Answer:
Option B
Explanation:
Degree of ionisation (α)= ∧m∧∞
Let ∧m(HY)=x
⇒ ∧m(HX)=X10
⇒ ∧m(HX)∧m(HY)=110=α(HX)α(HY)
[∵∧∞(HX)=∧∞(HY)]
Also : Ka(HX)=(0.01)[α(HX)]2
Ka(HX)=(0.10)[α(HX)]2
=(0.10)[10α(HX)]2
=10[α(HX)]2 ........(ii)
⇒ Ka(HX)Ka(HY)=0.0110
⇒ logKa(HX)−logKa(HY)=−3
⇒ −logKa(HX)−[−logKa(HY)]=3
⇒ pKa(HX)−pKa(HY)=3