Radioactive tracers are useful in studying very low concentrations of chemical species. A chemist has a sample of HgI₂ in which part of the iodine is the radioactive nuclide of mass 131, so that the count rate is 5.0 * 10¹¹ counts per minute per mole of I. The solid mercuric iodide is placed in water and allowed to come to equilibrium. Then 100 mL of the solution is withdrawn, and its radioactivity is measured and found to give 22 counts per minute. What is the molar concentration of iodide ion in the solution? A) 1.1*10^-¹¹ B) 1.1 * 10^-⁹ C) 1.1 *10^-¹⁰ D) 4.4 * 10^-¹⁰ E) 4.4* 10^-¹¹

Question

Quizplus · Accepted Answer

First, we need to set up the equilibrium reaction:

HgI2(s) ⇌ Hg2+(aq) + 2I-(aq)

Since we know that HgI2 is a solid, we can assume that the concentration of the solid is constant and does not affect the equilibrium constant. Therefore, we can write the equilibrium constant expression as:

K = [Hg2+][I-]2

We are given the count rate, which is proportional to the concentration, of the radioactive nuclide of mass 131 as 5.0 x 10^10 counts per minute per mole of I. This means that:

count rate ∝ [I-] ∝ [Hg2+]

We can use this information to find the equilibrium concentrations of Hg2+ and I-:

count rate = 5.0 x 10^10 counts/min/mole of I
= 5.0 x 10^10 counts/min/[I-]   (since the count rate is proportional to [I-])
22 counts/min = 5.0 x 10^10 counts/min/[I-]
[I-] = 1.1 x 10^-10 moles/L

Now we can use this information to find the equilibrium constant and solve for [Hg2+]:

K = [Hg2+][I-]2
= [Hg2+](1.1 x 10^-10)^2
[Hg2+] = K/(1.1 x 10^-10)^2

We don't have the exact value of K, but we can use the fact that the count rate is also proportional to [Hg2+] to find the ratio of the count rate to [Hg2+] at equilibrium:

count rate ∝ [I-] ∝ [Hg2+]
count rate/[Hg2+] = 5.0 x 10^10 counts/min/mole of I/[Hg2+]
= 5.0 x 10^10 counts/min/([Hg2+]/2)
= 1.0 x 10^11 counts/min/[Hg2+]

At equilibrium, this ratio should be equal to the count rate we measured from the withdrawn solution. Solving for [Hg2+], we get:

[Hg2+] = (1.0 x 10^11 counts/min)/22 counts/min
= 4.54 x 10^9 moles/L

Finally, we can use the equilibrium expression to find [I-]:

K = [Hg2+][I-]2
4.54 x 10^9 = (1.1 x 10^-10)^2[I-]2
[I-] = 4.4 x 10^-4 moles/L

Therefore, the molar concentration of iodide ion in the solution is 4.4 x 10^-4 moles/L, which corresponds to choice D.

Radioactive Tracers Are Useful in Studying Very Low Concentrations of Chemical