Question

Tritium ${}_1^3\text{T}$ (an isotope of H) combine with fluorine to form a weak acid TF which ionises to given $T^{+}$. Tritium is radioactive and is a $\beta$-emitter. A freshly prepared dilute aqueous solution of TF has a pT (the equivalent of pH) of 1.7 and freezes at $-{0.372}^{\circ }C$. If 600 mL of freshly prepared solution were allowed to stand for 24.8 years, then ionisation constant of TF and total charge emitted (in F) will be :

• A. $2\times {10}^3,0.54F$
• B. $2.5\times {10}^{-3},0.054F$
• C. $5\times {10}^{-3},0.32F$
• D. $2.5\times {10}^3,1F$

Answer 1

Answer: (B)

$2.5\times {10}^{-3},0.054F$

$TF\rightleftharpoons T^{+}+F^{-}$
1 0 0
$(1-\alpha )$ $\alpha$ $\alpha$

Given, $pT=1.7\left(likepH\right)$
$\therefore \left[T^{+}\right]=0.02=C\alpha$
Since, $\Delta T=0-(-0.372)=0.372$ for solution of TF

$\Delta T_f=K_f\times (1+\alpha )\times molality=K_f(1+\alpha )\times C$
$(\because Molality=molarityfordilutesolution)$

$\therefore \Delta T_f=K_f(C+C\alpha )$
$0.372=1.86(C+0.02)$
$\therefore C=0.2-0.02=0.18M$

Thus, $K_a=\frac{\left[T^{+}\right]\left[F^{-}\right]}{\left[TF\right]}=\frac{0.02\times 0.02}{(0.18-0.02)}=\underset{–}{2.5\times {10}^{-3}}$ Option B.

Also, 0.18 mole of TF contain 0.18 mole of T (including $T^{+}$) per litre.

Thus, $moleofTFin600mL=\frac{0.18\times 600}{1000}=0.108$

$Thashalf$-$life=12.4year;N_0=0.108$

$\therefore Amountleftin24.8year=\frac{N_0}{2}=0.054$

or 0.05 mole of tritium undergoes $\beta$-emission. Since one tritium atom emits one $\beta$-particle.

$No.of\beta$-$emitted=[0.108-0.054]\times 6.023\times {10}^{23}$

$=3.25\times {10}^{22}$

$\therefore Totalchargeemitted=\frac{3.25\times {10}^{22}\times 1.602\times {10}^{-19}}{96500}Faraday$

$=0.054Faraday$

Option B is correct.