Question

A fish is placed in a freshwater environment with a low NaCl concentration environment. The pressure tolerance of fish cells membrane is 5 atm, the surronding osmotic pressure on the cell membrane is 28.6 atm. Luckily, cell membranes have ion channels which allow salt concentration to change. If the following "reaction" follows zero order kinetics, with a reaction constant of 10 mM/hr, how long does the fish need to acclimised to its new salt concentration?

Na+ (inside) + Cl- (inside) ---> Na+ (outside) + Cl- (outside)

Answer 1

Step I

Rate of Reaction for Zero Order Kinetics is given by -

$\mathbb{R}=K = \frac{\partial C}{\partial t}$

Where C is concentration and t is time.

Now for the given question, reaction constant K is given as 10 mM/hr. Hence

$\mathbb{R} = 10$

Stage II

Osmotic Pressure is given by -

$\pi = CRT$

Here T is the temperature which is not changing. We have to bring the pressure from 28.6atm (outside) to 5 atm (inside).

$\mathbb{R} = 10 = \frac{C2-C1}{t2-t1}$

Therefore, $\mathbb{R} = \frac{(P2-P1)RT}{t2-t1}$

Stage - III

Assuming T = 20 degree celsius = 293.15 K for freshwater environment and R = 10mM/hr = 0.01 M/hr, substituting these values we get -

$0.01 = \frac{(28.6-5)\times 0.0821\times 293.15}{t}$

Hence, $t = 1.76\times (10^{-5}) hr$

or, we know 1 hr = 3600 sec, which gives

So, the fish need only 0.06336 sec to acclimised to its new salt concentration at the above assumed temperature of the environment. If you change the season or temperature, there will be difference in the calculation but the procedure will remain the same.