$$\begin{array} { | l | l | l | }
\hline & \begin{array} { l }
\bf{ Extracellular-concentration } \
( \bf { mM } )
\end{array} & \begin{array} { l }
\bf { Intracellular-concentration } \
( \bf { mM } )
\end{array} \
\bf{ Inn }\
\hline \mathbf { Na } ^ { + } & 300 & 50 \
\hline \mathbf { K } ^ { + } & 40 & 350 \
\hline
\end{array}$$
-Calculate the equilibrium potential for sodium. Assume a temperature of 37°C.
A) +48.2 mV
B) 0.0 mV
C) -58.4 mV
D) -80.0 mV

Question

$$\begin{array} { | l | l | l | } 
\hline & \begin{array} { l } 
\bf{ Extracellular-concentration } \
( \bf { mM } )
\end{array} & \begin{array} { l } 
\bf { Intracellular-concentration } \
( \bf { mM } )
\end{array} \
\bf{ Inn }\
\hline \mathbf  { Na } ^ { + } & 300 & 50 \
\hline \mathbf { K } ^ { + } & 40 & 350 \
\hline
\end{array}$$
-Calculate the equilibrium potential for sodium. Assume a temperature of 37°C.
A) +48.2 mV
B) 0.0 mV
C) -58.4 mV
D) -80.0 mV

Quizplus · Accepted Answer

The equilibrium potential for sodium can be calculated using the Nernst equation: $ E_{Na} = \frac{RT}{zF} \ln \left( \frac{[Na^+]_{outside}}{[Na^+]_{inside}} \right) $. At 37°C, this simplifies to approximately $ E_{Na} = 61.5 \log \left( \frac{300}{50} \right) $, which equals +48.2 mV.

-Calculate the Equilibrium Potential for Sodium