Most Important Selected Qs for JEE AdvancedChemistryIonic Equilibrium
5.35 g of a salt ACl (of weak base AOH ) is dissolved in 250 ml of solution. The pH of the resultant solution was found to be 4.827 . Find the ionic radius of A ⁺ and Cl ⁻ if ACl forms CsCl type crystal having 2.2 ~g / cm ^3 . Given K _ b ( AOH ) =1.8 10⁻⁵, r ⁺ r ⁻ =0.732 for this cell unit (mark the anser in Å )
Correct answer
1.715
Step-by-step solution
ACl ( s ) A ⁺( aq )+ Cl ⁻( aq ) A ⁺( aq )+ H ₂ O AOH ( aq )⁻+ H ⁺( aq ) At equilibrium C (1- ) C C For salts of weak base and strong acid, = K _ w K _ b [ H ⁺ ]= C = K _ w C K _ b 10^ -4.827 = 1 10⁻⁴ 1.8 10⁻⁵ 5.35 1000 M 250 M=53.5 For CsCl type structure = z M a^3 6.023 10²³ a = [ 1 53.5 2.2 6.023 10²³ ]^ 1 / 3 =3.43 År_A⁺+r_ C r = 3 2 a=0.866 a =0.866 3.43 Å=2.97 Å r_ A⁺ r_ C- =0.732 On solving r_ A⁺ =1.255 Å r_ C l⁻ = 1 . 7 1 5 Å