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A conductivity cell has two platinum plates fixed in a glass tube…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A conductivity cell has two platinum plates fixed in a glass tube. To find its cell constant, it is first filled with $0.0100\ \text{mol L}^{-1}$ potassium chloride solution at $298\ \text{K}$ , whose conductivity NCERT gives as $0.141\ \text{S m}^{-1}$ . The cell then shows a resistance of $742\ \Omega$ . The same cell is rinsed and filled with $0.0200\ \text{mol L}^{-1}$ acetic acid at the same temperature, and now the resistance is $4.61 \times 10^{3}\ \Omega$ . The conductivity of the water used is small enough to ignore. The plates are coated with platinum black so that they do not polarise, and the cell sits in a water bath that keeps every reading at the same temperature. Assume that the cell constant does not change between the two fillings. Use NCERT's limiting molar ionic conductivities: $\lambda^\circ(\text{H}^+) = 349.6\ \text{S cm}^2\ \text{mol}^{-1}$ and $\lambda^\circ(\text{CH}_3\text{COO}^-) = 40.9\ \text{S cm}^2\ \text{mol}^{-1}$ . Take the degree of dissociation of the acid as $\alpha = \Lambda_m / \Lambda_m^\circ$ . Give conductivities in S per cm where a unit is asked for.

What is the cell constant?
What is the molar conductivity of the acetic acid solution?
What is the degree of dissociation of the acid?
What is the dissociation constant $K_a$ of acetic acid?
The acid is diluted to $0.00500\ \text{mol L}^{-1}$ and put into the same cell. What resistance should the cell now show? Keep $K_a$ from (d), and find the new $\alpha$ from $K_a = \dfrac{c\alpha^2}{1 - \alpha}$ by solving the quadratic in full.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Calibrate the cell, turn the resistance of the acid into $\Lambda_m$ , then use Ostwald's dilution law both ways.

(a) Cell constant. $\kappa(\text{KCl}) = 0.141\ \text{S m}^{-1} = 1.41 \times 10^{-3}\ \text{S cm}^{-1}$ , so $G^* = \kappa R = 1.41 \times 10^{-3} \times 742\ \text{cm}^{-1} = 1.046\ \text{cm}^{-1}$ .

(b) Molar conductivity. $\kappa = \dfrac{G^*}{R} = \dfrac{1.0462}{4610}\ \text{S cm}^{-1} = 2.269 \times 10^{-4}\ \text{S cm}^{-1}$ . Then $\Lambda_m = \dfrac{1000 \times 2.269 \times 10^{-4}}{0.0200}\ \text{S cm}^2\ \text{mol}^{-1} = 11.35\ \text{S cm}^2\ \text{mol}^{-1}$ .

(c) Degree of dissociation. $\Lambda_m^\circ = 349.6 + 40.9 = 390.5\ \text{S cm}^2\ \text{mol}^{-1}$ , so $\alpha = \dfrac{11.35}{390.5} = 0.0291$ .

(d) Dissociation constant. $K_a = \dfrac{c\alpha^2}{1 - \alpha} = \dfrac{0.0200 \times (0.0291)^2}{1 - 0.0291}\ \text{mol L}^{-1} = 1.74 \times 10^{-5}\ \text{mol L}^{-1}$ , close to the value NCERT lists for acetic acid.

(e) After dilution. Solve $\dfrac{0.00500\,\alpha^2}{1 - \alpha} = 1.739 \times 10^{-5}$ : the quadratic gives $\alpha = 0.0573$ . Then $\Lambda_m = 0.0573 \times 390.5 = 22.36\ \text{S cm}^2\ \text{mol}^{-1}$ , $\kappa = \dfrac{22.36 \times 0.00500}{1000} = 1.118 \times 10^{-4}\ \text{S cm}^{-1}$ and $R = \dfrac{1.0462}{1.118 \times 10^{-4}}\ \Omega = 9.36 \times 10^{3}\ \Omega$ .

Watch out. A quarter of the acid does not mean four times the resistance ( $1.84 \times 10^{4}\ \Omega$ ). A weak acid ionises more when it is diluted, so the resistance grows only 2.03 times. And using $\kappa$ in S per metre would make the cell constant $104.6$ , a hundred times too large.

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