Question

Quiz 3 Bios 240 1. Two solutions are separated by a semipermeable membrane that is permeable to sodium but not chloride or potassium ions. Solution A on the left contains a solution that is 100 mM NaCI, while Solution B on the right has a concentration of sodium chloride of 1 mM. There is an electrode in both solutions, and a comparative assessment of voltages between the two compartments is being made. The investigator doing the study decides to call side A ground potential (or, in other words, defines it as 0). What is the value for the sodium equilibrium potential in this condition? (show how you calculate this) (2) In one sentence, describe why it has a positive or negative value. (1) What would be the value of the chloride equilibrium potential? (show how you calculate this) (2) What is the voltage of side B relative to side A? (2) Does the chloride equilibrium potential contribute significantly to the voltage actually being measured by the investigator? Why or why not (one sentence)? (2)

Answer 1

Ans1.

a). The equilibration potential of the sodium ions can be calculated using the Nernst's equation,

$V_{Na^{+}}=(RT/zF)ln([{Na^{+}_{out}}]/[Na^{+}_{in}])$

Where,

R = Gas Constant= 8.31446 JK^-1mol^-1

F= Faraday's Constant = 96485 J volt^-1 gram^-1

T= Temperature is not given so as per STP we take it 25^oC= 25 +273= 298K

z = charge = +1

[Na⁺_out]= Concentration of Na⁺ on left side = 100mM = 0.1 M

[Na⁺_in]= Concentration of Na⁺ on right side = 1mM = 0.001M

$V_{Na^{+}}=((2477.71)/96485)ln(100)$

b). The value of the V_Na⁺ is positive because the concentration of the positive ions is higher on the left side.

c). The equilibration potential of the Cl ions can be calculated using the Nernst's equation,

$V_{Cl^{-}}=(RT/zF)ln([{Cl^{-}_{in}}]/[Cl^{-}_{out}])$

Where,

R = Gas Constant= 8.31446 JK^-1mol^-1

F= Faraday's Constant = 96485 J volt^-1 gram^-1

T= 25^oC= 25 +273= 298K

z = charge = +1

[Cl^-_out]= Concentration of Cl^- left side = 100mM = 0.1 M

[Cl^-_in]= Concentration of Cl^- on the right side = 1mM = 0.001M

$V_{Cl^{-1}}=((2477.71)/96485)ln(0.01)$

d). The voltage at side B is -118.26 mV wrt the side A.