The diffusion coefficient for a Ca2+ ion in aqueous solution at T = 25 oC is 790

Question

The diffusion coefficient for a Ca2+ ion in aqueous solution at T = 25 oC is 790 ?m2/sec. Suppose a phosphate group is bathed in water and is held fixed by an immobilizing particle. If you release a Ca2+ ion 566 ?m away from the phosphate group, what is the approximate amount of time it will be before the Ca2+ ion locates the phosphate group? (You may assume that the calcium ion behaves as a randomly moving particle, that the charges of the ions are not relevant to ionic mobility, and that the immobilizing particle does not get in the way of the diffusion calcium ion.)

Select one:

A: 0.119 second

B: 67.6 seconds

C: 22.5 seconds

D: 203 seconds

E: 45.1 seconds

Explanation / Answer

Here, the diffusion coefficient for a Ca2+ ion in aqueous solution at T = 25 oC is 790 m2/sec

The distance from Ca2+ ion to the phosphate group = 566 m

The approximate amount of time it will be before the Ca2+ ion locates the phosphate group can be calculated as follows.

t = 1/2 {(distance of Ca2+ ion away from the phosphate group)2/diffusion coefficient of Ca2+ ion}

i.e. t = 1/2 {(566 m)2/(790 m2/sec)}

i.e. t ~ 203 sec

Therefore, the correct answer is the option D.

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