Equilibrium swelling of multi-stimuli-responsive superabsorbent hydrogels

Abstract

Superabsorbent gels (lightly cross-linked copolymer gels with water absorption capacity up to several thousands g/g) have recently attracted substantial attention due to their novel applications in agriculture, environmental management, and civil engineering. As superabsorbent hydrogels (SAHs) are conventionally prepared by copolymerization of polyelectrolyte and temperature-sensitive neutral monomers, their response is strongly affected by temperature, pH, and ionic strength of solutions. A characteristic feature of SAHs is an anomalous increase in their elastic modulus with degree of swelling.

Constitutive equations are derived for the mechanical response and equilibrium swelling of SAHs. An advantage of the model is that it involves only six material constants. These quantities are found by fitting experimental data in equilibrium swelling tests and uniaxial compressive tests on a series of N-isopropylacrylamide-co-2-acrylamido-2-methylpropane sulfonic acid gels with various molar fractions of ionic comonomers. An acceptable agreement is demonstrated between the observations and results of simulation. The model is applied to examine the effects of temperature, pH, and molar fraction of monovalent salt in aqueous solutions on equilibrium swelling of SAHs with various molar fractions of comonomers in the feed.

Appendices

Appendix : 1

The specific free energy of a gel is determined by Eq. 37. Differentiation of Eq. 33 for the specific mechanical energy W with respect to time implies that

$$ \dot{W} = W_{,1}\dot{I}_{\text{e 1}}+W_{,2}\dot{I}_{\text{e 2}} +W_{,3}\dot{I}_{\text{e 1}}, $$

(A-1)

where

$$ W_{,m}=\frac{\partial W}{\partial I_{\mathrm{e}m}} \qquad (m=1,2,3). $$

(A-2)

The derivatives of the principal invariants I_e1, I_e2, and I_e3 of the Cauchy–Green tensor B_e with respect to time are given by

$$ \dot{I}_{\text{e1}}=2 \mathbf{B}_{\mathrm{e}}: \mathbf{D},\qquad \dot{I}_{\text{e2}}=2 \left( I_{\text{e2}} \mathbf{I}-I_{\text{e3}} \mathbf{B}_{\mathrm{e}}^{-1}\right): \mathbf{D},\qquad \dot{I}_{\text{e3}}=2 I_{\text{e3}} \mathbf{I}: \mathbf{D}, $$

(A-3)

where the rate-of-strain tensor D is determined by Eq. 40. Combination of Eqs. A-1 and A-3 results in

$$ \dot{W}=2\mathbf{K}_{\text{mech}}:\mathbf{D} $$

(A-4)

with

$$ \mathbf{K}_{\text{mech}} = W_{,1}\mathbf{B}_{\mathrm{e}} -I_{\text{e3}} W_{,2}\mathbf{B}_{\mathrm{e}}^{-1} +\left( I_{\text{e2}} W_{,2}+I_{\text{e3}} W_{,3}\right)\mathbf{I}. $$

(A-5)

Differentiation of Eq. 20 for the specific energy of the electric field W_el with respect to time implies that

$$ \dot{W}_{\text{el}}=\frac{1}{2\epsilon J}\left( 2{\mathbf{H}}\cdot {\mathbf{C}}\cdot \dot{\mathbf{H}} +{\mathbf{H}}\cdot \dot{\mathbf{C}}\cdot {\mathbf{H}} -\frac{\dot{J}}{J}{\mathbf{H}}\cdot {\mathbf{C}}\cdot {\mathbf{H}}\right). $$

This relation together with Eq. 16 yields

$$ \dot{W}_{\text{el}}={\mathbf{E}} \cdot \dot{\mathbf{H}} +\frac{1}{2\epsilon J}\left( {\mathbf{H}}\cdot \dot{\mathbf{C}}\cdot {\mathbf{H}} -\frac{\dot{J}}{J}{\mathbf{H}}\cdot {\mathbf{C}}\cdot {\mathbf{H}}\right). $$

(A-6)

It follows from Eqs. 17, 40, and 41 that

$$ \dot{\mathbf{C}}=2\mathbf{F}^{\top}\cdot{\mathbf{D}}\cdot{\mathbf{F}}. $$

Combination of this equality with Eq. 15 yields

$$ \frac{1}{2\epsilon J} {\mathbf{H}}\cdot \dot{\mathbf{C}}\cdot {\mathbf{H}} =\frac{1}{\epsilon J} {\mathbf{H}}\cdot {\mathbf{F}}^{\top}\cdot {\mathbf{D}} \cdot {\mathbf{F}}\cdot {\mathbf{H}} =\frac{J}{\epsilon}{\mathbf{h}}\cdot {\mathbf{D}}\cdot {\mathbf{h}} =\frac{J}{\epsilon}(\mathbf{h}\otimes \mathbf{h}): \mathbf{D}, $$

