Question

The molar conductivity of $0.007M$ acetic acid is $20Sc{m}^{2}mo{l}^{–1}$. What is the dissociation constant of acetic acid? Choose the correct option.

$[{\Lambda }_{H^{+}}^{\circ }=350Sc{m}^{2}mo{l}^{-1}{\Lambda }_{C{H}_{3}CO{O}^{-}}^{\circ }=50Sc{m}^{2}mo{l}^{-1}]$

• A. $2.50\times {10}^{-5}mol{L}^{-1}$
• B. $1.75\times {10}^{-4}mol{L}^{-1}$
• C. $2.50\times {10}^{-4}mol{L}^{-1}$
• D. $1.75\times {10}^{-5}mol{L}^{-1}$

Answer 1

The correct option is D $1.75\times {10}^{-5}mol{L}^{-1}$

${\Lambda }_m^{\circ }=20Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_{mC{H}_{3}COOH}^{\circ }={\Lambda }_{C{H}_{3}CO{O}^{-}}^{\circ }+{\Lambda }_{m{H}^{+}}^{\circ }$

$=50+350=400Sc{m}^{2}mo{l}^{-1}$

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{20}{400}=\frac{1}{20}$

$K_a=\frac{C{\alpha }^2}{1-\alpha }=C{\alpha }^2=7\times {10}^{-3}\times {\left(\frac{1}{20}\right)}^2$

$=7\times {10}^{-3}\times \frac{1}{4}\times {10}^{-2}$

$=1.75\times {10}^{-5}mol{L}^{-1}$