Redox Reactions And Electrochemistry Question 674
Question: A hydrogen electrode placed in a buffer solution of $ CH_3COONa $ and acetic acid in the ratio x : y and y : x has electrode potential value $ E_1 $ volts and $ E_2 $ volts, respectively at $ 25{}^\circ C $ . The $ pK_{a} $ value of acetic acid is ( $ E_1 $ and $ E_2 $ are oxidation potential):
Options:
A) $ \frac{E_1+E_2}{0.118} $
B) $ \frac{E_2-E_1}{0.118} $
C) $ -\frac{E_1+E_2}{0.118} $
D) $ \frac{E_1-E_2}{0.118} $
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Answer:
Correct Answer: A
Solution:
[a] $ E_1=E^{O}-\frac{0.059}{1}\log {{[ {H^{+}} ]}_1} $ $ E_2=E^{O}-\frac{0.059}{1}\log {{[ {H^{+}} ]}_2} $ on adding (also $ E_H^{o}=0 $ )
$ E_1+E_2=-\frac{0.059}{1}[ \log {{[ {H^{+}} ]} _1}+\log {{[ {H^{+}} ]} _2} ] $
Now for $ CH_3COOHCH_3CO{O^{-}}+{H^{+}} $ $ [ {H^{+}} ]=\frac{K _{a}[ CH_3COOH ]}{[ CH_3CO{O^{-}} ]} $
$ \therefore $ $ {{[ {H^{+}} ]} _1}=K _{a}\frac{y}{x} $ $ ,{{[ {H^{+}} ]} _2}=K _{a}\frac{x}{y} $
$ \therefore $ $ E_1+E_2=-\frac{0.059}{1}[ \log \frac{K_{a}y}{x}+\log \frac{k_{a}x}{y} ] $ $ =-0.059[ 2\log K_{a} ] $
$ \log K_{a}=\frac{E_1+E_2}{2\times ( -0.059 )} $ $ \log K_{a}=-\frac{E_1+E_2}{0.118} $ or $ pK_{a}=\frac{E_1+E_2}{0.118} $
BETA
