The investigator applies a brief electrical stimulus that causes the potassium channels to open. there are no hco3- channels, thus pk+ = 1 and phco3- = 0. first, what is the potential difference between compartments a and b after the system reaches equilibrium? second, to estimate the potential difference do we need to be concerned about a change in potassium concentration as the concentration gradient drives k+ from a to b? why or why not?

1- The pottential difference between compartments A and B after the system reaches equilibrium is 0.

2- NO.

Explanation:

1- The membane pottential is produced by the electric charge of the ions K+ and HCO3-. Also the different concentration of those ions creates a concentration gradient that it could be modified by altering the polarity of the membrane that separates both compartments. The membrane polarity is fundamentaly the specific membrane channels. When the system reaches equilibrium there is not a pottential difference between compartments, that is zero. That is because the potentials are equalaized and therefore their net difference is zero.

2- V = ΔEp /q

Where V indicates de electric potential difference

           ΔEp indicates potential energy

and      q indicates electric charge

The potential difference depends on the change of the potential energy (Ep) of the charge (q) and the magnitude of the charge.

The potential energy changes when an eletric charge (K+ in this case) moves from one point to another. The potential energy magnitude is higher if the mass is higher, that is to say, if the concentration of potassium is higher.

On the other hand, the potential difference does not depend on the magnitude of the electric charge (q). So resuming, we do not have to be conerned about the potassium concentration in order to estimate the potential difference. The only thing that really matters to calculate the potential difference is the electric field intensity and the location or direction of the initial and final points: concentration gradient drives K+ from A to B.