Colligative Properties of Electrolytes

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Two beakers of water. In the first, zero point one two molal sodium chloride, our common salt. In the second, zero point one molal sodium sulphate. The first one has more solute per kilogram of water. So the sodium chloride solution boils at the higher temperature, right? Hmm, not so fast. Today we see how electrolytes change the colligative properties, how to compare two solutions in one line, and how a freezing point can even give you the acid constant of a weak acid. Questions like this come up often in JEE, so let's begin.

0:41Electrolytes: count the particles

So, a quick recap from the last video. For an electrolyte, every colligative formula gets one extra factor: small i, the van 't Hoff factor. Delta T b equals i times K b times m; the little triangle is the Greek letter delta, for change, the small b below is for boiling, K b is the solvent's ebullioscopic constant, and m is the molality. Delta T f equals i times K f times m, with the small f for freezing, and K f the cryoscopic constant. And the osmotic pressure pi equals i CRT, with C the molarity, R the gas constant and T the temperature in kelvin. Okay? And the conditions, as always: a dilute solution, and a non-volatile solute. Right? Now think. One formula unit of sodium chloride goes into water and dissociates completely. How many particles do you get? Two, yes, exactly! One sodium ion and one chloride ion. So for complete dissociation, i equals n, the number of ions from one formula unit. Clear? So, sodium chloride: i is two. Potassium sulphate, K two S O four, with the small two and four below: two potassium ions and one sulphate ion, so i is three. And urea or glucose? They do not ionise at all, so i is just one. Got it?

2:08Comparing solutions: look at i×m

Now, the main idea of today. Colligative properties depend on the number of solute particles, not on what they are. And the number of particles goes as i times m, or i times C when we use molarity. So, to compare two solutions, compare the product i times m, not m alone. With me? Now tell me. One solution has the larger i times m. Does its freezing point go up, or down? Down, correct! More particles, a bigger delta T f, so it freezes lower. And more particles also lower the vapour pressure more, so its vapour pressure is lower too. Okay? And the other two go up. Larger i times m: higher boiling point, and higher osmotic pressure. Two go down, two go up. Really simple, right? Last, isotonic solutions: equal osmotic pressure. At the same temperature, R T is the same on both sides, so it cancels; the double arrow means so. Isotonic means i one C one equals i two C two, with the small ones and twos below for the two solutions. Clear?

3:17Weak electrolytes: Ka from colligative data

Now, weak electrolytes. Take a weak acid, H A, which dissociates only partly, into H plus and A minus; the two half arrows mean an equilibrium. The fraction that dissociates is alpha, the Greek letter: the degree of dissociation. Got it? Now think. One H A gives how many ions, and what is i here? Two ions, so n is two, yes! From the last video, i equals one, plus n minus one, times alpha. With n equal to two, i equals one plus alpha. So if we know i, alpha is just i minus one. Okay? Now, at equilibrium, the square brackets mean concentration. H plus and A minus are each m times alpha, and the H A left over is m times one minus alpha. Right? Put these into the acid dissociation constant, K a, capital K with a small a below. K a equals m alpha squared, over one minus alpha. One condition: we take molality as roughly equal to molarity; the wavy equals sign means approximately. With me? So the whole plan, in one line: measure delta T f, get i, then alpha, then K a. Each arrow is just the next step. Example two will do exactly this. Clear?

4:42Example 1: Na2SO4 or NaCl, which boils higher

Alright, example one, and it is our puzzle from the start. Which has the higher boiling point: zero point one molal sodium sulphate, or zero point one two molal sodium chloride? Both dissociate completely. The small m after a number means molal. Pause the video, and try it first. Done? Okay, let's check. Zero point one two is bigger than zero point one. So sodium chloride boils higher, right? No, no, careful! That is the trap. Count the particles first. Sodium sulphate, N a two S O four, gives two sodium ions, N a plus, and one sulphate ion, S O four two minus; the plus and the two minus written up top are the charges, and the arrow means it dissociates into these. Three ions, so i is three. Okay? Sodium chloride gives one sodium ion, N a plus, and one chloride ion, C l minus. Two ions, so i is two. Right? Now multiply. Sodium sulphate: three times zero point one, that is zero point three zero. Sodium chloride: two times zero point one two, that is zero point two four. Clear? Zero point three zero is greater than zero point two four; that is the greater-than sign. So zero point one molal sodium sulphate boils higher! Puzzle solved: less solute, but more particles. Always compare i times m. Got it?

6:17Example 2: alpha and Ka of a weak acid from freezing point

Now a harder one, example two. Zero point one molal solution of a weak monobasic acid freezes at minus zero point two zero four six degrees Celsius; the small circle means degrees. K f is one point eight six. Monobasic means one H plus per molecule, just like our H A acid. Find its degree of dissociation, and its K a. Pause and try. Hint: find i first. Done? Okay, let's check. Water freezes at zero degrees. So the observed delta T f is zero minus minus zero point two zero four six, which is zero point two zero four six. Okay? Now, to get i, we need a second number. If the acid did not dissociate at all, what would delta T f be? Just K f times m, yes, exactly! One point eight six times zero point one: zero point one eight six. Right? So i is observed over calculated: zero point two zero four six, over zero point one eight six. That is one point one. Bigger than one, so the acid does dissociate, but only a little. With me? Now, i equals one plus alpha. So alpha is zero point one: ten percent dissociated. That is the first answer. Clear? Now K a equals m alpha squared, over one minus alpha. On top: zero point one, times alpha squared, zero point zero one. At the bottom: one minus zero point one, zero point nine. Keep that one minus alpha; alpha is ten percent here, not tiny. Okay? Work it out: K a is about one point one one times ten to the power minus three. See? From one freezing point, we got the K a of an acid. Got it?

8:14Quick revision: compare i × m, isotonic, Ka of a weak acid

So, quick revision. To compare solutions, compare i times m, not m alone: larger i m, lower freezing point, higher boiling point. Isotonic means i one C one equals i two C two. And for a weak acid, i equals one plus alpha, and K a equals m alpha squared, over one minus alpha. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!