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Solve the following.
The specific conductance of 2·5 × 10⁻⁴ M formic acid is 5·25 × 10⁻⁵ ohm⁻¹ cm⁻¹. Calculate its molar…
(a)[2.5]
The specific conductance of $2\cdot5 \times 10^{-4}\text{ M}$ formic acid is $5\cdot25 \times 10^{-5}\text{ ohm}^{-1}\text{ cm}^{-1}$. Calculate its molar conductivity and degree of dissociation.
Given $\lambda^\circ_{\text{H}^+} = 349\cdot5\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1}$ and $\lambda^\circ_{\text{HCOO}^-} = 50\cdot5\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1}$.
(b)[2.5]
Calculate the time taken to deposit $1\cdot27\text{ g}$ of copper at cathode when a current of $2\text{ amp.}$ is passed through the solution of $\text{CuSO}_4$.
(Atomic weight of $\text{Cu} = 63\cdot5\text{ g mol}^{-1}$)
Answer
Answer (a)
Molar conductivity: $\Lambda_m=\dfrac{\kappa\times1000}{C}=\dfrac{5.25\times10^{-5}\times1000}{2.5\times10^{-4}}=210\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1}$
$\Lambda_m^\circ=\lambda^\circ_{\mathrm{H^+}}+\lambda^\circ_{\mathrm{HCOO^-}}=349.5+50.5=400\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1}$
Degree of dissociation: $\alpha=\dfrac{\Lambda_m}{\Lambda_m^\circ}=\dfrac{210}{400}=0.525$
Final answer: 210 ohm^-1 cm^2 mol^-1
Answer (b)
Formula used: Faraday's first law: w = M*Q/(n*F) with F = 96500 C mol^-1, I = 2 A, M = 63.5 g mol^-1, n = 2, w = 1.27 g, so t = 1930 s
$\mathrm{Cu^{2+}+2e^-\rightarrow Cu}$, so $n=2$.
$w=\dfrac{M\,I\,t}{nF}$, so $t=\dfrac{w\,n\,F}{M\,I}=\dfrac{1.27\text{ g}\times2\times96500\text{ C mol}^{-1}}{63.5\text{ g mol}^{-1}\times2\text{ A}}=1930\text{ s}$
Final answer: 1930 s