Question

The conductivity of 0.001028 mol $ L^{-1}$ acetic acid is 4.95 × $10^{-5} S cm^{-1}.$ Calculate its dissociation constant if $\wedge_{m}^{∘}$ for acetic acid is $390.5 S cm^{2} mol^{-1}.$

• A. $1.78 \times 10^{-5} mol L^{-1}$
• B. $1.87\times10^{-5} mol L^{-1}$
• C. $0.178\times10^{-5} mol \, L^{-1}$
• D. $0.0178\times10^{-5} mol \, L^{-1}$

Answer 1

Answer: (A)

$\Lambda_{m} =\frac{k}{c}=\frac{4.95\times10 ^{-5}S cm^{-1}}{0.001028 mol L^{-1}} \times\frac{1000 cm^{3}}{L}$
$=48.15 S cm^{2} mol^{-1}$
$\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{∘}}=\frac{48.15}{390.5}=0.1233$
$k=\frac{c \alpha^{2}}{\left(1-\alpha\right)}=\frac{0.001028\times\left(0.1233\right)^{2}}{1-0.1233}=1.78\times10^{-5} mol \, L^{-1}$