Answer to Question #68522 in General Chemistry for Natasha

Question

A Beaker is divided into two chambers, A and B, by a semi-permeable membrane lying across the middle of the beaker. Chamber A contains the following concentration of ions:
{Na+} = 48 mM, {K+} = 100 mM, {Cl-} = 110 mM
Chamber B contains the following concentration of ions:
{Na+} = 125 mM, {K+} = 10 mM, {Cl-} = 110 mM
If the membrane is permeable only to Na+, what is the transmembrane potential difference of Chamber A with respect to Chamber B?

Answer 1

According to Nernst equilibrium state for a single ion:
E_Na=1/Z_Na RT/F ln([〖Na〗^+ ]_A/[〖Na〗^+ ]_B )
E_Na=25.6/(+1) ln(48/125)=-24.5 (mV)

Answer:-24.5 mV.