(A-7)

where ⊗ stands for the tensor product. Keeping in mind that

$$ \dot{J}=J \mathbf{I}:\mathbf{D}, $$

(A-8)

and using Eq. 17, we conclude that

$$ \frac{\dot{J}}{2\epsilon J^{2}}\mathbf{H}\cdot {\mathbf{C}}\cdot {\mathbf{H}} =\frac{1}{2\epsilon J}{\mathbf{H}}\cdot \mathbf{F}^{\top}\cdot {\mathbf{F}}\cdot {\mathbf{H}} (\mathbf{I}:\mathbf{D}) =\frac{J}{2\epsilon} (\mathbf{h}\cdot \mathbf{h})\mathbf{I}:\mathbf{D}. $$

(A-9)

Substitution of Eqs. A-7 and A-9 into Eq. A-6 implies that

$$ \dot{W}_{\text{el}}={\mathbf{E}} \cdot \dot{\mathbf{H}} +2\mathbf{K}_{\text{el}}:\mathbf{D}, $$

(A-10)

where

$$ \mathbf{K}_{\text{el}}=\frac{J}{2\epsilon} \left[ (\mathbf{h}\otimes \mathbf{h})-\frac{1}{2} (\mathbf{h}\cdot \mathbf{h})\mathbf{I}\right]. $$

(A-11)

Differentiation of the other terms in Eq. 37 is straightforward. The derivative of expression Eq. 37 with respect to time is determined by Eq. 43 with

$$ \begin{array}{@{}rcl@{}} && {\Theta}_{C} = \mu^{0}+k_{\mathrm{B}}T \left[ \ln \frac{C v}{1+C v}+\frac{1}{1+C v} +\frac{\chi}{(1+C v)^{2}} -\frac{C_{\mathrm{H}^{+}}+C_{\text{Na}^{+}}+C_{\text{OH}^{-}}+C_{\text{Cl}^{-}}}{C}\right],\\ && {\Theta}_{\mathrm{H}^{+}} = {\mu}_{\mathrm{H}^{+}}^{0} +k_{\mathrm{B}}T \ln \frac{C_{\mathrm{H}^{+}}}{C},\qquad {\Theta}_{\text{Na}^{+}} = {\mu}_{\text{Na}^{+}}^{0} +k_{\mathrm{B}}T \ln \frac{C_{\text{Na}^{+}}}{C},\\ && {\Theta}_{\text{OH}^{-}} = {\mu}_{\text{OH}^{-}}^{0} +k_{\mathrm{B}}T \ln \frac{C_{\text{OH}^{-}}}{C},\qquad {\Theta}_{\text{Cl}^{-}} = {\mu}_{\text{Cl}^{-}}^{0} +k_{\mathrm{B}}T \ln \frac{C_{\text{Cl}^{-}}}{C}. \end{array} $$

(A-12)

Appendix : 2

Under unconstrained swelling of a gel, the deformation gradient for macrodeformation reads

$$ \mathbf{F}=(1+C v)^{\frac{1}{3}} \mathbf{I}. $$

(B-1)

Substitution of Eqs. B-1 and 10 into Eq. 11 implies that

$$ \mathbf{F}_{\mathrm{e}}=\left( \frac{1+C v}{1+C_{0} v}\right)^{\frac{1}{3}} \mathbf{I}. $$

(B-2)

It follows from Eqs. B-2 and 12 that

$$ \mathbf{B}_{\mathrm{e}}=\left( \frac{1+C v}{1+C_{0} v}\right)^{\frac{2}{3}} \mathbf{I}. $$

Insertion of this relation into Eq. 54 yields

$$ \mathbf{T}=T\mathbf{I}, $$

(B-3)

where

$$ T=-{\Pi}+\frac{G}{1+C v} \left[\frac{1}{V} \left( \frac{1+C v}{1+C_{0}v} \right)^{\frac{2}{3}}-1\right] $$

(B-4)

and

$$ V=1+\frac{3}{K} \left[ \left( \frac{1+C v}{1+C_{0}v} \right)^{\frac{2}{3}}-1\right]. $$

(B-5)

