Chapter 13, Problem 5P

Interpretation Introduction

Interpretation:

The equilibrium potentials of two ions at the given temperature needs to be determined.

Concept introduction:

The equilibrium potential of two ions is calculated as

$V_{eq}=-\left(\frac{RT}{zF}\right)\ln \left(\frac{{\left[X\right]}_{in}}{{\left[X\right]}_{out}}\right)$

Here, ${\left[X\right]}_{in}$ represents the ion concentration of the inside cell and ${\left[X\right]}_{out}$ represents the ion concentration of the outside cell.

Answer to Problem 5P

The equilibrium potential of calcium ions is $0.122V$

The equilibrium potential of chloride ions is $-0.0968V$

Explanation of Solution

Given Information:

Intracellular concentration of chloride ion, ${\left[C{l}^{-}\right]}_{in}$ = 4mM

Extracellular concentration of chloride ion, ${\left[C{l}^{-}\right]}_{out}$ = 150mM

Intracellular concentration of calcium ion, ${\left[C{a}^{2+}\right]}_{in}$ = 0.2 $\mu M$

Extracellular concentration of chloride ion, ${\left[C{a}^{2+}\right]}_{out}$ = 1.8Mm

Temperature, T = 35ºC

Calculation:

The equilibrium potential of chloride ions is calculated by

$V_{eq}=-\left(\frac{RT}{zF}\right)\ln \left(\frac{{\left[C{l}^{-}\right]}_{in}}{{\left[C{l}^{-}\right]}_{out}}\right)$

Here,

$\begin{array}{l}R=8.314J{K}^{-1}mo{l}^{-1}(gasconstant)\\ T={37}^{\circ }C=310K\\ F=96.5J{V}^{-1}mo{l}^{-1}(faraday\text{'}s\text{ constant})\\ z=-1(charge\text{ of chloride ion})\end{array}$

$\begin{array}{l}{V}_{eq}=-\left(\frac{(8.314(310))}{(-1)(96,500)}\right)\ln \left(\frac{4}{150}\right)\\ V_{eq}=(-0.0267)ln(0.0267)\\ V_{eq}=-0.0968V\end{array}$

The equilibrium potential of calcium ions is calculated by

$V_{eq}=-\left(\frac{RT}{zF}\right)\ln \left(\frac{{\left[C{a}^{2+}\right]}_{in}}{{\left[C{a}^{2+}\right]}_{out}}\right)$

Here,

$\begin{array}{l}R=8.314J{K}^{-1}mo{l}^{-1}(gasconstant)\\ T={37}^{\circ }C=310K\\ F=96.5J{V}^{-1}mo{l}^{-1}(faraday\text{'}s\text{ constant})\\ z=+2(charge\text{ of calcium ion})\end{array}$

$\begin{array}{l}{V}_{eq}=-\left(\frac{(8.314(310))}{(+2)(96,500)}\right)\ln \left(\frac{0.2}{1800}\right)\\ V_{eq}=(0.01335)ln(1.11\times {10}^{-4})\\ V_{eq}=0.122V\end{array}$

Conclusion

The equilibrium potential of calcium ions is $0.122V$.

The equilibrium potential of chloride ions is $-0.0968V$.