Dissociation Behavior of l(+)-Lactic Acid in Aqueous Solutions of (1-Alkyl-4-methylpyridinium bromide + Poly (ethyleneglycol)) at T = (288.15–318.15) K

Abstract

The effect of 1-alkyl-4-methylpyridinium based ionic liquids on the conductivity behavior of l(+)-lactic acid (LaH) was studied in Poly(ethylene glycol) (PEG) aqueous solutions. The molar conductivities of LaH in the aqueous solutions of PEG, (PEG + 1-hexyl-4-methylpyridinium bromide) and (PEG + 1-octyl-4-methylpyridinium bromide) were measured over the temperature ranges of 288.15–318.15 K. The molar conductivity data were analyzed by applying the Quint–Viallard (QV) conductivity equation to determine the limiting molar conductivities (Λ ⁰) and dissociation constants ($K_{\text{D}}$). The results show that the values of limiting molar conductivity increased as the temperature increased, which indicates that the dissociation process is endothermic. The $K_{\text{D}}$ values were also used to calculate the dissociation standard thermodynamic functions ($\Delta G_{\text{D}}^{0}$, $\Delta S_{\text{D}}^{0}$ and $\Delta H_{\text{D}}^{0}$). The results revealed that the dissociation process of LaH is controlled by entropy at all temperatures.

Appendix

The coefficients of the Quint—Viallard conductance equation were calculated as follows [24, 41]:

$$ S_{{}} = \alpha \varLambda^{0} + \beta_{{}} $$

(13)

$$ E = E_{1} \varLambda^{0} - E_{2} $$

(14)

$$ J_{1} = \sigma_{1} \varLambda^{0} + \sigma_{2} $$

(15)

$$ J_{2} = \sigma_{3} \varLambda^{0} + \sigma_{4} $$

(16)

$$ \alpha_{{}} = \frac{{2.8012 \times 10^{6} }}{{(DT)^{3/2} }}\left| {z_{1} } \right|\left| {z_{2} } \right|\frac{q}{1 + \sqrt q }, \, q = 1/2 $$

(17)

$$ \beta_{{}} = \frac{41.243}{{\eta (DT)^{1/2} }}\left| z \right| $$

(18)

$$ E_{1} = \frac{{5.8851 \times 10^{12} }}{{(DT)^{3} }}z_{1}^{2} z_{2}^{2} q $$

(19)

$$ E_{2} = \frac{{4.3324 \times 10^{7} }}{{\eta (DT)^{2} }}\left| {z_{1} } \right|\left| {z_{2} } \right|Q_{1} $$

(20)

$$ Q{}_{1} = \left\{ {\frac{{\left( {\left| {z_{1} } \right| + \left| {z_{2} } \right|} \right)q\lambda_{j}^{0} }}{{\lambda_{1}^{0} + \lambda_{2}^{0} }} + q\left| {z_{j} } \right| - \frac{{2z_{j} \left| {z_{j} } \right|(z_{1} + z_{2} )}}{{z_{1} z_{2} }}} \right\} $$

(21)

$$ \sigma_{1} = \frac{{1.17702 \times 10^{13} }}{{(DT)^{3} }}z_{1}^{2} z_{2}^{2} Q_{2} $$

(22)

$$ Q_{2} = \gamma + \ln \xi + R_{1} - R_{2} + R_{3} $$

(23)

$$ \xi = 5.0291 \times 19^{9} {a \mathord{\left/ {\vphantom {a {(DT)^{{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}}} }}} \right. \kern-0pt} {(DT)^{{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0pt} 2}}} }}, \, \gamma = 0.5772156649 $$

(24)

$$ R_{1} = \frac{{(1 - q)^{3} \ln (1 + \sqrt q ) + q(q^{2} - q + 2)\ln (2 + \sqrt q ) + 2q(1 - 2q)\ln (1 + 2\sqrt q )}}{2q(1 - q)} $$

(25)

$$ R_{2} = \frac{{6 + 15\sqrt q + 30q + 23q^{3/2} - 6q^{2} }}{{12\sqrt q (1 + \sqrt q )^{2} }} $$

(26)

$$ R_{3} = \frac{{2z_{1}^{2} z_{2}^{2} b^{2} + 2\left| {z_{1} } \right|\left| {z_{2} } \right|b - 1}}{{\left| {z_{1} } \right|^{3} \left| {z_{2} } \right|^{3} b^{3} }} $$

(27)

$$ b = \frac{{1.6671 \times 10^{ - 3} }}{aDT} $$

(28)

$$ \sigma_{2j} = \frac{{8.6648 \times 10^{7} }}{{\eta (DT)^{2} }}\left| {z_{1} } \right|\left| {z_{2} } \right|qQ_{3} $$

(29)

$$ Q_{3} = \left\{ { - \frac{{\left( {\left| {z_{1} } \right| + \left| {z_{2} } \right|} \right)\lambda_{j}^{0} Q_{4} }}{{\lambda_{1}^{0} + \lambda_{2}^{0} }} - \left( {\left| {z_{j} } \right| + \frac{{2z_{j} \left| {z_{j} } \right|(z_{1} + z_{2} )}}{{q\left| {z_{1} } \right|\left| {z_{2} } \right|}}} \right)Q_{5} + \left| {z_{j} } \right|Q_{6} } \right\} $$

(30)

$$ Q_{4} = \gamma + \ln \xi + \frac{2}{{3\left| {z_{1} } \right|\left| {z_{2} } \right|b}} + R_{4} - R_{5} $$

(31)

$$ Q_{5} = \gamma + \ln 2 + \ln \xi $$

(32)

$$ Q_{6} = - \ln 2 + \frac{3}{2} + \frac{1}{{z_{1}^{2} z_{2}^{2} b^{2} }} + \frac{8 - 3q}{{2qb\left| {z_{1} } \right|\left| {z_{2} } \right|}} + R_{6} $$

(33)

$$ R_{4} = \frac{{(1 - q)^{2} \ln (1 + \sqrt q ) - q(q - 4)\ln (2 + \sqrt q )}}{2q} $$

(34)

$$ R_{5} = \frac{6 + 13\sqrt q - 6q}{12\sqrt q } $$

(35)

$$ R_{6} = \frac{(1 + q)\ln (1 + \sqrt q ) - 2q\ln 2 - \sqrt q (1 - \sqrt q )}{(1 - q)} $$

(36)

$$ \sigma_{3} = \frac{{16E_{1} E_{2} }}{\beta }\frac{{0.8047b^{2} + 2.2137b + 1.9024}}{{b^{3} }} $$

(37)

$$ \sigma_{4} = 2E_{1} \beta \frac{8.5 - 2.8383b}{{b^{2} }} $$

(38)