When pressure $\bar {\Pi }$ is the bath is disregarded, the surface of a gel is traction-free. It follows from this condition and the equilibrium equation for the Cauchy stress tensor that T = 0. Combining this equality with Eq. B-3 and using Eq. B-4, we find that

$$ {\Pi}=\frac{G}{1+C v} \left[ \frac{1}{V} \left( \frac{1+C v}{1+C_{0}v} \right)^{\frac{2}{3}}-1\right]. $$

(B-6)

Substitution of Eqs. 48 and 55 for the chemical potentials of mobile ions into Eq. 56 results in

$$ \begin{array}{@{}rcl@{}} && \ln \frac{C_{\mathrm{H}^{+}}}{C} = \ln \frac{\bar{c}_{\mathrm{H}^{+}}}{\bar{c}} -\frac{e}{k_{\mathrm{B}}T}\left( {\Phi}-\bar{{\Phi}}\right),\quad \ln \frac{C_{\text{Na}^{+}}}{C} = \ln \frac{\bar{c}_{\text{Na}^{+}}}{\bar{c}} -\frac{e}{k_{\mathrm{B}}T}\left( {\Phi}-\bar{{\Phi}}\right),\\ && \ln \frac{C_{\text{OH}^{-}}}{C} = \ln \frac{\bar{c}_{\text{OH}^{-}}}{\bar{c}} +\frac{e}{k_{\mathrm{B}}T}\left( {\Phi}-\bar{\Phi}\right),\quad \ln \frac{C_{\text{Cl}^{-}}}{C} = \ln \frac{\bar{c}_{\text{Cl}^{-}}}{\bar{c}} +\frac{e}{k_{\mathrm{B}}T}\left( {\Phi}-\bar{\Phi}\right). \end{array} $$

(B-7)

It follows from Eq. B-7 that

$$ \begin{array}{@{}rcl@{}} && \frac{C_{\mathrm{H}^{+}}}{C} \frac{C_{\text{OH}^{-}}+C_{\text{Cl}^{-}}}{C} =\frac{\bar{c}_{\mathrm{H}^{+}}}{\bar{c}} \frac{\bar{c}_{\text{OH}^{-}}+\bar{c}_{\text{Cl}^{-}}}{\bar{c}}, \end{array} $$

(B-8)

$$ \begin{array}{@{}rcl@{}} && \frac{C_{\mathrm{H}^{+}}+C_{\text{Na}^{+}}}{C} \frac{C_{\text{OH}^{-}}+C_{\text{Cl}^{-}}}{C} =\frac{\bar{c}_{\mathrm{H}^{+}}+\bar{c}_{\text{Na}^{+}}}{\bar{c}} \frac{\bar{c}_{\text{OH}^{-}}+\bar{c}_{\text{Cl}^{-}}}{\bar{c}}. \end{array} $$

(B-9)

The electro-neutrality conditions for the gel and for the bath read

$$ \begin{array}{@{}rcl@{}} C_{\text{OH}^{-}}+C_{\text{Cl}^{-}} &=& C_{\mathrm{H}^{+}}+C_{\text{Na}^{+}}-C_{\text{b}^{-}}, \end{array} $$

(B-10)

$$ \begin{array}{@{}rcl@{}} \bar{c}_{\text{OH}^{-}}+\bar{c}_{\text{Cl}^{-}} &=& \bar{c}_{\mathrm{H}^{+}}+\bar{c}_{\text{Na}^{+}}. \end{array} $$

(B-11)

Inserting expression Eq. B-11 into Eq. B-9 and using Eqs. 3, 5, and 6, we find that

$$ \frac{C_{\mathrm{H}^{+}}+C_{\text{Na}^{+}}}{C} \frac{C_{\text{OH}^{-}}+C_{\text{Cl}^{-}}}{C} =\frac{1}{\kappa^{2}}\left( 10^{-\text{pH}}+\theta\right)^{2}. $$

(B-12)

Using Eq. B-10, we present Eq. B-12 in the form

$$ X\left( X-\frac{C_{\text{b}^{-}}}{C}\right) =\frac{1}{\kappa^{2}}({10}^{-\text{pH}}+\theta)^{2}, $$

(B-13)

where

$$ X=\frac{C_{\mathrm{H}^{+}}+C_{\text{Na}^{+}}}{C}. $$

(B-14)

