# Question #55869

##### 1 Answer

Here's what I got.

#### Explanation:

The idea here is that you need to figure out how many tritium atoms will be left in solution after

The pH of the solution will be determined by the concentration of the *remaining* tritium atoms.

So, the equation that allows you to calculate the amount of a radioactive substance that's left *undecayed* after a period of time

#color(blue)(A(t) = A_0 * (1/2)^(t/t_("1/2"))) " "# , where

Now, since molarity is defined as moles of solute, which in your case is hydrochloric acid, divided by liters of solution, and you didn't provide a *volume* for the container, let's assume that you have a volume of

The *number of moles* of hydrochloric acid present in solution will be equal to

#color(blue)(c = n/V implies n = c * V)#

#n_"TCl" = "0.0100 M" * v" L" = "0.0100v moles TCl"#

Here

Now, since hydrochloric acid dissociates completely to form hydrogen cations, which in your case will be tritium cations, and chloride anions, **every mole** of hydrochloric acid will produce **one mole** of tritium cations.

Therefore, your solution will contain

#n_("T"^(+)) = n_"TCl" = 0.0100v " moles T"^(+)#

Tritium has a molar mass of about

#0.0100vcolor(red)(cancel(color(black)("moles T"^(+)))) * "3.016 g"/(1color(red)(cancel(color(black)("mole T"^(+))))) = 0.03016v " g T"^(+)#

The mass of tritium left in solution **after**

#A(t) = 0.03016v * (1/2)^( ("55 years")/("12.3 years"))#

#A(t) = 0.001359v " g T"^(+)#

The *number of moles* of tritium ions will be

#0.001359v color(red)(cancel(color(black)("g T"^(+)))) * "1 mole T"^(+)/(3.016color(red)(cancel(color(black)("g T"^(+))))) = 0.0004506v " moles T"^(+)#

The volume of the solution remains **unchanged**, so the molarity of the tritium ions after

#["T"^(+)] = (0.0004506color(red)(cancel(color(black)(v)))"moles")/(color(red)(cancel(color(black)(v)))"L") = "0.0004506 M"#

The pH of the solution will thus be

#"pH" = -log( ["T"^(+)])#

#"pH" = - log(0.0004506) = color(green)(3.35)#