Question

Now let us refine our model by noting that there is a second source of Na+ ions in the cell: NaI. Suppose the outside of the cell has a concentration of NaI of 0.04 mM and the inside has a NaI concentration of 4 mM. How will the presence of these ions change the Na+ Nernst potential across the membrane? (Assume the NaI is fully ionized in solution and give your answer in mV.)

Answer 1

`E=55mV`

Step-by-step explanation:

Hello,

Considering the given information, the new concentration of Na+inside will be (13.6 + 4) mM = 17.6 mM, and outside will be (150 + 0.04) mM = 150.04 mM due to the previous specified conditions.

Now, the recalculation of the Nerst potential at the supposed temperature of 25 °C (which is modifiable) is done via:

`E=(RT)/(zF) ln((C_(Na^+,outside))/(C_(Na^+,outside)) )\\\\E=(8.314(J)/(mol*K)*298K)/(1*9.65x10^4(C)/(mol) ) *ln((150.04)/(17.6) )=0.055V*(1x10^3mV)/(1V) \\E=55mV`

Best regards.