Conductivity and Molar Conductivity

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Two students measure the resistance of the very same salt solution, at the same temperature. One gets fifty ohms. The other gets one hundred ohms. Same solution, different answers. So who made the mistake? Hmm, maybe nobody! By the end of this video you will see exactly why. And you will meet the two numbers that really describe a solution: conductivity, and molar conductivity. Questions on these come up often in JEE electrochemistry, so let's begin.

0:36From resistance to conductivity

So, start with something you already know from physics. R, the resistance, equals rho times small l, over capital A, right? Rho, the Greek letter you see on screen, is the resistivity, a property of the material. For a solution, small l is the distance between the two electrodes, and capital A is their area. Okay? Now tell me. Same solution, but the electrodes are now twice as far apart. What happens to the resistance? It doubles, yes! So resistance depends on the cell, not only on the solution. Maybe that is what happened to our two students, right? Now, conductance, capital G, is simply one over R, nothing more. Its unit is the siemens, written as a capital S, okay? So we need a number for the solution itself. That is kappa, the Greek letter that looks like a small k. Kappa is the conductivity, also called specific conductance. It equals one over rho, which is G times small l over capital A, where the dot just means times. Change the cell, and the resistance changes, but kappa stays the same. With me? And its unit? Siemens per centimetre, or siemens per metre in S I units. The little minus one written up top means per. Clear?

1:58The cell constant

Now look at the ratio, small l over capital A, okay? It depends only on the cell, so we give it a name: the cell constant, G star, that is G with a little star at the top. Its unit is per centimetre. Got it? But how do we find it? We fill the cell with a potassium chloride solution, K C l, whose kappa is already known, and we measure its resistance. Now, kappa equals G times G star. So how do you get G star? Yes, exactly! G is one over R, so G star simply equals kappa times R, that's it! Right? And once the cell constant is known, put any other solution in that same cell, measure its resistance, and kappa equals G star over R, simple! Now, our puzzle! The two students used cells with different cell constants. Twice the cell constant, twice the resistance, but the same kappa. Nobody was wrong! Clear?

2:57Molar conductivity

Alright, kappa tells us how well the solution as a whole conducts. But to compare electrolytes fairly, we often want the conductivity per mole of electrolyte. So what should we divide kappa by? By the concentration, small c. Correct! That gives the molar conductivity: capital lambda, the Greek letter, with a small m below it. Capital lambda m equals kappa times one thousand, over c, with kappa in siemens per centimetre, and c in moles per litre. Okay? But where does this one thousand come from? From the litre! One litre is one thousand cubic centimetres. So c over one thousand is the concentration in moles per cubic centimetre, and now everything is in centimetres. The unit comes out as siemens centimetre squared per mole. Got it? In S I units it is cleaner: capital lambda m equals just kappa over c, but now c must be in moles per cubic metre. Then the unit is siemens metre squared per mole. With me? And to switch between them: one siemens centimetre squared per mole equals ten to the power minus four siemens metre squared per mole, because one square centimetre is ten to the power minus four square metres. Right? One more. If you use normality, capital N, in place of molarity, you get the equivalent conductivity: capital lambda with a small e q below it, equal to kappa times one thousand, over N, okay?

4:31Example 1: kappa and molar conductivity, 0.2 M at 50 ohm

Now, Example one. We have a cell with cell constant one point five per centimetre, filled with a zero point two molar solution; the capital M means molar. Its resistance is fifty ohms, and the horseshoe symbol, capital omega, means ohms. Find kappa and capital lambda m. Pause the video, and try it first. Done? Okay, let's check. Kappa first. Cell constant one point five, resistance fifty. Multiply them: seventy five siemens per centimetre, right? No, no, careful! That is the trap. Kappa equals G star over R, so more resistance means less conductivity. One point five over fifty: zero point zero three zero siemens per centimetre. Okay? Now molar conductivity. Kappa is in siemens per centimetre and c is in moles per litre, so we use the formula with the thousand. Zero point zero three zero times one thousand is thirty, and thirty over zero point two is one hundred and fifty. So capital lambda m is one hundred and fifty siemens centimetre squared per mole. That's it!

5:44Example 2: cell constant from KCl, molar conductivity in SI

Now a harder one, JEE level. One conductivity cell. Filled with zero point one molar K C l, whose kappa is one point two nine siemens per metre, it shows one hundred ohms. Filled with a zero point zero two molar solution of another salt, it shows five hundred and twenty ohms. Find capital lambda m of the salt, in siemens metre squared per mole. Pause and try. Hint: stay in S I units all the way. Done? Okay, let's check. Step one, the cell constant, from the K C l reading. G star equals kappa times R: one point two nine times one hundred, which is one hundred and twenty nine per metre. Okay? Step two, the same cell with the salt. Kappa equals G star over R: one hundred and twenty nine over five hundred and twenty, about zero point two four eight siemens per metre. Topper's tip: in one cell, kappa times R stays fixed, so kappa is inversely proportional to the resistance: double the resistance, and kappa is halved. Right? Now step three. Capital lambda m equals kappa over c. And c is zero point zero two, so we divide by zero point zero two, right? Oh no, careful! This is the trap. In S I units, c must be in moles per cubic metre. One cubic metre is one thousand litres, so zero point zero two moles per litre is twenty moles per cubic metre. Got it? Now divide. Zero point two four eight over twenty is zero point zero one two four. So capital lambda m is one point two four times ten to the power minus two siemens metre squared per mole. Divide by zero point zero two instead, and your answer is a thousand times too big. See how one unit slip changes everything?

7:43Quick revision: kappa, cell constant, molar conductivity

So, quick revision. One: kappa equals G star over R, and it belongs to the solution, not to the cell. Two: the cell constant, G star, equals kappa times R, found with K C l of known kappa. Three: capital lambda m equals kappa times one thousand over c, in siemens centimetre squared per mole, or kappa over c in S I units, with c in moles per cubic metre. Now go and try the practice problems on this page. All the best!