Reaghan asked in Science & MathematicsBiology · 1 month ago

# If the concentration of Sodium in the extracellular fluid is increased to 180mM while the concentration of Sodium in the excitable cell sta-?

If the concentration of Sodium in the extracellular fluid is increased to 180mM while the concentration of Sodium in the excitable cell stays at 10mM, what would be the resting potential of the cell if you assuming all Sodium channels are open, and the cell retains normal osmotic pressure? (in mV)

Relevance
• 1 month ago

We know that Equilibrium potential or the nerst potential for an ION is the the resting membrane potential.

Since, concentration of other ions are not given.

So, Considering among all the ions moving across the membrane, Na+ makes the largest proportion(Actually it is K+)

Under the given conditions, we have the following equation:

Vm = 61.5 log([ion]out/[ion]in) (for Na+like atoms)

↑

cutting

membrane  [ion]out → for conc. outside the wall

potential.     [ion]in   → for conc. inside the wall

Given that [ion]out = 180mM

[ion]in.   = 10mM

.

.   .  Vm = 61.5 log(180/10)

Vm = 61.5 log(18)

= 77.199 mV

x= 77.2mM