A `500mL` of an equilibrium mixture of gaseous `N_(2)O_(4)` and `NO_(2)` at `25^(@)C` and `753mm` of `Hg` was allowed to react with enough water to make `250mL` of solution at `25^(@)C`. Assume that all the dissolved `N_(2)O_(4)` is converted to `NO_(2)` which disproportionates in water yielding a solution of nitrous acid and nitric acid. aAsume further that disproportionation reaction goes to completion and that none of the nitrous acid disproportionates. The equilibrium constant `(K_(p))` for `N_(2)O_(4)(g)hArr 2NO_(2)(g) 0.113` at `25^(@)C`. `K_(a)` for `HNO_(2)` is `4.5 xx 10^(-4) at 25^(@)C`.
a. Write balanced equation for disproportionation.
b. What is the molar concentration of `NO_(2)` and `pH` of the solution?
c. What is osmotic pressure of solution?
d. How many grams of lime `(CaO)` would be required to neutralise the solution?

Explanation

a. `N_(2)O_(4) hAtt 2NO_(2)`
Let `a mol` of `N_(2)O_(4)` and `bmol` of `NO_(2)` were present in equilibrium mixture.
`:. (a+b) = (PV)/(RT) = (653 xx 0.5)/(760 xx 0.0821 xx 298)`
`K_(p) = ((n_(NO_(2)))^(2))/((n_(N_(2)O_(4))))xx[(P)/((n_(NO_(2))+n_(N_(2)O_(4))))] ..(i)`
`:. 0.113 = (b^(2))/(a) xx [(753)/(760 xx (a+b))]`
`:. (b^(2))/(a(a+b)) = 0.114 ...(ii)`
From equations, (i) and (ii), we get
`(b^(2))/(a) = 0.114 xx 0.020 = 2.3 xx 10^(-3)`
`b^(2) = 2.3 xx 10^(-3) a = 2.3 xx 10^(-3) xx (0.02 -b)`
`b^(2) + 0.0023 b - 4.6 xx 10^(-5) = 0`
`:. b = - (0.0023+-sqrt((0.0023)^(2)+4xx4.6xx10^(-5)xx1))/(2xx1)`
`=(-0.0023+-sqrt(1.90 xx 10^(-4)))/(2)`
`b = 5.73 xx 10^(-3)`
`:.` By equaiton (i) `a = 0.014`
b. `2NO_(2)(aq) + 2H_(2) (l) rarr NHO_(2)(aq) + H_(3)O^(o+)(aq) + NO_(3)^(Theta) (aq)`
Total `NO_(2)` moles `=underset((From NO_(2)))(5.73xx10^(-3))+2xx` moles of `N_(2)O_(4)`
`= 5.73 xx 10^(-3) + 2 xx 0.014`
`= 0.0337`
`{:(2NO_(2)(aq)+,2H_(2)O(l)rarr,HNO_(2)+,H_(3)O^(o+)(aq)+,NO_(3)^(o+)(aq)),(0.0337,,0,0,0),(0,,(0.0337)/(2),(0.0337)/(2),(0.0337)/(2)):}`
`:. [HNO_(2)] = (0.337 xx 1000)/(2xx250) = 0.0674 M`,
`[H^(o+)] = (0.337xx1000)/(2xx250) = 0.0674M`,
Due to common ion effect `(H_(3)O^(o+)` furnished by `HNO_(3))`, the dissociation of `HNO_(2)` is suppressed.
`K_(a) = ([H^(o+)][NO_(2)^(Theta)])/([HNO_(2)]) = (0.067 xx [NO_(2)^(Theta)])/(0.0674) = 4.5 xx 10^(-4)`
`:. [NO_(2)^(Theta)] = 4.5 xx 10^(-4)`
Also `pH =- log [H^(o+)] =- log 0.0674 = 1.17`
c. `pi = iERT` (where `i =` Number of particles present in solution or it is Vant Hoff factor)
`= 0.0674 xx 0.0821 xx 298 xx3`
`(2NO_(2)` furnishes three particles)
`= 4.95 atm`
d. `Eq` of `CaO` required `= Eq of HNO_(2) + Eq` of `HNO_(3)`
`= (0.0337)/(2) xx 1 + (0.0337)/(2) xx1 = 0.0337`
`:.` Mole of `CaO` required `= (0.0337)/(2)or (W)/(56) = (0.0337)/(2)`
`W_(CaO) = 0.924g`