Inserting Eqs. B-10 and B-11 into Eq. B-8, we find that

$$ \frac{C_{\mathrm{H}^{+}}}{C} \frac{C_{\mathrm{H}^{+}}+C_{\text{Na}^{+}}-C_{\text{b}^{-}}}{C} =\frac{\bar{c}_{\mathrm{H}^{+}}}{\bar{c}} \frac{\bar{c}_{\mathrm{H}^{+}}+\bar{c}_{\text{Na}^{+}}}{\bar{c}}. $$

Using Eqs. 3, 5, 6, and B-14 and introducing the notation

$$ X_{1}=\frac{C_{\mathrm{H}^{+}}}{C}, $$

(B-15)

we present this relation in the form

$$ X_{1}\left( X-\frac{C_{\text{b}^{-}}}{C}\right) =\frac{1}{\kappa^{2}}10^{-\text{pH}}\left( 10^{-\text{pH}}+\theta\right). $$

(B-16)

Dividing Eq. B-16 by Eq. B-13, we find that

$$ X_{1}=\frac{10^{-\text{pH}}}{{10}^{-\text{pH}}+\theta} X. $$

(B-17)

Introducing pH in the gel by means of Eqs. 3 and 5,

$$ \text{pH}_{\mathrm{g}}=-\log \left( \kappa \frac{C_{\mathrm{H}^{+}}}{C}\right), $$

(B-18)

and using Eqs. B-15 and B-17, we arrive at Eq. 61.

Substitution of expression Eq. B-6 into Eqs. 48 and 49 for the chemical potential of water molecules in the gel results in

$$ \begin{array}{@{}rcl@{}} \mu &=& \mu^{0}+k_{\mathrm{B}}T \left[ \ln \frac{C v}{1+C v}+\frac{1}{1+C v} +\frac{\chi}{(1+C v)^{2}} +\frac{g}{1+C v} \left( \frac{1}{V} \left( \frac{1+C v}{1+C_{0}v} \right)^{\frac{2}{3}}-1\right)\right.\\ &&\left. -\frac{C_{\mathrm{H}^{+}}+C_{\text{Na}^{+}}+C_{\text{OH}^{-}}+C_{\text{Cl}^{-}}}{C}\right], \end{array} $$

(B-19)

where g = Gv/(k_BT) is the dimensionless shear modulus. Using Eqs. B-10 and B-14, we present Eq. B-19 in the form

$$ \begin{array}{@{}rcl@{}} \mu &=& \mu^{0}+k_{\mathrm{B}}T \left[ \ln \frac{C v}{1+C v}+\frac{1}{1+C v} +\frac{\chi}{(1+C v)^{2}} +\frac{g}{1+C v} \left( \frac{1}{V} \left( \frac{1+C v}{1+C_{0}v} \right)^{\frac{2}{3}}-1\right)\right.\\ &&\left.-\left( 2X -\frac{C_{\mathrm{b}^{-}}}{C}\right)\right]. \end{array} $$

(B-20)

Applying Eqs. 3, 5, and B-11, we transform Eq. 55 for the chemical potential of water molecules in the bath as follows:

$$ \bar{\mu} = \mu^{0}-\frac{2k_{\mathrm{B}}T}{\kappa} \left( 10^{-\text{pH}}+\theta\right). $$

(B-21)

Substitution of Eqs. B-20 and B-21 into Eq. 56 implies that

$$ \begin{array}{@{}rcl@{}} && \ln \frac{C v}{1+C v}+\frac{1}{1+C v} +\frac{\chi}{(1+C v)^{2}} +\frac{g}{1+C v} \left( \frac{1}{V} \left( \frac{1+C v}{1+C_{0}v} \right)^{\frac{2}{3}}-1\right)\\ && +\frac{C_{\text{b}^{-}}}{C} -2 \left[ X-\frac{1}{\kappa}\left( 10^{-\text{pH}}+\theta \right)\right]=0. \end{array} $$

(B-22)

It follows from Eq. B-13 that

$$ \left( X-\frac{1}{\kappa}\left( 10^{-\text{pH}}+\theta\right)\right) \left( X+\frac{1}{\kappa}\left( 10^{-\text{pH}}+\theta\right) \right) =\frac{C_{\text{b}^{-}}}{C}X , $$

which implies that

$$ X-\frac{1}{\kappa}\left( 10^{-\text{pH}}+\theta\right) =\frac{C_{\text{b}^{-}}}{C}X \left( X+\frac{1}{\kappa}\left( 10^{-\text{pH}}+\theta\right) \right)^{-1}. $$

(B-23)

Combining Eqs. B-22 and B-23 and using Eq. 59, we arrive at Eqs. 57 and 58.

