Question

The potassium-ion concentration in blood plasma is about \(5.0 \times 10^{-3} M,\) whereas the concentration in muscle-cell fluid is much greater \((0.15 \mathrm{M}) .\) The plasma and intracellular fluid are separated by the cell membrane, which we assume is permeable only to \(\mathrm{K}^{+} .\) (a) What is \(\Delta G\) for the transfer of 1 \(\mathrm{mol}\) of \(\mathrm{K}^{+}\) from blood plasma to the cellular fluid at body temperature \(37^{\circ} \mathrm{C} ?\) (b) What is the minimum amount of work that must be used to transfer this \(\mathrm{K}^{+} ?\)

Short answer

The minimum work required to transfer 1 mol of K+ ions from blood plasma to muscle-cell fluid at body temperature \(37^{\circ}C\) is approximately 12000 Joules per mole.

Step-by-step solution

Write down the given information.

We have the following information: - Potassium-ion concentration in blood plasma: \(5.0 \times 10^{-3} M\) - Potassium-ion concentration in muscle-cell fluid: \(0.15 M\) - Body temperature: \(37°C\) We need to convert the temperature from Celsius to Kelvin by adding 273.15 to the Celsius value.

Convert the temperature to Kelvin.

To convert the temperature to Kelvin, we add 273.15 to 37: \[T = 37 + 273.15 = 310.15 K\]

Calculate the change in concentration.

The change in potassium-ion concentration can be calculated using the initial concentration in blood plasma and the final concentration in muscle-cell fluid: \[\Delta concentration = 0.15 M - 5.0 \times 10^{-3} M = 0.145 M\]

Calculate the Gibbs free energy change (ΔG).

We will use the following equation for calculating ΔG: \[\Delta G = RT \ln \frac{C_{2}}{C_{1}}\] where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, \(C_{1}\) is the initial concentration, and \(C_{2}\) is the final concentration. Substituting the values and solving for ΔG, we get: \[\Delta G = (8.314 \frac{J}{mol \cdot K})(310.15 K) \ln \frac{0.15 M}{5.0 \times 10^{-3} M}\] \[\Delta G \approx 12000 J/mol\] So, the Gibbs free energy change (ΔG) for transferring one mole of K+ from blood plasma to the muscle-cell fluid is approximately 12000 J/mol.

Calculate the minimum work required for transferring the ions.

The work to transfer one mole of K+ ions can be calculated using the ΔG value we found. As the minimum work required is equal to ΔG, we already have the answer: Minimum work required = ΔG Minimum work required ≈ 12000 J/mol The minimum work required to transfer 1 mol of K+ ions from blood plasma to muscle-cell fluid is approximately 12000 Joules per mole.

Key concepts

Potassium-Ion Concentration

Potassium-ion concentration plays a crucial role in various physiological processes within the body. It is essential for maintaining the electrical balance across cell membranes.
The concentration of potassium ions, \( \text{K}^+ \), differs significantly between the intracellular and extracellular environments. In blood plasma, potassium concentration is typically around \(5.0 \times 10^{-3} \text{M}\), while inside muscle cells, it increases to approximately \(0.15 \text{M}\).
This disparity is critical for the maintenance of the resting membrane potential, which is vital for nerve transmission and muscle contraction. The potassium concentration gradient is utilized by cells to perform work, such as transmitting signals in neurons. If the gradient were to change dramatically, it could disrupt cell functions and lead to serious physiological issues.
Understanding these concentration differences helps us appreciate the necessity of potassium in cellular activities and its regulation through diet and medication.

Cell Membrane Permeability

Cell membrane permeability refers to the cell membrane's ability to allow substances to pass in and out. The permeability of the membrane is a critical factor influencing the movement of potassium ions. In this context, the membrane is permeable only to \(\text{K}^+ \), meaning that these ions can move freely across the membrane.
The selective permeability is achieved through protein channels and pumps embedded in the membrane. These structures allow the cell to maintain its internal environment distinct from the extracellular fluid. For potassium ions, this permeability is managed primarily by potassium channels, which enable the passive movement of \(\text{K}^+ \) ions.By controlling the permeability, cells can:- Regulate their internal potassium concentration.- Maintain the electric potential needed for vital functions, like muscle contraction and nerve impulse conduction.
Through this regulation, cells effectively sustain life by ensuring a stable environment that supports biochemical processes.

Thermodynamics Calculations

Thermodynamics calculations in this context refer to the computation of the Gibbs free energy change \(\Delta G\) for moving potassium ions across the cell membrane. Gibbs free energy is a measure of the maximum reversible work done by a thermodynamic process at constant temperature and pressure.
To calculate \( \Delta G \), we use the formula:\[ \Delta G = RT \ln \frac{C_{2}}{C_{1}} \]where:- \( R \) is the gas constant, \(8.314 \frac{J}{mol \cdot K}\),- \( T \) is the temperature in Kelvin,\( 310.15 \text{ K} \) for body temperature in this case,- \( C_{2} \) and \( C_{1} \) are the final and initial concentrations of potassium ions.This equation shows how the concentration gradient influences \( \Delta G \). A significant gradient, like the one between blood plasma and muscle-cell fluid, results in a greater \( \Delta G \). In our calculation, we found \(\Delta G \approx 12000 \text{ J/mol} \), highlighting the energy associated with moving potassium ions against their concentration gradient.
Understanding these calculations is fundamental for biochemical and physiological studies, helping to explore how energy changes affect cellular processes.