Question

The conductivity of $0.001028mol{L}^{-1}$ acetic acid is $4.95\times 10^{-5}Sc{m}^{-1}$. Calculate its dissociation constant, if ${\Lambda }_m^{\circ }$ for acetic acid is $390.5Sc{m}^{2}mo{l}^{-1}$

• A. $1.78\times 10^{-5}mol{L}^{-1}$
• B. $1.87\times 10^{-5}mol{L}^{-1}$
• C. $0.178\times 10^{-5}mol{L}^{-1}$
• D. $0.0178\times 10^{-5}mol{L}^{-1}$

Answer 1

Answer: (A)

${\Lambda }_m=\frac{\kappa }{C}=\frac{4.95\times 10^{-5}Sc{m}^{-1}}{0.001028mol{L}^{-1}}\times \frac{1000c{m}^3}{L}$

$=48.15Sc{m}^{2}mo{l}^{-1}$

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^0}=\frac{48.15}{390.5}=0.1233$

$K_a=\frac{C{\alpha }^2}{(1-\alpha )}=\frac{0.001028\times (0.1233{)}^2}{1-0.1233}=1.78\times 10^{-5}mol{L}^{-1}$