Question

Calculate the chemical potential difference when intracellular [Na+] = 10 mM and extracellular [Na+] = 150 mM at 37°C. (b) What would the electrochemical potential be if the membrane potential were -60 mV (inside negative)?

Short answer

Therefore, in the solution (a), the free energy change for glucoseto enter into the cells at the concentration is $-7.0KJmo{l}^{-1}$.

Similarly, in the solution (b), the electrochemical potential for membrane potential is $-12.8KJmo{l}^{-1}$.

Step-by-step solution

Introduction

The amount of free energy required to move a substance across a membrane is determined by the concentrations on both sides of the membrane. The free energy change of ions is determined by the membrane potential. Because the free energy change is minimized when all concentrations are equal, a random distribution of molecules is an energy-rich condition. The chemical potential difference is caused by a difference in the concentrations of chemicals on opposite sides of the membrane.

Explanation for (a)

The intracellular concentration of $\left[N{a}^{+}\right]=0.01M$

The extracellular concentration of $\left[N{a}^{+}\right]=0.15M$

$\Delta G=RT\ln {\left[N{a}^{-}\right]}_{in}{\left[N{a}^{-}\right]}_{out}$

Then,

$\begin{array}{l}\mathrm{\Delta G}=\left(8.3145{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\right)\ln \left[0.01\right]\left[0.15\right]\\ =7.0{\mathrm{kJmol}}^{-1}\end{array}$

$\Delta G=RT\ln {\left[N{a}^{-}\right]}_{in}{\left[N{a}^{-}\right]}_{out}$

Explanation for (b)

The intracellular concentration of $\left[{\mathrm{Na}}^{+}\right]=0.01\mathrm{M}$

The extracellular concentration of $\left[{\mathrm{Na}}^{+}\right]=0.15\mathrm{M}$

$\Delta G=RT\ln {\left[N{a}^{-}\right]}_{in}{\left[N{a}^{-}\right]}_{out}$

Then,

$\begin{array}{l}\mathrm{\Delta G}=-6980+\left(1\right)\left(96485{\mathrm{JVmol}}^{-1}\right)\left(-0.06\mathrm{V}\right)\\ =12.8{\mathrm{kJmol}}^{-1}\end{array}$

Conclusion

Therefore, in the solution (a), the free energy change for glucose to enter into the cells at the concentration is $-7.0{\mathrm{kJmol}}^{-1}$.

Similarly, in the solution (b), the electrochemical potential for membrane potential is $-12.8{\mathrm{kJmol}}^{-1}$.