Question

Equilibrium constant of ${{{T}}_2}{{O}}$ (${{T}}$ or ${_1^3}{{H}}$ is an isotope of ${}_1^1{{H}}$) and ${{{H}}_2}{{O}}$ are different at ${{298K}}$. Let at ${{298K}}$ pure ${{{T}}_2}{{O}}$ has ${{pT}}$ (like ${{pH}}$) is $7.62$. The ${{pT}}$ of a solution prepared by adding $10{{mL}}$ of $0.2{{M}}$ ${{TCl}}$ to ${{15mL}}$ of $0.25{{M}}$ ${{NaOT}}$ is:
A. $2 - \log 7$
B. $14 + \log 7$
C. $1.324 - \log 7$
D. $13.24 + \log 7$

Answer 1

Hint: It is known that tritium is an isotope of hydrogen in which there are two neutrons and one proton in tritium. While in protium, there is only one proton. Similar to neutralization reaction of ${{HCl}}$ and ${{NaOH}}$, ${{TCl}}$ and ${{NaOT}}$ reacts together to form ${{{T}}_2}{{O}}$ and ${{NaCl}}$.

Complete step by step answer:
We know that when an acid is reacted with base, it forms corresponding salt and water molecule as products. This reaction is similar with the given reaction.
The chemical reaction of ${{TCl}}$ and ${{NaOT}}$ is given below:
${{TCl}} + {{NaOT}} \to {{NaCl}} + {{{T}}_2}{{O}}$
It is given that the molarity of ${{TCl}}$, ${{{M}}_{{{TCl}}}} = 0.2{{M}}$
Volume of ${{TCl}}$, ${{{V}}_{{{TCl}}}} = 10{{mL}}$
Molarity of ${{NaOT}}$, ${{{M}}_{{{NaOT}}}} = 0.25{{M}}$
Volume of ${{NaOT}}$, ${{{V}}_{{{NaOT}}}} = 15{{mL}}$
${{pT}}$ of ${{{T}}_2}{{O}}$ is $7.62$
${{{M}}_{{{TCl}}}}{{ = 0}}{{.2mol}}{{{L}}^{ - 1}} = \dfrac{{{{0}}{{.2mol}}}}{{1000{{mL}}}}$
The number of moles of ${{TCl}}$ in ${{1000mL}}$ is $0.2$
So the number of moles of ${{TCl}}$ in $10{{mL}}$ is $\dfrac{{0.2}}{{1000}} \times 10 = \dfrac{{0.2}}{{100}}$
Number of millimoles of ${{TCl}}$ in $10{{mL}}$ is $\dfrac{{0.2}}{{100}} \times 1000 = 2{{mmol}}$
Similarly, number of millimoles of ${{NaOT}}$ in ${{15mL}}$ is $0.25 \times 15 = 3.75{{mmol}}$
Remaining ${{NaOT = 1}}{{.75mmol}}$
Concentration of ${{O}}{{{T}}^ - }$, $\left[ {{{O}}{{{T}}^ - }} \right] = \dfrac{{1.75}}{{10 + 15}} = \dfrac{{1.75}}{{25}} = 7 \times {10^{ - 2}}{{M}}$
${{pOT}}$ can be calculated by taking a negative algorithm of concentration of ${{O}}{{{T}}^ - }$.
i.e. ${{pOT = - log}}\left[ {{{O}}{{{T}}^ - }} \right]$
Substituting the value of $\left[ {{{O}}{{{T}}^ - }} \right]$, we get
${{pOT = - log}}\left( {7 \times {{10}^{ - 2}}{{M}}} \right)$
This can be expressed as ${{pOT = 2 - log7}}$
For pure water, ${{pT + pOT}} = 7.62 \times 2 = 15.24$
Now let’s find the ${{pT}}$ of the solution. From the above equation, we get
${{pT}} = 15.24 - {{pOT}}$
By substituting the value of ${{pOT}}$, we get
${{pT}} = 15.24 - \left( {2 - \log 7} \right)$
On simplification, we get
${{pT}} = 15.24 - 2 + \log 7 = 13.24 + \log 7$
So the ${{pT}} = 13.24 + \log 7$

So the correct option for this answer is option D.

Note: This calculation is very much similar to the calculation of ${{pH}}$ in neutralization reaction of ${{HCl}}$ and ${{NaOH}}$. In the given reaction, hydrogen is substituted with tritium. Moreover, ${{pH}}$ or ${{pT}}$ measures the strength of acid while ${{pOT}}$ or ${{pOH}}$ measures the strength of bases.