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}} $



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