Question

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

• 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)

$\Lambda_{m}=\frac{\kappa}{C}=\frac{4.95 \times 10^{-5}\, S\, cm ^{-1}}{0.001028 \,mol\, L ^{-1}} \times \frac{1000 \,cm ^{3}}{ L }$

$=48.15 \,S \,cm ^{2} \,mol ^{-1}$

$\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{0}}=\frac{48.15}{390.5}=0.1233$

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