Redox Reactions And Electrochemistry Ques 639
Question: Specific conductance of 0.1 MHA is $ 3.75\times {10^{-4}}oh{m^{-1}}c{m^{-1}} $ . If $ {{\lambda }^{\infty }}(HA)=250,oh{m^{-1}}cm^{2}mo{l^{-1}} $ , the dissociation constant $ K_{a} $ of HA is:
Options:
A) $ 1.0\times {10^{-5}} $
B) $ 2.25\times {10^{-4}} $
C) $ 2.25\times {10^{-5}} $
D) $ 2.25\times {10^{-13}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ {\lambda_{m}}=\frac{1000\kappa }{0.1}=\frac{1000\times 3.75\times {10^{-4}}}{0.1}=3.75; $
$\alpha =\frac{{\lambda_{m}}}{\lambda _m^{\infty }}=\frac{3.75}{250}=1.5\times {10^{-2}} $ ;
$ K _{a}=C{{\alpha }^{2}}=0.1\times {{(1.5\times {10^{-2}})}^{2}}=2.25\times {10^{-5}} $
