Variation with Concentration; Kohlrausch's Law

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Take a beaker of salt solution, and keep pouring in water. Now there are fewer ions in every cubic centimetre, so the solution should conduct less, right? But chemists say that on dilution, the molar conductivity goes up! Falling and rising together? Strange! Today we sort this out, and then we meet Kohlrausch's law. This comes up often in JEE electrochemistry. So, let's begin.

0:30Dilution: one quantity falls, one rises

Recap first. Kappa, the Greek letter like a small k, is the conductivity, in siemens per centimetre. And the upside down V is capital lambda. With the small m below it, we read capital lambda m, the molar conductivity. Capital lambda m equals kappa times one thousand, over c, where c is the concentration in moles per litre. Okay? Now, we add water. Tell me, what happens to kappa? It goes down, yes! That is the down arrow. After dilution, there are fewer ions per unit volume to carry the current, so kappa falls. Clear? So, capital lambda m also falls, right? No, no, careful! This is the trap. Capital lambda m rises on dilution: the up arrow. So kappa falls, capital lambda m rises: both true! But why? The reason differs for strong and weak electrolytes. With me?

1:30Strong versus weak electrolytes

On the graph, capital lambda m goes up, and root c, the square root of concentration, goes across. The dark blue line is sodium chloride, a strong electrolyte. For it, in dilute solution, capital lambda m equals capital lambda m naught, minus capital A times root c. The little circle, read naught, means infinite dilution, so capital lambda m naught is the limiting molar conductivity. And capital A is a constant. Okay? So, capital lambda m against root c is a straight line. Extend it back to root c equal to zero, and the intercept is capital lambda m naught. Right? As we dilute, we move left, and the blue line rises, but only a little. Why only a little? Because a strong electrolyte is already completely ionised. Dilution only pulls the ions further apart, so the interionic attraction weakens. That's it! Now the orange curve: acetic acid, a weak electrolyte. The label says it rises steeply only as c tends to zero; that little arrow means tends to. Why? Because alpha, the degree of dissociation, increases on dilution. More molecules split into ions. Got it? So, can we extend the orange curve back to the axis, like the blue line, and read capital lambda m naught? No! Near zero it shoots up, so there is no straight line to extend. Capital lambda m naught cannot be found by extrapolation. So how did we get the value at the top, three hundred and ninety point seven? By Kohlrausch's law. Clear?

3:13Kohlrausch's law: each ion counts on its own

Kohlrausch's law says: at infinite dilution, every ion makes its own contribution to the molar conductivity, independently of its partner ion. Okay? So, capital lambda m naught equals nu plus times small lambda plus naught, plus nu minus times small lambda minus naught. Small lambda naught, with a plus or a minus below it, is the cation's or the anion's own share. And nu, the Greek letter like a v, counts how many of those ions one formula unit gives. With me? For sodium chloride, both nu values are one. So capital lambda m naught is small lambda naught of the sodium ion, plus small lambda naught of the chloride ion. Right? For a weak electrolyte, we add and subtract the values of strong electrolytes, so that only the ions we want are left. That is how the acetic acid value on the graph was found. We will do one together in Example one.

4:09Two more uses: dissociation and solubility

Now, two more uses. First, the degree of dissociation of a weak electrolyte: alpha equals capital lambda m over capital lambda m naught. Then, for a weak acid, K with a small a below it, read K a, the acid dissociation constant, equals c alpha squared over one minus alpha. Okay? Next, a sparingly soluble salt, like silver chloride. What is special about its saturated solution? Yes, exactly! So little dissolves that the solution is extremely dilute. So capital lambda m is approximately capital lambda m naught: that is the wavy equals sign. Got it? Put c equal to the solubility, s, and rearrange: s equals kappa times one thousand, over capital lambda m naught. But careful! Water itself conducts a little, so first subtract water's conductivity from the measured kappa. Clear?

5:09Example 1: limiting molar conductivity of NH4OH

Now, Example one. The limiting molar conductivities, in siemens centimetre squared per mole, are: ammonium chloride, one hundred and twenty nine point eight; sodium hydroxide, two hundred and forty eight point one; and sodium chloride, one hundred and twenty six point four. Find capital lambda m naught of ammonium hydroxide. Pause the video and try it. Done? Okay, let's check. Ammonium hydroxide is a weak base, so we use Kohlrausch's law, not extrapolation. We need small lambda naught of the ammonium ion, plus small lambda naught of the hydroxide ion. Okay? So, which values do we add, and which one do we subtract? Capital lambda naught of ammonium chloride brings the ammonium ion, and that of sodium hydroxide brings the hydroxide ion, so add those two. But they also bring sodium and chloride ions we do not want: exactly sodium chloride. So, subtract it, and those two cancel. Topper's trick: write the ions you need first, then pick the salts. Got it? One hundred and twenty nine point eight, plus two hundred and forty eight point one, is three hundred and seventy seven point nine. Minus one hundred and twenty six point four: two hundred and fifty one point five siemens centimetre squared per mole. Done!

6:36Example 2: solubility and Ksp of AgCl from conductivity

Now a harder one. The conductivity of saturated silver chloride, after subtracting that of water, is one point eight six times ten to the power minus six siemens per centimetre, at two hundred and ninety eight kelvin. Capital lambda m naught of silver chloride is one hundred and thirty eight point three siemens centimetre squared per mole. Find the solubility, and K s p, the solubility product. Pause and try. Done? Let's check. See, the water is already subtracted for us. The solution is saturated, but very dilute, so capital lambda m is about capital lambda m naught: one hundred and thirty eight point three. Okay? So, s equals one point eight six times ten to the power minus six, times one thousand, over one hundred and thirty eight point three. Now, why times one thousand? Because kappa over capital lambda m naught gives moles per cubic centimetre, but we want s in moles per litre, and one litre is one thousand cubic centimetres. So s is one point three four times ten to the power minus five molar; capital M means moles per litre. Forget it, and you are a thousand times off! Clear? Now, K s p. Silver chloride gives one silver ion and one chloride ion. So in terms of s, what is K s p? Each ion has concentration s, so K s p is s times s: s squared. The square brackets mean concentration in moles per litre. Right? One point three four times ten to the power minus five, squared, is one point eight times ten to the power minus ten. That's it! Both answers from one conductivity reading.

8:31Quick revision: dilution, Kohlrausch law, alpha, solubility

So, quick revision. On dilution, kappa falls, but capital lambda m rises. For a strong electrolyte, extrapolate the straight line to get capital lambda m naught; for a weak one, use Kohlrausch's law. And alpha is capital lambda m over capital lambda m naught, while the solubility is kappa times one thousand, over capital lambda m naught. Now go and try the practice problems on this page. All the best!