# Calculate the work required to transfer 1 mol of K+ from the blood to muscle cells?

## Potassium ions are moved across cell membranes by “ion pumps,” which make it possible for the cell to expend energy needed to maintain different concentrations of ions on either side of the membrane. In humans, the concentration of potassium ions in the blood is 5.0x10-3 M, while the concentration inside of muscle cells is 0.15M. Calculate the work required to transfer 1 mol of K+ from the blood to muscle cells.

Nov 19, 2016

The work required is 0.09 kJ.

#### Explanation:

A typical formula for the energy of ion transfer is

color(blue)(bar(ul(|color(white)(a/a) ΔG = RTln(["K"^(+)]_text(in)/["K"^(+)]_text(out)) + ZFΔψ color(white)(a/a)|)))

where

ΔG = the free energy change
$R$ = the Ideal Gas Law Constant
$T$ = the temperature
${\left[{\text{K}}^{+}\right]}_{\textrm{\in}}$ and ${\left[{\text{K}}^{+}\right]}_{\textrm{o u t}}$ are the potassium ion concentrations inside and outside the cell
$Z$ = the charge on the ion
$F$ = the Faraday constant
Δψ = the membrane potential

The first term in the expression represents the energy involved in crossing the concentration gradient.

The second term represents the energy involved in crossing the electrical gradient (the membrane potential).

In a typical skeletal muscle cell, the membrane potential for $\text{K"^"+}$ ions is -90 mV.

The negative sign indicates that the inside of the cell is negative with respect to the surrounding extracellular fluid.

In this problem,

R = 8.314 × 10^"-3" color(white)(l)"kJ·K"^"-1""mol"^"-1"
$T = \text{37 °C" = "310.15 K}$
["K"^(+)]_text(in) = "0.15 mol/L"
["K"^(+)]_text(out) = 5.0 × 10^"-3"color(white)(l) "mol/L"
$Z = + 1$
$F = \text{96 485 C·mol"^"-1" = "96.485 kJ·V"^"-1""mol"^"-1}$
Δψ = "-90 mV" = "-0.090 V"

ΔG = RTln(["K"^(+)]_text(in)/["K"^(+)]_text(out)) + ZFΔψ

= 8.314 × 10^"-3"color(white)(l) "kJ"·color(red)(cancel(color(black)("K"^"-1")))·"mol"^"-1"× 310.15 color(red)(cancel(color(black)("K"))) × ln((0.15 color(red)(cancel(color(black)("mol/L"))))/(5.0 × 10^"-3" color(red)(cancel(color(black)("mol/L"))))) + (+1)( "96.485 kJ"·color(red)(cancel(color(black)("V"^"-1")))"mol"^"-1")(-0.090 color(red)(cancel(color(black)("V"))))

$= 2.443 \ln 30 \text{kJ·mol"^"-1" - "6.75 kJ·mol"^"-1}$

$= \text{(8.77 - 8.68) kJ·mol"^"-1}$

$= \text{0.09 kJ·mol"^"-1}$

The transfer of 1 mol of potassium ions will require 0.09 kJ.

Under the specified conditions, the transfer of potassium ions to the muscle is almost energetically neutral.