Question

Tritium, T (an isotope of H) combines with fluorine to form weak acid TF, which ionizes to give Tritium is radioactive and is $\alpha \beta -emitter.$ A freshly prepared aqueous solution of TF has pT (an equivalent of pH) of 1.5 and freezes at $-{0.372}^{\circ }C.$ If 600 ml of freshly prepared solution were allowed to stand for 24.8 years. Calculate (i) ionization constant of TF. (ii)Number of\beta- particles emitted?(Given $K_{f}forwater=1.86kgmol{K}^{-1},t_{1/2}$ for tritium$=12.4years)$

• A. $\left(i\right)K_a=7.3\times {10}^{-3}\left(ii\right)4.55\times {10}^{22}$
• B. $\left(i\right)K_a=7.3\times {10}^{-5}\left(ii\right)4.55\times {10}^{24}$
• C. $\left(i\right)K_a=14.3\times {10}^{-3}\left(ii\right)9.1\times {10}^{22}$
• D. $\left(i\right)K_a=14.3\times {10}^{-5}\left(ii\right)9.1\times {10}^{24}$

Answer 1

The correct option is A $\left(i\right)K_a=7.3\times {10}^{-3}\left(ii\right)4.55\times {10}^{22}$

Given, $pT=1.5$

$T^{+}=0.0316=c\alpha$

$0.372=1.86\times C\times \begin{array}{cc}1+& \alpha \end{array}$

$\therefore C+C\alpha =0.2$

Solving these two equations

$C=0.1684$ and $\alpha =0.1876$

$k_a=\frac{C{\alpha }^2}{1-\alpha }$

$k_a=k_a=\frac{0.168{\left(0.1876\right)}^2}{1-0.1876}=7.3\times {10}^{-3}$

In $600$ ml solution,

$\left[TF\right]=C-C\alpha =0.1368$ mole per litre

$\therefore moles=0.1368\times 0.6=0.08208$

Moles after $24.8$ years $=\frac{0.08208}{4}=0.02052$

$\therefore$ moles disintegrated $=0.08208-0.02052=0.06156$

Moles of $\beta$ particle emitted $=0.06156$

No. of $\beta$ particle emitted $=0.06156\times 6.023\times {10}^{23}=3.7\times {10}^{22}$