To determine the electrochemical potential for each ion and predict their movement direction, we need to consider both the concentration gradient and the electrical membrane potential. We’ll use the Nernst equation to calculate the equilibrium potential for each ion and compare it with the membrane potential to decide the direction of ion flow.

Given Data:

• Intracellular concentration of Na⁺ ([Na⁺]_in): 10 mM
• Extracellular concentration of Na⁺ ([Na⁺]_out): 250 mM
• Intracellular concentration of Cl⁻ ([Cl⁻]_in): 6 mM
• Extracellular concentration of Cl⁻ ([Cl⁻]_out): 250 mM
• Membrane potential (Vm): -70 mV (inside of the cell is more negative)
• Temperature (T): 25 °C or 298 K

Step 1: Calculate the Equilibrium Potential Using the Nernst Equation

The Nernst equation is:

[ E_{\text{ion}} = \frac{RT}{zF} \ln \left( \frac{[\text{ion}]{\text{out}}}{[\text{ion}]{\text{in}}} \right) ]

Where:

• $R$ is the gas constant, approximately $8.314 \, \text{J/mol·K}$
• $T$ is the temperature in Kelvin (298 K at 25 °C)
• $F$ is the Faraday constant, approximately $96485 \, \text{C/mol}$
• $z$ is the charge of the ion (+1 for Na⁺ and -1 for Cl⁻)

We can simplify this to:

[ E_{\text{ion}} = \frac{61.5 , \text{mV}}{z} \ln \left( \frac{[\text{ion}]{\text{out}}}{[\text{ion}]{\text{in}}} \right) ]

a) Calculating the Equilibrium Potential for Na⁺:

For Na⁺ ($z = +1$):

$E_{\text{Na}^+} = \frac{61.5 \, \text{mV}}{1} \ln \left( \frac{250 \, \text{mM}}{10 \, \text{mM}} \right)$

1. Calculate the ratio: $\frac{250}{10} = 25$
2. Take the natural logarithm: $\ln(25) \approx 3.2189$
3. Multiply by 61.5 mV: $E_{\text{Na}^+} = 61.5 \times 3.2189 \approx 198 \, \text{mV}$

So, $E_{\text{Na}^+} \approx +198 \, \text{mV}$.

b) Calculating the Equilibrium Potential for Cl⁻:

For Cl⁻ ($z = -1$):

$E_{\text{Cl}^-} = \frac{61.5 \, \text{mV}}{-1} \ln \left( \frac{250 \, \text{mM}}{6 \, \text{mM}} \right)$

1. Calculate the ratio: $\frac{250}{6} \approx 41.67$
2. Take the natural logarithm: $\ln(41.67) \approx 3.73$
3. Multiply by 61.5 mV: $E_{\text{Cl}^-} = -61.5 \times 3.73 \approx -229 \, \text{mV}$

So, $E_{\text{Cl}^-} \approx -229 \, \text{mV}$.

Step 2: Predicting the Direction of Ion Movement

Now, compare each equilibrium potential with the membrane potential (-70 mV):

• For Na⁺:

  • The equilibrium potential for Na⁺ is +198 mV, which is more positive than the membrane potential (-70 mV).
  • Na⁺ will move inward because it is attracted to the negative interior and has a concentration gradient driving it into the cell.

• For Cl⁻:

  • The equilibrium potential for Cl⁻ is -229 mV, which is more negative than the membrane potential (-70 mV).
  • Cl⁻ will tend to move outward since the cell interior is less negative than its equilibrium potential, and it will follow its concentration gradient outward.

Summary:

• Na⁺ ions will move inward to bring the membrane potential closer to their equilibrium potential.
• Cl⁻ ions will move outward for the same reason.

Would you like further clarification on any part of this solution?

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Follow-up Questions:

1. What is the impact on membrane potential if Na⁺ and Cl⁻ reach their equilibrium potentials?
2. How would changing the external NaCl concentration affect the movement of Na⁺ and Cl⁻?
3. What would happen if the temperature were increased or decreased significantly?
4. How would adding