Molar conductivity at infinite dilution for NH4Cl, NaOH and NaCl solution at 298 K are respectively 129.8, 218.4 and 108.9 Scm2 mol-1 and ^m for 10-2 M solution of NH4OH is 9.33 Scm2 mol-1. Calculate the degree of dissociation of nh4oh at this dilution?
Answers
Answer:
The degree of dissociation of NH_4OHNH
4
OH is 0.0389.
Explanation: To calculate the molar conductivity at infinite dilution of NH_4OHNH
4
OH , we use Kohlrausch's Law.
\Lambda ^o_{NH_4OH}=\Lambda ^o_{NH_4Cl}+\Lambda ^o_{NaOH}-\Lambda ^o_{NaCl}Λ
NH
4
OH
o
=Λ
NH
4
Cl
o
+Λ
NaOH
o
−Λ
NaCl
o
\begin{gathered}\Lambda ^o_{NH_4Cl}=129.8Scm^2mol^{-1}\\\Lambda ^o_{NaOH}=218.4Scm^2mol^{-1}\\\Lambda ^o_{NaCl}=108.9Scm^2mol^{-1}\end{gathered}
Λ
NH
4
Cl
o
=129.8Scm
2
mol
−1
Λ
NaOH
o
=218.4Scm
2
mol
−1
Λ
NaCl
o
=108.9Scm
2
mol
−1
Putting the values in above equation, we get
\begin{gathered}\Lambda ^o_{NH_4OH}=129.8+218.4-108.9\\\Lambda^o_{NH_4OH}=239.9Scm^2mol^{-1}\end{gathered}
Λ
NH
4
OH
o
=129.8+218.4−108.9
Λ
NH
4
OH
o
=239.9Scm
2
mol
−1
We are given molar conductivity of NH_4OHNH
4
OH ,
\Lambda^m_{NH_4OH}=9.33Scm^2mol^{-1}Λ
NH
4
OH
m
=9.33Scm
2
mol
−1
Now, to calculate the degree of association, we use the formula:
\alpha =\frac{\Lambda ^m_{NH_4OH}}{\Lambda ^o_{NH_4OH}}α=
Λ
NH
4
OH
o
Λ
NH
4
OH
m
Putting the values in above equation, we get
\begin{gathered}\alpha =\frac{9.33}{239.9}\\\alpha =0.0389\end{gathered}
α=
239.9
9.33
α=0.0389