Question

The conductivity of $0.001028mol{L}^{-1}$ acetic acid is $4.95\times {10}^{-5}Sc{m}^{-1}$. Find out its dissociation constant if ${\Lambda }_m$ for acetic acid is $390.5Sc{m}^{-1}mo{l}^{-1}$.

• A. $2.18\times {10}^{-5}mo{l}^{-1}{L}^{-1}$
• B. $1.78\times {10}^{-5}mo{l}^{-1}{L}^{-1}$
• C. $3.72\times {10}^{-4}mo{l}^{-1}{L}^{-1}$
• D. $2.37\times {10}^{-4}mo{l}^{-1}{L}^{-1}$

Answer 1

The correct option is B $1.78\times {10}^{-5}mo{l}^{-1}{L}^{-1}$

${\Lambda }_m=\frac{1000\kappa }{C}$
${\Lambda }_m=\frac{1000\times 4.95\times {10}^{-5}Sc{m}^{-1}}{0.001028mol{L}^{-1}}=48.15Sc{m}^{2}mo{l}^{-1}$
$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^0}=\frac{48.15Sc{m}^{2}mo{l}^{-1}}{390.5Sc{m}^{2}mo{l}^{-1}}=0.123$
$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}=\frac{C\alpha \times C\alpha }{C(1-\alpha )}$
$K_a=\frac{C{\alpha }^2}{1-\alpha }$
$K_a=\frac{0.001028\times (0.123{)}^2}{1-0.123}=1.78\times {10}^{-5}$ mol/L