A) 4
B) 7
C) 10
D) 14
Correct Answer: A
Solution :
\[{{X}^{-}}+{{H}_{2}}O\]⇌\[O{{H}^{-}}+HX\] \[{{K}_{b}}=\frac{[O{{H}^{-}}]\,\,[HX]}{[{{X}^{-}}]}\] HX ⇌ \[{{H}^{+}}+{{X}^{-}}\] \[{{K}_{a}}=\frac{[{{H}^{+}}]\,\,[{{X}^{-}}]}{[HX]}\] \[\therefore \,{{K}_{a}}\times {{K}_{b}}=[{{H}^{+}}]\,\,[O{{H}^{-}}]={{K}_{w}}={{10}^{-14}}\] Hence \[{{K}_{a}}={{10}^{-4}}\] Now as \[[{{X}^{-}}]=[HX],\,\,pH=p{{K}_{a}}=4\].You need to login to perform this action.
You will be redirected in
3 sec