Question

Reset Help a positive value For scenario (a), there of materials, because AG of transport will be and thus zero the concentration of ions of materials, because AG of transport will be For scenario (b), there won't be in equilibrium and thus the concentrations of ions will equalize of materials, because AG of transport will be For scenario (c), there a negative val ue and thus will be no transport will be a transport

The equilibrium membrane potentials to be expected across a membrane at 37 ∘ C, with a NaCl concentration of 0.10 M on the right and 0.01 M on the left, given the following conditions.

1. Membrane permeable only to Na+: Δψ = +0.0615 V.
2. Membrane permeable only to Cl−: Δψ = −0.0615 V.
3. Membrane equally permeable to both ions: Δψ = 0 V.

Will any appreciable transport of material take place in establishing the membrane potential? Briefly explain each answer.

Match the items in the left column to the appropriate blanks in the sentences on the right.

Answer 1

(a) No, because the concentration of ions won't be in equilibrium

(b) No, because the concentration of ions won't be in equilibrium

(c) Yes, because the concentrations of ions will equalize

Answer 2

For scenario (a), there will be no transport For scenario (a), there will be no transportof materials, because $\mathrm{\Delta }G$of transport will be will be no transportand thus will be no transport.  of materials, because ΔG$\mathrm{\Delta }G$ of transport will be a positive value For scenario (a), there will be no transportof materials, because $\mathrm{\Delta }G$of transport will be will be no transportand thus will be no transport.  and thus the concentration of ions won't be in equilibrium For scenario (a), there will be no transportof materials, because $\mathrm{\Delta }G$of transport will be a positive valueand thus a positive value. . For scenario (b), there will be no transport For scenario (b), there will be no transportof materials, because $\mathrm{\Delta }G$of transport will be will be no transportand thus will be no transport.  of materials, because ΔG$\mathrm{\Delta }G$ of transport will be a positive value For scenario (b), there will be no transportof materials, because $\mathrm{\Delta }G$of transport will be will be no transportand thus will be no transport.  and thus the concentration of ions won't be in equilibrium For scenario (b), there will be no transportof materials, because $\mathrm{\Delta }G$of transport will be a positive valueand thus a positive value. . For scenario (c), there will be a transport For scenario (c), there will be a transportof materials, because $\mathrm{\Delta }G$of transport will be will be a transportand thus will be a transport.  of materials, because ΔG$\mathrm{\Delta }G$ of transport will be zero For scenario (c), there will be a transportof materials, because $\mathrm{\Delta }G$of transport will be will be a transportand thus will be a transport.  and thus the concentrations of ions will equalize Explanation: As you are given a membrane, which influences the distribution of ions, a membrane potential is maintained. The free energy change (ΔG$\mathrm{\Delta }G$) for transport across such a membrane is given by the equation: ΔG=RTln[Cin][Cout]+ZFΔψ(1)$\mathrm{\Delta }G=RT\mathrm{l}\mathrm{n}\frac{[C_{\mathrm{i}\mathrm{n}}]}{[C_{\mathrm{o}\mathrm{u}\mathrm{t}}]}+ZF\mathrm{\Delta }\psi (1)$ where R$R$ is the gas constant, T$T$ is the temperature in K$\mathrm{K}$, Z$Z$ is the charge on the ion, F$F$ is the Faraday constant (96.5 kJmol−1V−1$\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}{\mathrm{V}}^{-1}$), and Δψ$\mathrm{\Delta }\psi$ is the change in membrane potential calculated as  Δψ=Δψfinal−Δψinitial$\mathrm{\Delta }\psi =\mathrm{\Delta }{\psi }_{\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}}-\mathrm{\Delta }{\psi }_{\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}$. In scenario (a) the charge on Na+$\mathrm{N}{\mathrm{a}}^{+}$ ions is +$+$1 and Δψ$\mathrm{\Delta }\psi$ is a positive value. Therefore, the expression ZFΔψ$\mathrm{Z}\mathrm{F}\mathrm{\Delta }\psi$ will be positive. You can calculate the value for ΔG$\mathrm{\Delta }G$ from Equation (1)$(1)$ as: