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

Answer:

$E=55mV$

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:

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

Best regards.