Equilibrium Question 762
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}} $
BETA