Appendix : 3

To calculate the Young modulus of a gel in the fully swollen state, we analyze rapid uniaxial tension of a sample, where the strain rate under loading exceeds strongly the rates of diffusion for water molecules and mobile ions. The deformation gradient F for macrodeformation reads

$$ \mathbf{F}=\mathbf{F}_{2}\cdot{ \mathbf{F}}_{1}, $$

(C-1)

where

$$ \mathbf{F}_{1}=(1+Q)^{\frac{1}{3}}\mathbf{I} $$

(C-2)

describes isotropic extension of the sample under swelling, and

$$ \mathbf{F}_{2}=\lambda \boldsymbol{i}_{1}\otimes\boldsymbol{i}_{1} +\lambda^{-\frac{1}{2}}(\boldsymbol{i}_{2}\otimes\boldsymbol{i}_{2}+\boldsymbol{i}_{3}\otimes\boldsymbol{i}_{3}) $$

(C-3)

characterizes uniaxial tension along the axis i₁ with elongation ratio λ (i₁,i₂,i₃ denote unit vectors of a Cartesian frame, and the coefficient $\lambda ^{-\frac {1}{2}}$ is calculated from Eq. 8).

Inserting expressions Eq. C-2 and C-3 into Eq. C-1 and using Eq. 12, we find that

$$ \mathbf{B}_{\mathrm{e}}=\left( \frac{1+Q}{1+Q_{0}}\right)^{\frac{2}{3}} \left[ \lambda^{2} \boldsymbol{i}_{1}\otimes\boldsymbol{i}_{1} +\lambda^{-1}(\boldsymbol{i}_{2}\otimes\boldsymbol{i}_{2}+\boldsymbol{i}_{3}\otimes\boldsymbol{i}_{3})\right]. $$

(C-4)

Equations (C-4) and (54) result in the following expression for the Cauchy stress tensor:

$$ \mathbf{T}= T_{1} \boldsymbol{i}_{1}\otimes\boldsymbol{i}_{1} +T_{2}(\boldsymbol{i}_{2}\otimes\boldsymbol{i}_{2}+\boldsymbol{i}_{3}\otimes\boldsymbol{i}_{3}), $$

where

$$ T_{1}=-{\Pi}+\frac{G}{1+Q}\left[ \frac{1}{V} \left( \frac{1+Q}{1+Q_{0}}\right)^{\frac{2}{3}}\lambda^{2} -1 \right], \quad T_{2}=-{\Pi}+\frac{G}{1+Q}\left[ \frac{1}{V} \left( \frac{1+Q}{1+Q_{0}}\right)^{\frac{2}{3}}\frac{1}{\lambda}-1 \right] $$

(C-5)

with

$$ V=1-\frac{1}{K} \left[ \left( \frac{1+Q}{1+Q_{0}}\right)^{\frac{2}{3}}\left( \lambda^{2}+\frac{2}{\lambda}\right) -3 \right]. $$

(C-6)

Bearing in mind that the lateral surface of a sample is traction-free, we set T₂ = 0. Excluding π from this equality and Eq. C-5, we conclude that

$$ T_{1}=\frac{G}{(1+Q)V}\left( \frac{1+Q}{1+Q_{0}}\right)^{\frac{2}{3}} \left( \lambda^{2} -\frac{1}{\lambda} \right). $$

(C-7)

The engineering tensile stress σ = T₁/λ is given by

$$ \sigma=\frac{G}{(1+Q)V}\left( \frac{1+Q}{1+Q_{0}}\right)^{\frac{2}{3}} \left( \lambda -\frac{1}{\lambda^{2}} \right). $$

(C-8)

At small strains, when λ = 1 + 𝜖 and 𝜖 ≪ 1, Eq. C-8 is simplified,

$$ \sigma=\alpha E \epsilon, $$

(C-9)

where E = 3G stands for the Young modulus of the polymer network, and

$$ \alpha=\frac{1}{(1+Q)^{\frac{1}{3}}(1+Q_{0})^{\frac{2}{3}}} \left[1-\frac{3}{K}\left( \left( \frac{1+Q}{1+Q_{0}}\right)^{\frac{2}{3}}-1\right)\right]^{-1}. $$

(C-10)

When results in tensile tests are compared on fully swollen samples (with equilibrium degree of swelling Q) and as-prepared samples (with degree of swelling Q_prep), it follows from Eqs. C-9 and C-10 that the Young moduli measured in these tests, E and E_prep, are connected by Eq. 62.