Reading equilibrium pot 1

Here are the answers for Figure 6:

• To maintain bulk electroneutrality, the 50 mM of positive sodium ions inside the cell must be balanced by 50 mM of negatively charged chloride ions. Therefore, $\left[\text{Cl}^-\right]_\text{in} = 50\ \text{mM}$.
• We can now sum up the total concentration of solute particles within the cell: 50 mM sodium ions, 50 mM chloride ions, and 100 mM uncharged proteins, for a total concentration of $200\ \text{milliosmolar}$ (note that concentration of particles is measured in osmolar rather than just in molar).
• To maintain bulk electroneutrality on the outside, the concentration of positively charged sodium ions has to equal the concentration of negatively charged chloride ions, or $\left[\text{Na}^+\right]_\text{out} = \left[\text{Cl}^-\right]_\text{out}.$ At the same time, to assure osmotic neutrality on the inside and the outside, the concentration of solute particles within the cell has to equal the concentration of solute particles outside of the cell, or $200\ \text{milliosmolar} = \left[\text{Na}^+\right]_\text{out} + \left[\text{Cl}^-\right]_\text{out} = 2\ \left[\text{Cl}^-\right]_\text{out}.$ Therefore, $\left[\text{Cl}^-\right]_\text{out} = 100\ \text{mM}$.
• Although water molecules are permeable to the membrane, we have balanced the concentration of solute particles on each side, so that the concentration of water is the same on each side; this means there will be no net movement of water to either side, and that the volumes are stable.
• We now know the concentration of chloride ions outside of the cell, which is 100 mM, and inside the cell, which is 50 mM, so we have the information we need to compute the equilibrium or Nernst potential across the membrane:

$E_{\text{Cl}^-} = 58\ \text{mV} \times \log_{10} \frac{50\ \text{mM}}{100\ \text{mM}} = -17.5\ \text{mV}.$