Question

At 291 K, the equivalent conductivities at infinite dilution of $HCl,NaCl$ and $C{H}_{3}COONa$ are 369.3, 101.7 and 71.7 $Sc{m}^{2}e{q}^{-1}$ respectively. If equivalent conductivity of centinormal solution of acetic acid is 14.3 $Sc{m}^{2}e{q}^{-1}$, The degree of dissociation (in %) of $C{H}_{3}COOH$ is:

Answer 1

${\Lambda }_{\infty }\left(C{H}_{3}COOH\right)={\Lambda }_{\infty }\left(C{H}_{3}COONa\right)+{\Lambda }_{\infty }\left(HCl\right)-{\Lambda }_{\infty }\left(NaCl\right)$

$=71.7+369.3-101.7=339.3Sc{m}^{2}e{q}^{-1}$

The degree of dissociation $\alpha =\frac{\Lambda \left(C{H}_{3}COOH\right)}{{\Lambda }_{\infty }\left(C{H}_{3}COOH\right)}=\frac{14.3Sc{m}^{2}e{q}^{-1}}{339.3Sc{m}^{2}e{q}^{-1}}=0.04$
This corresponds to 4%.