The van't Hoff Factor

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Common salt, sodium chloride, has a molar mass of fifty eight point five grams per mole. Now a student finds its molar mass from a colligative property of a zero point one molal solution. And the experiment says: thirty point eight grams per mole! Almost half. So, is the experiment wrong? No, the experiment is fine! And benzoic acid in benzene does the opposite: its molar mass comes out too high. Today, one small number explains both: the van 't Hoff factor, i. It comes up often in JEE, so let's begin.

0:42More particles, or fewer?

So, first, a quick reminder. Colligative properties count the solute particles: how many, not what kind. More particles, bigger effect. Okay? Now take sodium chloride, N a C l, in solution. It splits into ions, N a plus and C l minus. The arrow means gives, and the small plus and minus on top are the charges. One unit, two particles. This is dissociation. Right? Now acetic acid in benzene. Its molecules pair up: two of them join into one. So tell me, more particles, or fewer? Fewer, exactly! Two molecules of acetic acid, C H three COOH, give one double molecule, written in brackets with a small two below. We call it a dimer. Two particles become one. This is association. With me?

1:36The van 't Hoff factor, i

So we need a correction factor. Its name: the van 't Hoff factor, small i. i equals the observed colligative property, over the calculated one. Calculated means what we expect with no dissociation and no association. The long bar means divided by. Clear? Another way to see it: i equals the moles of particles after dissolving, over the moles of solute taken. Sodium chloride: one mole gives two moles of ions, so i is two. Okay? Now the molar mass. More particles means a bigger colligative effect. So the molar mass we calculate from that effect: does it come out bigger, or smaller? Smaller, correct! In the molar mass formulas, the measured effect sits at the bottom, so a bigger effect gives a smaller molar mass. That is our salt puzzle, half solved! So for molar mass, i is the normal value over the observed, abnormal value. Look carefully: normal is on top here. Got it?

2:38Which way does i point?

Now, students often get confused here. i greater than one, the sign on screen, means dissociation: more particles, and the observed molar mass is lower than normal. Right? And if i is less than one? Dissociation, or association? Association, yes! Fewer particles, so the observed molar mass comes out higher than normal, just like benzoic acid in our puzzle. Clear? And non-electrolytes, such as glucose and urea, give no ions, so i is exactly one. With me?

3:12Dissociation: i = 1 + (n - 1)α

Now, how big is i? First, dissociation. Take one mole of solute, the m o l on screen, where each unit gives n ions. And alpha, this Greek letter, is the degree of dissociation: the fraction that breaks up. Okay? So, one minus alpha moles stay whole. And alpha moles break up, each into n ions, so n times alpha moles of ions. Right? Add them: one minus alpha, plus n alpha. So i equals one plus, n minus one in brackets, times alpha. Write this down. Clear? Turn it around, and alpha equals i minus one, over n minus one. And for complete dissociation, alpha is one, so i simply equals n, the number of ions. Got it? Quick one. Aluminium sulphate: A l two, then S O four in brackets, taken three times. Complete dissociation. What is i? Five, very good! Two aluminium ions plus three sulphate ions, five in all. The same way, N a C l gives two, K two S O four gives three, and A l C l three gives four. With me?

4:27Association: i = 1 - (1 - 1/n)α

Now, association. Here n molecules join into one bigger particle, and alpha is now the degree of association: the fraction that joins. Okay? From one mole, one minus alpha moles stay single. And alpha moles join up, n at a time, so they make alpha over n moles of big particles. Right? Add them up: i equals one minus alpha, plus alpha over n. That is one minus, bracket, one minus one over n, bracket closed, times alpha. Clear? Now, complete dimerisation: n is two, and alpha is one. What is i? One half, yes! Half the particles, half the colligative effect, so the observed molar mass doubles. You will see this with acetic acid or benzoic acid in benzene. Got it?

5:20Old formulas, now times i

So, how do we use i? Simple: every colligative formula gets multiplied by i. Relative lowering of vapour pressure: delta p over p nought is about i times n B over n A, the moles of solute B and solvent A, for a dilute solution. The capital Greek delta means change, the small circle, read nought, means pure solvent, and the wavy equals sign means approximately. Okay? Boiling point: delta T b equals i times K b times m, the molality. Freezing point: delta T f equals i times K f times m. And osmotic pressure, the Greek letter pi, equals i times CRT: C is the molarity, R the gas constant, and T the temperature. And i takes care of dissociation and association. Right?

6:13Example 1: degree of dissociation of NaCl from i = 1.9

Now, Example one. Remember our salt puzzle? Let's finish it. Zero point one molal sodium chloride shows i equal to one point nine; the small m means molal. Find the apparent molar mass of sodium chloride, and its degree of dissociation. Pause the video, and try it. Done? Okay, let's check. i is normal molar mass over observed, so the apparent molar mass, capital M, is the normal one over i. Fifty eight point five over one point nine: about thirty point eight grams per mole, written g mol to the power minus one. There it is, our puzzle number! The salt splits into ions, so more particles. Very good! Now alpha. Sodium chloride gives two ions, so n is two. Alpha equals one point nine minus one, over two minus one: zero point nine. So ninety percent dissociated. Not one hundred percent, so i is a little below two. Clear?

7:18Example 2: degree of association of benzoic acid

Now the other half of our puzzle, a little harder. The molar mass of benzoic acid is one hundred and twenty two. But from the freezing point of its solution in benzene, it comes out as two hundred grams per mole. Find the degree of association. Pause and try. Hint: which way does i point? Done? Let's check. i equals normal over observed: one hundred and twenty two over two hundred, that is zero point six one. Less than one, so association, as expected. Okay? Now many students grab alpha equals i minus one, over n minus one. Zero point six one minus one, over two minus one: minus zero point three nine. Negative alpha? What went wrong? No, no, careful! That formula is only for dissociation. Benzoic acid pairs up, so n is two, and we use the association formula: zero point six one equals one minus one half alpha. So half alpha is zero point three nine, and the double arrow means therefore: alpha equals zero point seven eight. Seventy eight percent of the benzoic acid is dimerised. That's it!

8:33Quick revision: i for dissociation and association

So, quick revision. i is observed over calculated, or normal molar mass over observed. Dissociation: i equals one plus, n minus one, times alpha. Association: i equals one minus, one minus one over n, times alpha. i greater than one, the molar mass looks smaller; i less than one, it looks larger; and glucose has i equal to one. Now go and try the practice problems on this page. All the best!