Given that the intracellular concentration of potassium is 150 mEq/L, how would the potassium equilibrium potential be affected if the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L

The potassium equilibrium potential would increase, meaning that more K+ would be leaving the cell.

Explanation:

Let us assume that the only ion transported through the cell membrane is K+. We need to use the Nernst equation to know the destiny of the ion.

Nernst equation:

E = 58 millivolts/z. [Log₁₀ (C-out/C- in)

Where,    

•E = Equilibrium potential

•58 millivolts/z = Constant

•z = Ion charge + positive or negative symbol

•C-out = Ion concentration out of the cell

•C-In = Ion concentration inside the cell

By convenience, in the Nerts equation, the bigger concentration value corresponds to the numerator and the smaller concentration value to the denominator.

Now let us see the provided values,

•z = Ion charge + positive or negative symbol ⇒ +1 ⇒ K+

•C-out = Ion concentration out of the cell ⇒ 5 mEq/L

•C-In = Ion concentration inside the cell ⇒ 150 mEq/L

E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)

E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 5 mEq/L)

E = 58 millivolts (Log₁₀ 30)

E = 58 millivolts (1.477)

E = 85.67 millivolts

85.7 mV is the absolute value of equilibrium potential.

E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)

E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 3.5 mEq/L)

E = 58 millivolts (Log₁₀ 42.85)

E = 58 millivolts (1.63)

E = 94.65 millivolts

94.7 mV is the absolute value of equilibrium potential.

If the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L, there will be an increase in the membrane potential from 85.7 to 94.7 mV. The increase in the equilibrium potential will result in more potassium diffusing out of the cell, turning the cell interior less positive than before.  

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