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Exercises · 2.9

Q.The resistance of a conductivity cell containing 0.001 M0.001\ M KCl solution at 298 K is 1500 Ω1500\ \Omega. What is the cell constant if conductivity of 0.001 M0.001\ M KCl solution at 298 K is 0.146×10−3 S cm−10.146 \times 10^{-3}\ S\ cm^{-1}?

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The cell constant is the product of measured resistance and known conductivity. Using G∗=κ×RG^* = \kappa \times R, we get G∗=0.146×10−3×1500=0.219 cm−1G^* = 0.146 \times 10^{-3} \times 1500 = 0.219\ \text{cm}^{-1}.

The key idea here is simple: a conductivity cell has a fixed geometry — the distance between electrodes divided by their area — which we call the cell constant (G∗G^*). You can’t measure it directly with a ruler inside the solution, but you can find it electrically.

Conductivity (κ\kappa) is an intrinsic property of the solution. Resistance (RR) is what you measure. They are related by:

κ=1R×G∗\kappa = \frac{1}{R} \times G^*

Rearranging:

G∗=κ×RG^* = \kappa \times R

That’s the whole backbone. Once you know the conductivity of a standard solution (here, 0.001 M0.001\ M KCl at 298 K, whose conductivity is tabulated and given), and you measure its resistance in your cell, the cell constant follows immediately.

Let’s walk through it.

  1. Write down what’s given.

    Resistance, R=1500 ΩR = 1500\ \Omega

    Conductivity, κ=0.146×10−3 S cm−1\kappa = 0.146 \times 10^{-3}\ \text{S cm}^{-1}

    (Notice the units: Siemens per centimetre — that’s the clue that our cell constant will come out in cm−1\text{cm}^{-1}.)

  2. Apply the relation.

G∗=κ×R=(0.146×10−3)×1500G^* = \kappa \times R = (0.146 \times 10^{-3}) \times 1500

  1. Do the arithmetic.

0.146×10−3=1.46×10−40.146 \times 10^{-3} = 1.46 \times 10^{-4}

Multiply by 15001500:

1.46×10−4×1500=1.46×10−4×1.5×103=1.46×1.5×10−1=2.19×10−1=0.2191.46 \times 10^{-4} \times 1500 = 1.46 \times 10^{-4} \times 1.5 \times 10^{3} = 1.46 \times 1.5 \times 10^{-1} = 2.19 \times 10^{-1} = 0.219

  1. Attach the unit. …
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