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 $



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