Equilibrium Question 805

Question: The first and second dissociation constants of an acid $ H_2A $ are $ 1.0\times {10^{-5}} $ and $ 5.0\times {10^{-10}} $ respectively. The overall dissociation constant of the acid will be

Options:

A) $ 0.2\times 10^{5} $

B) $ 5.0\times {10^{-5}} $

C) $ 5.0\times 10^{15} $

D) $ 5.0\times {10^{-15}} $

Show Answer

Answer:

Correct Answer: D

Solution:

$ H_2A\rightarrow {H^{+}}+H{A^{-}}~ $

$ \therefore K_1=1.0\times {10^{-5}}=\frac{[{H^{+}}][H{A^{-}}]}{[H_2A]} $ (Given) $ H{A^{-}}\xrightarrow{{}}{H^{+}}+{A^{2-}} $

$ \therefore K_2=5.0\times {10^{-10}}=\frac{[{H^{+}}][{A^{2-}}]}{[H{A^{-}}]} $ (Given) $ K=\frac{{{[{H^{+}}]}^{2}}[{A^{2-}}]}{[H_2A]}=K_1\times K_2 $

$ =( 1.0\times {10^{-5}} )\times ( 5\times {10^{-10}} )=5\times {10^{-15}} $



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