Redox Reactions And Electrochemistry Question 664
Question: Given the following molar conductivities at $ 250^{o}C;HCl,425,{{\Omega }^{-1}}cm^{2}mo{l^{-1}}; $ $ NaCl125{{\Omega }^{-1}} $ $ cm^{2}mo{l^{-1}} $ NaC (sodium crotonate), $ 82,{{\Omega }^{-1}}cm^{2}mo{l^{-1}}; $ what is the ionisation constant of crotonic acid- If the conductivity of a 0.001 M crotonic acid solution is $ 3.82\times {10^{-4}}{{\Omega }^{-1}}c{m^{-1}} $ -
Options:
A) $ 1.5\times {10^{-4}} $
B) $ 1.11\times {10^{-5}} $
C) $ 2.11\times {10^{-5}} $
D) $ 1.8\times {10^{-5}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Idea This problem can be solved by using the concept of Kohlrausch law of independent migration by using which we can easily calculate the value of molar conductivity of crotonic acid after determining the $ {\wedge_{m(HC)}} $
calculate degree of dissociation $ (\alpha ),\alpha =\frac{{\wedge _{m(HC)}}}{\wedge _m^{\infty } (HC)} $ Molar conductivity of the dissociated crotonic acid.
$ {\wedge_{m(HC)}}={\wedge_{m(HCl)}}+{\wedge_{m(NaC)}}-{\wedge_{(m)NaCl}} $
$ =(425+82-125){{\Omega }^{-1}}cm^{2}mo{l^{-1}} $
$ =382,{{\Omega }^{-1}}cm^{2}mo{l^{-1}} $
Molar conductivity of HC $ {\wedge_{m(HC)}}=\frac{K}{C}=\frac{3.82\times {10^{-4}}}{0.001}\times 1000 $ $ =38.2{{\Omega }^{-1}}cm^{2},mo{l^{-1}} $ The degree of dissociation
$ \alpha =\frac{{\wedge_{m(HC)}}}{{\wedge_{{m^{\infty }}(HC)}}}=\frac{38.2,{{\Omega }^{-1}},cm^{2},mo{l^{-1}}}{382,{{\Omega }^{-1}}cm^{2},mo{l^{-1}}}=0.1 $
$ K_{a}=\frac{C{{\alpha }^{2}}}{1-\alpha }=\frac{{10^{-3}}{{(0.1)}^{2}}}{1-0.1}=1.11\times {10^{-5}} $
BETA
