Redox Reactions And Electrochemistry Question 560
Question: Equivalent conductance at infinite dilution, $ {{\lambda }^{{}^\circ }} $ of $ NH_4Cl,,NaOH $ and $ NaCl $ are 128.0,217.8 and $ 109.3oh{m^{-1}}cm^{2}e{q^{-1}} $ respectively. The equivalent conductance of $ 0.01NNH_2OH $ is $ 9.30oh{m^{-1}},cm^{2}e{q^{-1}} $ then the degree of ionization of $ NH_4OH $ at this temperature would be
Options:
A) 0.04
B) 0.1
C) 0.39
D) 0.62
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \Lambda _eq^{\infty }(NH_4OH)=\Lambda _eq^{\infty }(NH_4Cl)+\Lambda _eq^{\infty }(NaOH)-\Lambda _eq^{\infty }(NaCl) $
$ =129.8+217.8-109.3=238.3oh{m^{-1}}cm^{2}e{q^{-1}} $
$ \alpha =\frac{{\Lambda_{eq}}}{\Lambda _eq^{\infty }}=\frac{9.30}{238.3}=0.04 $
BETA
