Question

Calculate ${\lambda }_m^{\infty }$ for acetic acid.
${\lambda }_m^{\infty }$ $\left(HCl\right)$ $=426$ $Sc{m}^{2}mo{l}^{-1}$
${\lambda }_m^{\infty }$ $\left(NaCl\right)$ $=126$ $Sc{m}^{2}mo{l}^{-1}$
${\lambda }_m^{\infty }\left(C{H}_{3}COONa\right)=91$ $Sc{m}^{2}mo{l}^{-1}$

• A. $420Sc{m}^{2}mo{l}^{-1}$
• B. $329Sc{m}^{2}mo{l}^{-1}$
• C. $391Sc{m}^{2}mo{l}^{-1}$
• D. None of these

Answer 1

Answer: (C)

$391Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_m^{\infty }C{H}_{3}COOH=?$

${\Lambda }_m^{\infty }\left(NaCl\right)=126Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_m^{\infty }\left(C{H}_{3}COONa\right)=91Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_m^{\infty }\left(HCl\right)={\Lambda }_{H^{+}}^{\infty }+{\Lambda }_{C{l}^{-}}^{\infty }=426$.....(i)

${\Lambda }_m^{\infty }\left(NaCl\right)={\Lambda }_{N{a}^{+}}^{\infty }+{\Lambda }_{C{l}^{-}}^{\infty }=126$.....(ii)

${\Lambda }_m^{\infty }\left(C{H}_{3}COONa\right)={\Lambda }_{C{H}_{3}CO{O}^{-}}^{\infty }+{\Lambda }_{N{a}^{+}}^{\infty }=91$.....(iii)

Add equation (i) and (iii) and subtract equation (ii)

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

$={\Lambda }_{C{H}_{3}CO{O}^{-}}^{\infty }+{\Lambda }_{H^{+}}^{\infty }$

${\Lambda }_m^{\infty }C{H}_{3}COOH=426+91-126$

$=391Sc{m}^{2}mo{l}^{-1}$