Equilibrium Potentials II Answer 2

Here are the answers for Figure 8, when sodium is not permeable:

• Solving for the concentration of the chloride ions outside of the cell (the right-hand side) can be done easily by applying the rule of bulk electroneutrality.
• The total concentration of positive charges on the outside is 120 mM + 5 mM = 125 mM, so the concentration of chloride ions must also be $\left[\text{Cl}^-\right]_\text{out} = 125\ \text{mM}$.
• The total number of particles on the outside is thus 120 mOsM + 5 mOsM + 125 mOsM = 250 mOsM.
• In turn, this implies a constraint on the number of particle inside the cell, which must equal $$250\ \text{mOsm} = \left[\text{Na}^+\right]_\text{in} + \left[\text{K}^+\right]_\text{in} + 5\ \text{mOsm} + 110\ \text{mOsm},$$ which in turn implies that $$250\ \text{mOsm} - 5\ \text{mOsm} - 110\ \text{mOsm} = 135\ \text{mOsm} = \left[\text{Na}^+\right]_\text{in} + \left[\text{K}^+\right]_\text{in}.$$
• We also have Equation 7, $$\left[\text{Cl}^-\right]_\text{in}\ \left[\text{K}^+\right]_\text{in} = \left[\text{K}^+\right]_\text{out}\ \left[\text{Cl}^-\right]_\text{out}.$$
• Since we know the concentrations of the potassium and chloride ions outside the cell, and the concentration of chloride ions inside the cell, this becomes $$5\ \text{mM} \times \left[\text{K}^+\right]_\text{in} = 5\ \text{mM} \times 125\ \text{mM},$$ so that $$\left[\text{K}^+\right]_\text{in} = 125\ \text{mM}.$$
• Electroneutrality provides another equation: $$\left[\text{Na}^+\right]_\text{in} + \left[\text{K}^+\right]_\text{in} = 5\ \text{mM} + 1.2 (110\ \text{mM}) = 137\ \text{mM},$$ which implies that $$\left[\text{Na}^+\right]_\text{in} = 137\ \text{mM} - 125\ \text{mM} = 12\ \text{mM}.$$
• Now we have a problem: the number of particles on the inside of the cell is 12 mOsm + 125 mOsm + 5 mOsm + 110 mOsm = 252 mOsm; but we already established that the number on the outside of the cell is 250 mOsm!
• So, this implies that water will move from where it is more concentrated (i.e., on the outside) to where it is is less concentrated (i.e., to the inside, where the concentration of particles is higher).
• Assuming initial equal volumes of 1 liter, the change in volume will be (using Equation 3) $(250 \times 1 - 252 \times 1)/(252 + 250) = -2/502$, or a change in volume on either side of about 0.4 percent (an increase in volume on the inside, and a corresponding decrease on the outside). We will therefore neglect this, although if we wanted a precise answer, we would need to include this change.
• An estimate of the membrane potential is provided by solving the Nernst equation for the potassium ion, or $$E_{\text{K}^+} = 58\ \text{mV} \times \log_{10} \frac{5\ \text{mM}}{125\ \text{mM}} = -81\ \text{mV}.$$