Solveeit Logo
Login

Question

Chemistry Question on Conductance

The conductivity of 0.001028 mol L−1 L^{-1} acetic acid is 4.95 ? 10−5Scm−1.10^{-5} S cm^{-1}. Calculate its dissociation constant if \wedge_{m}^{?? for acetic acid is 390.5Scm2mol−1.390.5 S cm^{2} mol^{-1}.

A

1.78×10−5molL−11.78 \times 10^{-5} mol L^{-1}

B

1.87×10−5molL−11.87\times10^{-5} mol L^{-1}

C

0.178×10−5mol L−10.178\times10^{-5} mol \, L^{-1}

D

0.0178×10−5mol L−10.0178\times10^{-5} mol \, L^{-1}

Answer

1.78×10−5molL−11.78 \times 10^{-5} mol L^{-1}

Explanation

Solution

Λm=kc=4.95×10−5Scm−10.001028molL−1×1000cm3L\Lambda_{m} =\frac{k}{c}=\frac{4.95\times10 ^{-5}S cm^{-1}}{0.001028 mol L^{-1}} \times\frac{1000 cm^{3}}{L} =48.15Scm2mol−1=48.15 S cm^{2} mol^{-1} \alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{??}=\frac{48.15}{390.5}=0.1233 k=cα2(1−α)=0.001028×(0.1233)21−0.1233=1.78×10−5mol L−1k=\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}