1,2/ void  v3;1,2/ hyd
(13)
2.- Simulation of the thermodynamic basis
In this section the calculation procedure of the thermodynamic basis of the method will be
simulated using data of the work of Vela et al. [3]. The 3-component system will be
composed of the following components:
Diisopropyl Ether
2,2,4-Trimethylpentene
Methanol
component 1 (solvent)
component 2
component 3
From the fit function for the excess molar volume of the ternary liquid mixture and the
densities of the components given in the work of Vela et al. [3], the specific volume as
function of the mass fraction t1 and t3 was calculated:
v  v(t1 , t 3 )
(14)
A concentration of 3.473 g/L of component 2 was initially supposed in the volume cell of
126 mL. This solution was titrated with a stock solution of component 3 with concentration
c3s = 27 g/L. After the addition of a titration volume of Δv = 0.2 mL concentrations of
components 2 and 3 were calculated by the equations:
A thermodynamic method to study the interaction between NaOH and a highly carboxylate polymeric
particles
c(i2 1)  c(i)
2 e
c
(i 1)
3

v
VC
 c  (c  c ) e
s
3
s
3
(i)
3
(15)

v
VC
(16)
The variable ts3 was calculated in this way:
t s3 
c3
c 2  c3
(17)
In order to determine the specific volume in each titration it is necessary to calculate t 1 and
t3 (see equation (14) in this section). This calculation was carried out using the “Option
Solver” of Microsoft Excel in the following way. Supposing the same arbitrary value for m1
in all titrations, the values of m2 and m3 in each titration were calculated by the equations:
(i)
m(i)
2  c 2 VC
(18)
m3(i)  c3(i) VC
(19)
and values of t1 and t3 in each titration were calculated by:
t1(i) 
m1(i)
(i)
m  m (i)
2  m3
(20)
t 3(i) 
m3(i)
(i)
m  m (i)
2  m3
(21)
(i)
1
(i)
1
Using these values the specific volume was calculated by equation (14) of this section and
the total mass in the cell (m) was estimated by the equation:
m(i) 
VC
v(t1(i) , t 3(i) )
(22)
Additionally, the total mass in the cell was estimated by the equation:
(i)
m(i)  m1(i)  m(i)
2  m3
(23)
Logically the values of m and m’ are different. Calculating for any titration the amount (mm’)2 and adding the values of all titrations the value of the amount D was obtained. Using
Option Solver of Microsoft Exel, values of m1 such as the value of D is minimum (equal to
zero) were calculated automatically. With this, the value of the specific volume (v) in each
titration was determined. The apparent volume of the “complex solute” S was calculated by
the equation:
J. M. del Río, Santillán R., J.-P. E. Grolier, M. Corea
Journal of Solution Chemistry
vS;1 
v  t1 v1
tS
(24)
where tS = t2+t3. The apparent volume of component 2 was calculated by the equation:
v2;1  limts3 0 vS;1
(25)
In this case the value of ṽS;1 with ts3 = 0 was taken as ṽ2;1. The apparent volume of
component 3 was calculated by the equation:
v3,1  limts3 1 vS,1
(26)
This value was obtained fitting the ten latest points of ṽS;1 to a straight line and
extrapolating to ts3 = 1. Values of ΔvS;1 were calculated substituting the equation:
vS;1  vS;1   t s2 v 2;1  t s3 v3;1 
(27)
The approximated values of Δv2;1,3 and Δv3;1,2 using the proposed method were calculated
using the equations (19) and (20) of the paper. The derivatives were estimated using a
numerical scheme.
The exact valued of Δv2;1,3 and Δv3;1,2 were calculated in the following way. From the fit
function for the excess molar volume of the ternary liquid mixture of the components given
in the work of Vela et al. [3], the excess specific volume as function of t1 and t3 was
obtained. Specific partial volumes of components 2 and 3 were calculated from vE(t1,t3) and
the values of the densities of the components using the following equations:
 v E (t , t ) 

 v E (t1 , t 3 ) 
1 3
v 2;1,3 (t1 , t 3 )  v E  
 t1  
  t 3   v2
t1
t 3


 t3

 t1
(28)
  v E (t , t ) 

 v E (t1 , t 3 ) 
1 3
v3;1,2 (t1 , t 3 )  v E    
t

 1 
 (t1  t 2 )   v3
t1
t 3
 

 t3

 t1
(29)
The partial volume of component of component 2 alone in solution (v2;1) was calculated by
the equation:
v 2;1  v 2;1,3 (t1 , 0)
(30)
where the true value of t1 was obtained using Solver Option of Microsoft Excel as in the
same way than above. In the same way, v3;1 was calculated by the equation
A thermodynamic method to study the interaction between NaOH and a highly carboxylate polymeric
particles
v3;1  v3;1,2 (t 2 , t 3 )
(31)
where t3 = 1-t1 and where the true value of t1 was obtained again by Solver Option of
Microsoft Excel.
References
[1] Santillán, R., Nieves, E., Alejandre, P., Pérez, E., del Río, J. M., Corea, M.;
Comparative thermodynamic study of functional polymeric latex particles with different
morphologies. Colloid Surface A. 444, 189-208 (2014)
[2] Chalikian, T.V., Sarvazyan, A.P., Breslauer, K.J.: Partial molar volumes,
expansibilities, and compressibilities of alpha, omega amino carboxylic acids in aqueous
solutions between 18 and 55◦C. J. Phys. Chem. 97, 13017–13026 (1993)
[3] Vela, M. P., Artal, M., Embid, J. M., Bravo, R., Otín, S.: J. Chem. Eng. Data. 57, 11391145 (2012)