ΔG=RTln[CNa+]in[CNa+]right+ZFΔψ(2)$\mathrm{\Delta }G=RT\mathrm{l}\mathrm{n}\frac{[C_{\mathrm{N}{\mathrm{a}}^{+}}{]}_{\mathrm{i}\mathrm{n}}}{[C_{\mathrm{N}{\mathrm{a}}^{+}}{]}_{\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}}+ZF\mathrm{\Delta }\psi (2)$ Replacing given values in Equation (2)$(2)$: ΔG=(0.008314kJmolK)(310K)(ln0.01M0.1M)+(+1)(96.5kJmolV)(+0.0615V)(3)$\mathrm{\Delta }G=\left(0.008314\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{K}}\right)\left(310\mathrm{K}\right)\left(\mathrm{l}\mathrm{n}\frac{0.01\mathrm{M}}{0.1\mathrm{M}}\right)+(+1)\left(96.5\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{V}}\right)(+0.0615\mathrm{V})(3)$  ΔG=(2.58kJmol)(−1)+(+1)(5.93kJmol)=3.35kJmol$\mathrm{\Delta }G=\left(2.58\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)(-1)+(+1)\left(5.93\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)=3.35\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}$ ΔG$\mathrm{\Delta }G$ is positive, suggesting unfavorable thermodynamic conditions due to a lack of ionic equilibrium between the two sides of the membrane. Therefore, no transport of materials would occur. In scenario (b) the charge on Cl−$\mathrm{C}{\mathrm{l}}^{-}$ ions is −$-$1 and Δψ$\mathrm{\Delta }\psi$ is a negative value. Therefore, the expression ZFΔψ$\mathrm{Z}\mathrm{F}\mathrm{\Delta }\psi$ will be positive. You can calculate the value for ΔG$\mathrm{\Delta }G$ from Equation (1)$(1)$ as: ΔG=RTln[CCl−]left[CCl−]right+ZFΔψ(4)$\mathrm{\Delta }G=RT\mathrm{l}\mathrm{n}\frac{[C_{\mathrm{C}{\mathrm{l}}^{-}}{]}_{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}}}{[C_{\mathrm{C}{\mathrm{l}}^{-}}{]}_{\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}}+ZF\mathrm{\Delta }\psi (4)$ Substituting known values into the equation gives ΔG=(0.008314kJmolK)(310K)(ln0.01M0.1M)+(−1)(96.5kJmolV)(−0.0615V)(5)$\mathrm{\Delta }G=\left(0.008314\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{K}}\right)(310\mathrm{K})\left(\mathrm{l}\mathrm{n}\frac{0.01\mathrm{M}}{0.1\mathrm{M}}\right)+(-1)\left(96.5\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{V}}\right)(-0.0615\mathrm{V})(5)$  ΔG=(2.58kJmol)(−1)+(+1)(5.93kJmol)=3.35kJmol$\mathrm{\Delta }G=\left(2.58\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)(-1)+(+1)\left(5.93\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)=3.35\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}$ ΔG$\mathrm{\Delta }G$ is positive, suggesting unfavorable thermodynamic conditions due to a lack of ionic equilibrium between the two sides of the membrane. Therefore, no transport of materials would occur. In scenario (c), you are given that the membrane is permeable to both Na+$\mathrm{N}{\mathrm{a}}^{+}$ and Cl−$\mathrm{C}{\mathrm{l}}^{-}$, and Δψ=0$\mathrm{\Delta }\psi =0$ V$\mathrm{V}$. Hence the charges on both Na+$\mathrm{N}{\mathrm{a}}^{+}$ and Cl−$\mathrm{C}{\mathrm{l}}^{-}$ ions will cancel out. Therefore, the expression ZFΔψ$ZF\mathrm{\Delta }\psi$, will be zero. Further, as the membrane is permeable to both ions, the concentration of ions will be equal on both sides. In this case, Cin$C_{\mathrm{i}\mathrm{n}}$and Cout$C_{\mathrm{o}\mathrm{u}\mathrm{t}}$ will be equal. The ratio [Cin][Cout]=1$\frac{[C_{\mathrm{i}\mathrm{n}}]}{[C_{\mathrm{o}\mathrm{u}\mathrm{t}}]}=1$, and the term [Cin][Cout]=log1=0$\frac{[C_{\mathrm{i}\mathrm{n}}]}{[C_{\mathrm{o}\mathrm{u}\mathrm{t}}]}=\mathrm{l}\mathrm{o}\mathrm{g}1=0$. Now ΔG$\mathrm{\Delta }G$ can be calculated as: ΔG=(2.58kJmol)(log1)+(96.5kJmolV)(0V)(6)$\mathrm{\Delta }G=\left(2.58\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)(\mathrm{l}\mathrm{o}\mathrm{g}1)+\left(96.5\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{V}}\right)(0\mathrm{V})(6)$  ΔG=(2.58kJmol)(0)+(0)=0kJmol(7)$\mathrm{\Delta }G=\left(2.58\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)(0)+(0)=0\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}(7)$ ΔG$\mathrm{\Delta }G$ is zero, suggesting favorable thermodynamic conditions due to the establishment of an ionic equilibrium between the two sides of the membrane. As a result, appreciable transport of material would take place in order to establish a membrane potential.