When fish are exposed to contaminated water, the accumulation of the toxic subst

Question

When fish are exposed to contaminated water, the accumulation of the toxic substance is controlled by the sorption-desorption process:

     (dC/dt=k_{0}C_{d}-K_{0}C)     

Where C is the concentration in the fish (mg/kg),

k0 is the rate constant for the uptake ((mg/kg-day)fish / (mg/kg)water),

Cd is the concentration of the toxic substance in the water (mg/kg),

K0 is the rate constant for desorption ((mg/kg-day)fish / (mg/kg)fish),

C is the concentration of the toxic chemical in the fish (mg/kg)

Derive the equation for the elimination C=f(t), assuming a contaminated fish is placed in fresh water.

Explanation / Answer

dC/dt=koCd - KoC

When fish is placed in fresh water, Cd is zero as the water is fresh and not contaminated.

Thus, dC/dt = -Ko*C

So, the elimination of toxin will be a first order process as the rate of elimination will be directly proportional to the concentration of toxin in the fish.

Solving above differentiall equation

dC / C = -K0*dt

Integrating both sides within limits: t=0, C=C0 to t=t, C=C(t)

Where Co is the starting concentration of toxin inside fish when it is placed in fresh water

ln C - ln Co = -Ko*t

or, ln (C / Co) = -Ko*t

or, C = Co*exp(-Ko*t)

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.