Question #55869

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Question #55869

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Answer 1 · 2015-11-22T13:48:14

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 $55$ $55$ years.

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 $t$ $t$ by using the substance's nuclear half-life looks like this

$A (t) = A_{0} \cdot {(\frac{1}{2})}^{\frac{t}{t_{1/2}}}$ $color(blue)(A(t) = A_0 * (1/2)^(t/t_("1/2"))) " "$ , where

$A (t)$ $A(t)$ - the amount left after $t$ $t$ years;
$A_{0}$ $A_0$ - the initial quantity of the substance that will undergo decay;
$t_{1/2}$ $t_("1/2")$ - the half-life of the decaying quantity.

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 $v$ $v$ liters.

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

$c = \frac{n}{V} \Rightarrow n = c \cdot V$ $color(blue)(c = n/V implies n = c * V)$

$n_{TCl} = 0.0100 M \cdot v L = 0.0100v moles TCl$ $n_"TCl" = "0.0100 M" * v" L" = "0.0100v moles TCl"$

Here $T$ $"T"$ is the chemical symbol for tritium.

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

$n_{T^{+}}^{+} = n_{TCl} = 0.0100 v {moles T}^{+}$ $n_("T"^(+)) = n_"TCl" = 0.0100v " moles T"^(+)$

Tritium has a molar mass of about $3.016 g/mol$ $"3.016 g/mol"$ , which means that your solution will contain a total of

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

The mass of tritium left in solution after $55$ years will be equal to

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

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

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 $55$ years will be

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

The pH of the solution will thus be

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

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

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Question #55869

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