The degree of dissociation of `HI` at a particualr temperature is `0.8`. Calculate the volume of `2 M Na_(2)S_(2)O_(3)` solution required to neutralise the iodine present in an equilibrium mixture of a reaction when 2 mol each of `H_(2)` and `I_(2)` are heated in a closed vessel of `2 L` capacity and the equilibrium mixture is freezed.

Explanation

Degree of disociation of `HI (alpha) = 0*8`
`{:(,2 HI,hArr,H_(2),+,I_(2)),("Before dissociation",1,,0,,0),("Moles after disso",1-alpha,,alpha//2,,alpha//2):} `
`K_(c) = ((alpha//2)(alpha//2))/(1 -alpha)^(2) = alpha^(2)/(4(1 -alpha)^(2) )=(0*8)/(4(1-0*8)^(2))= 4`
Step 2. Calculation of `I_(2)` in equilibrium mixture.
` {:(,H_(2),+,I_(2),hArr,2 HI),("Intial moles",2,,2,,0), ("Moles after reaction ",(2-x),,(2-x),,2x//2), ("Molar conc.",(2-x)//2,,(2-x)//2,,2x//2):}`
`K'_(c)= 1/K_(c)-(2x//2)^(2)/(((2-x)/2)((2-x)/2))=(4x^(2))/(2-x)^(2)`
` :. (4x^(2))/(2-x)^(2) = 1/4 or (2x)/(2-x) =1/2 or x=2/5 `
Thus, `I_(2) " left" = 2 - 2/5 = 8/5 "mole. "`
Step 3. Calculation of volume of hypo solution used .
`2 Na _(2) S_(2) O_(3) + I_(2) to Na_(2) S_(4) O_(6) + 2 Nal `
` :. "Moles of " N_(2)S_(2)O_(3) " reacted " = 2 xx 8/5 "mole" = 16/5 "mole " `
Volume of 2 M ` Na_(2)S_(2) O_(3) " reacted " = 1000/2 xx 16/5 = 1600 "mL " = 1*6 L`