Tritium, T(an isotope of H) combines with fluroine to form weak acid TF, which ionizes to give `T^(+)`. Tritium is radioactive and is a `beta-"emitter".` A freshly prepared aqueous solution of TF has p T (equivalent of pH) of 1.5 and freezes at `-0.372^(@)"C"`. if 600 ml of freshly prepared solution were allowed to stand for 24.8 year. Calculate (i) ionization constant of TF. (ii) Number of `beta`-particle emitted.
(Given` K_("f") "for water= 1.86 kg mol K"^(-1),t_(1//2)" for tritium=12.4 years")`

Answer

(i)`K_(a)=7.3xx10^(-3)` (ii)`4.55xx10^(22)`

Explanation

`"pT"(~~"pH")=1.5=-"log"["T"^(+)]`
`["T"^(+)]=0.0316`
`{:("TF",iff,"T"^(+),+,"F"^(-)),("at eq."x",,0.0316,,0.0316):}`
`therefore "conc. Of Particles =(x+.0316+0.0316)"`
`"=(x+0.0632)"`
`Delta"T"_("f")=K_("f")m`
`implies0.372=(1.86)("x"+0.0632)`
`"x=0.1368"`
`therefore"K"_(a)=((0.0316)^(2))/((0.1368))=7.3xx10^(-3)`
`"Also"[TF]_("initial")"=0.1368+0.0316=0.1684"`
As `t_(1//2)`=12.4 years, i.e. after 24.8 years (i.e. two half lifes),`3/4`th of initial moles of 'T' has undergone radiactive decay.
`therefore` no. of `beta` particles obtained
``=(3)/(4)xx(0.1684)xx0.6xx6.022xx10^(23)=4.55xx10^(22)`