Solution A and solution B are separated by a semipermeable membrane, the membran

Question

Solution A and solution B are separated by a semipermeable membrane, the membrane is permeable to all the ionized molecules listed in the table. When the equilibrium is established, the concentrations are shown in the table. (8 poinsts)

solution A IN

solution B OUT

NaCl

140mM

10mM

KCl

4mM

155mM

CaCl2

0.1mM

3mM

Use this Nernst equation to calculate following potentials.

the Na+ equilibrium potential. (1 points)   

the Cl- equilibrium potential. (1 points)

the K+ equilibrium potential.   (1 points)    

the Ca2+ equilibrium potential. (1 points)

which solution has higher electric potential. (hint: combined equilibrium potentials in solution A vs. combined equilibrium potentials in solution B) (4 points)

solution A IN

solution B OUT

NaCl

140mM

10mM

KCl

4mM

155mM

CaCl2

0.1mM

3mM

Explanation / Answer

Write the Nernst Equation   V eq.= RT/zF x (ln [X]0/[X]i

where -

R = universal gas constant 8.314 J.K1.mol1 (Joules per Kelvin per mole).

T =Temperature in Kelvin (K = °C + 273.15).

z is the valence of the ionic species ( z is +1 for Na+, +1 for K+, +2 for Ca2+, 1 for Cl, etc, and z is unitless.

F =Faraday's constant = 96,485 C.mol1 (Coulombs per mole).

Also consider -

[X]out is the concentration of the ionic species X in the extracellular fluid.

And -

[X]in is the concentration of the ionic species X in the intracellular fluid.

Here consider the temperature as 25°C

Now -

(a) The Na+ equilibrium potential.

Nernst Equation   V eq.= RT/zF x (ln [X]0/[X]i

V eq=(8.314 J.K1.mol1) x298/(1x96,485 C.mol1) x ln[140]/[10] = -67.77mv

(b) The Cl- equilibrium potential.

V eq=(8.314 J.K1.mol1) x298/(1x96,485 C.mol1) x ln[140]/[10] = +67.77mv

Likewise Cl- equilibrium potential in KCl is -93.91 and CaCl2 is 87.34

(c) The K+ equilibrium potential.

V eq=(8.314 J.K1.mol1) x298/(1x96,485 C.mol1) xl n [4]/[155] = 93.91mv

(d) The Ca2+ equilibrium potential

V eq=(8.314 J.K1.mol1) x298/(1x96,485 C.mol1) x ln [0.1]/[3] = 43.67mv

(e) Which solution has higher electric potential?

K+ is having highest equilibrium potential

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