InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
An aqueous solution of aniline of concentration 0.24 M is prepared. What concentraton of sodium hydroxide is needed in this solution so that anilinium ion concentration remains at `1xx10^(-8)` M? `(K_(a) "for" C_(6)H_(5)NH_(3)^(+)=2.4xx10^(-6)M)` |
|
Answer» In aqueous solution, aniline is hydrolysed as `C_(6)H_(5)NH_(2)+ H_(2)O hArr C_(6)H_(5)overset(+)N H_(3)+OH^(-)` When NaOH is added, hydrolysis will be supressed so that in the final solution, `[C_(6)H_(5) overset(+)NH_(3)]=10^(-8)M` (Given) In conc. of NaOH added is x mol `L^(-1)`, then at equilibrium `[C_(6)H_(5)NH_(2)]=0.24-10^(-8) ~= 0.24` `[C_(6)H_(5)overset(+)NH_(3)]=10^(-8)M` (Given) `[OH^(-)]=10^(-8)+x ~= x M` Hydrolysis constant, `K_(h) = ([C_(6)H_(5)overset(+)H_(3)][OH^(-)])/([C_(6)H_(5)NH_(2)])` ...(i) Further, we are given `C_(6)H_(5)overset(N)H_(3) hArr C_(6)H_(5)NH_(2)+H^(+)` `K_(a) = ([C_(6)H_(5)NH_(2)][H^(+)])/([C_(6)H_(5)overset(N)H_(3)])` ...(ii) Also `[H^(+)][OH^(-)]=K_(w)` ...(iii) Combining eqns. (i), (ii) and (iii) `k_(h)=(K_(w))/(K_(a))=(10^(-14))/(2.4xx10^(-6))=(10^(-8))/(2.4)` Substituting the values in eqn. (i), we get, `(10^(-8))/(2.4)=(10^(-8)xx x)/(0.24)` which gives `x=10^(-2)M` |
|