Question

Consider the following hypothetical problem. Potassium ions have a concentration of 115 x 10^-3mol/liter inside the cell and 4.5 x 10^-3 mol/liter outside the cell. The cell wall has a channel through which the potassium ions can travel and it is 25.0 nm in length and 4.00 nm in diameter. Take the diffusion coefficient of potassium to be 3.00 x 10^-9 m²/s.

A)What is the difference in concentration, Δc across the cell wall in molecules/m³?

B)What is the cross-sectional area of the potassium channel?

C)What is the rate of molecule transfer of the potassium ions?

Answer 1

Consider the following hypothetical problem. Potassium ions have a concentration of 115 x 10^-3mol/liter inside the cell and 4.5 x 10^-3 mol/liter outside the cell. The cell wall has a channel through which the potassium ions can travel and it is 25.0 nm in length and 4.00 nm in diameter. Take the diffusion coefficient of potassium to be 3.00 x 10^-9 m²/s.

A)What is the difference in concentration, Δc across the cell wall in molecules/m³?

B)What is the cross-sectional area of the potassium channel?

C)What is the rate of molecule transfer of the potassium ions?

A)

C_i = concentration inside the cell = 115 x 10^-3 mol/liter = 115 x 10^-3 mol/(10^-3 m³) = 115 mol/m³

C_o = concentration outside the cell = 4.5 x 10^-3 mol/liter = 4.5 x 10^-3 mol/(10^-3 m³) = 4.5 mol/m³

Difference in concentration is given as

$\Delta$C = C_o - C_i = 4.5 - 115 = - 110.5 mol/m³

we know that 1 mol = 6.023 x 10²³ molecules

$\Delta$C = - 110.5 x 6.023 x 10²³ molecules/m³

$\Delta$C = - 6.66 x 10²⁵ molecules/m³

B)

d = diameter of the channel = 4 nm = 4 x 10^-9 m

Area of cross-section of the channel is given as

A = (0.25) $\pi$d²

A = (0.25) (3.14) (4 x 10^-9)²

A = 1.256 x 10^-17 m²

c)

x = length = 25 x 10^-9 m

D = diffusion constant = 3 x 10^-9 m²/s

rate of transfer = D A $\Delta$C /x

rate of transfer = (3 x 10^-9) (1.256 x 10^-17) (6.66 x 10²⁵)/(25 x 10^-9)

rate of transfer = 1.004 x 10⁸