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Conductivity of 0.00241 \(\mathrm{M}\) acetic acid is \(7.896 \times 10^{-5} \mathrm{~S}\) \(\mathrm{cm}^{-1}\). Calculate its molar conductivity. If \(\Lambda_{\mathrm{m}}^0\) for acetic acid is \(390 \cdot 5 \mathrm{~S} \mathrm{~cm}{ }^2 \mathrm{~mol}^{-1}\), what is its dissociation constant?
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Verified Answer
\(\Lambda_{\mathrm{m}}^{\mathrm{c}}=\frac{\kappa \times 1000}{\text { Molarity }}\)
\(\begin{aligned}
&=\frac{\left(7 \cdot 896 \times 10^{-5} \mathrm{Scm}^{-1}\right) \times 1000 \mathrm{~cm}^3 \mathrm{~L}^{-1}}{0 \cdot 00241 \mathrm{~mol} \mathrm{~L}^{-1}} \\
&=32 \cdot 76 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\
\alpha &=\frac{\Lambda_{\mathrm{m}}^{\mathrm{c}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{32 \cdot 76}{390 \cdot 5}=8 \cdot 4 \times 10^{-2}
\end{aligned}\)
Dissociation constant is given as :
\(\begin{aligned}
K_a &=\frac{C \alpha^2}{1-\alpha}=\frac{0.0024 \times\left(8.4 \times 10^{-2}\right)^2}{1-0.084} \\
&=1.86 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}\)
\(\begin{aligned}
&=\frac{\left(7 \cdot 896 \times 10^{-5} \mathrm{Scm}^{-1}\right) \times 1000 \mathrm{~cm}^3 \mathrm{~L}^{-1}}{0 \cdot 00241 \mathrm{~mol} \mathrm{~L}^{-1}} \\
&=32 \cdot 76 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\
\alpha &=\frac{\Lambda_{\mathrm{m}}^{\mathrm{c}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{32 \cdot 76}{390 \cdot 5}=8 \cdot 4 \times 10^{-2}
\end{aligned}\)
Dissociation constant is given as :
\(\begin{aligned}
K_a &=\frac{C \alpha^2}{1-\alpha}=\frac{0.0024 \times\left(8.4 \times 10^{-2}\right)^2}{1-0.084} \\
&=1.86 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}\)
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