Equilibrium Question 807
Question: Given
(i) $ HCN(aq)+H_2O(l)\rightarrow $
$ H_3{O^{+}}(aq)+C{N^{-}}(aq)K_{a}=6.2\times {10^{-10}} $
(ii) $ C{N^{-}}(aq)+H_2O(l)\rightarrow $
$ HCN(aq)+O{H^{-}}(aq)K_{b}=1.6\times {10^{-5}}. $
These equilibria show the following order of the relative base strength,
Options:
A) $ O{H^{-}}>H_2O>C{N^{-}} $
B) $ O{H^{-}}>C{N^{-}}>H_2O $
C) $ H_2O>C{N^{-}}>O{H^{-}} $
D) $ C{N^{-}}>H_2O>O{H^{-}} $
Show Answer
Answer:
Correct Answer: B
Solution:
The more is the value of equilibrium constant, the more is the completion of reaction or more is the concentration of products i.e. the order of relative strength would be $ O{H^{-}}>C{N^{-}}>H_2O $