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Home » Conductivity of acetic acid is . Calculate its molar conductivity. If for acetic acid is , what is its dissociation constant?
Conductivity of 0.00241 M acetic acid is 7.896 \times 10^{-5} S cm ^{-1}. Calculate its molar conductivity. If 0 \wedge m for acetic acid is 390.5 S cm ^{2} mol ^{-1}, what is its dissociation constant?
CBSE | Chemistry | Class 12 | Electrochemistry | NCERT
Conductivity of 0.00241 M acetic acid is 7.896 \times 10^{-5} S cm ^{-1}. Calculate its molar conductivity. If 0 \wedge m for acetic acid is 390.5 S cm ^{2} mol ^{-1}, what is its dissociation constant?

Solution: Given, K =7.896 \times 10^{-5} S m ^{-1} c

    \[=0.00241 mol L ^{-1}\]


As we know, molar conductivity, \Lambda_{m}=\frac{k}{c}

Substituting the value of known parameters we have,

    \[\begin{aligned}&=\frac{7.896 \times 10^{-5} S cm ^{-1}}{0.00241 mol L^{-1}} \times \frac{1000 cm ^{3}}{L} \&=32.76 S cm ^{2} mol ^{-1}\end{aligned}\]


Also, given in question,

    \[\Lambda_{m}^{0}=390.5 S cm ^{2} mol ^{-1}\]


We know that,

    \[\begin{aligned}&\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{0}} \&=\frac{32.76 S cm ^{2} mol ^{-1}}{390.5 Scm ^{2} mol ^{-1}}\end{aligned}\]


    \[=0.084\]


We know that, Dissociation constant, K_{a}=\frac{c \alpha^{2}}{(1-\alpha)}

    \[\begin{aligned}&=\frac{\left(0.00241 mol L^{-1}\right)(0.084)^{2}}{(1-0.084)} \&=1.86 \times 10^{-5} mol L ^{-1}\end{aligned}\]

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