To `1L` of a `1.6xx10^(-3)M` aqueous solution of ethylene diamine `(K_(b_(1))=8xx10^(-5),K_(b_(2))=2.7xx10^(-8)),5xx10^(-4)` mole of `Ba(OH)_(2)` is added.Then:

Options

A. `pH_(f)-pH_(i)=0.5`
B. `alpha_(i)-alpha_(f)=0.12`
C. `([C_(2)N_(2)H_(10)^(2+)])/([C_(2)N_(2)H_(10)^(2+)]_(f))=0.128`
D. `([C_(2)N_(2)H_(9)^(+)]_(i))/([C_(2)N_(2)H_(9)^(2+)]_(f))=0.128`

Explanation

initial :`8xx10^(-5)=(1.6xx10^(-5)alpha^(2))/(1-alpha)" "rArr alpha=(1)/(5)`
`:.[OH^(-)]=3.2xx10^(-4)" "pH=10.5`
`8xx10^(-5)xx2.7xx10^(-8)=((3.2xx10^(-4))^(2)(C_(2)N_(2)H_(10)^(2+))_(i))/(1.28xx10^(-3))`
`rArr2.7xx10^(-8)=[C_(2)N_(2)H_(10)^(2+)]`_(i)
Final `pH=11," " 8xx10^(-5)=10^(-3)xxalpha rArr alpha=0.08`
`[C_(2)N_(2)H_(9)^(+)]=1.28xx10^(-4)`
`2.7xx8xx10^(-13)=((10^(-6))[C_(2)N_(2)H_(10)^(2+)])_(f)/(1.6xx10^(-3))rArr [C_(2)N_(2)H_(10)^(2+)]=1.28xx2.7xx10^(-9)M`
So A,B and D are correct but `([C_(2)N_(2)H_(10)^(2+)])/([C_(2)N_(2)H_(10)^(2+)])=(1)/(0.128)`