Part A: Mixed Problems

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. I dissolve a weak acid in water, one mole per kilogram. If it did not ionise at all, this one molal solution would boil at one hundred point five two degrees Celsius. But the thermometer shows one hundred point five seven two. And from that tiny extra, I can tell you how strong the acid is: its acid dissociation constant! How? Hold that thought; our last example cracks it. This is Part A of the unit problem bank on solutions: mixed problems of the kind that come up often in JEE Main. Three quick tools, then two full problems. So, let's begin.

0:44Tool 1: mole fraction and molality

So, first tool: concentration. Thirty six grams of glucose are dissolved in one hundred and sixty two grams of water. Find x B, the mole fraction of glucose, and m, the molality. The small subscript B stands for the solute, glucose. Okay? First, moles. n B: thirty six over one hundred and eighty, the molar mass of glucose. Zero point two. n A, with A for the solvent, water: one hundred and sixty two over eighteen, that is nine. Right? Now, x B equals n B, over n A plus n B, the total moles. Zero point two over nine point two: zero point zero two one seven. Clear? Now molality. The whole solution weighs one hundred and ninety eight grams, so we divide by zero point one nine eight kilogram, right? Oh no, careful! Molality uses the mass of the solvent, not of the solution. So m equals n B over w A, the mass of water in kilograms: zero point two over zero point one six two, one point two three. The small m after the number means molal. Got it?

1:57Tool 2: Henry's law for a dissolved gas

Second tool: gases in liquids. Carbon dioxide, C O two, the small two below being a subscript, dissolves in five hundred grams of water at a partial pressure p of two point five bar. K H, capital K with a small H, is Henry's law constant: one point six seven kilobar. Find the mass dissolved. With me? Henry's law: p equals K H times x, the mole fraction of the gas. So a larger K H means a less soluble gas. Okay? So x is two point five over one point six seven, about one point five, right? No, no, wait! Mole fractions are never more than one. Match the units: one point six seven kilobar is one thousand six hundred and seventy bar. So x is two point five over one thousand six hundred and seventy: one point five times ten to the power minus three. Clear? Now, x is so tiny that the moles of gas are about x times the moles of water; the wavy equals sign means approximately. n water, the moles of water: five hundred over eighteen, twenty seven point eight. So n is zero point zero four one six mole. Right? Times forty four, the molar mass of carbon dioxide: about one point eight three grams. Got it?

3:19Tool 3: same molality, different boiling points

Third tool. Four solutions, all one molal: glucose, potassium chloride, potassium sulphate and aluminium chloride. Which boils highest? Same molality, so the same boiling point, right? Oh no, careful! Colligative properties count particles. For a dilute solution of a non-volatile solute, delta T b, the rise in boiling point, equals i times K b times m. Delta, the little triangle, means change; K b is the boiling point constant of the solvent; and i, the van 't Hoff factor, counts the particles each formula unit gives. So compare i times m. Okay? Take the salts as fully dissociated. Glucose stays whole, so its i is one. Potassium chloride gives two ions, potassium sulphate three, and aluminium chloride four. So i is one, two, three and four. Clear? Same m, so the biggest i wins: one molal aluminium chloride. Remember this i; it cracks our opening puzzle. With me?

4:27Example 1: total pressure and mole fraction of A in vapour

Alright, example one. An ideal solution of one mole of liquid A and four moles of liquid B is in equilibrium with its vapour. p A nought, the vapour pressure of pure A, is two hundred and fifty; the little circle on top means pure. p B nought is fifty millimetres of mercury. Find the total pressure, capital P, and y A, the mole fraction of A in the vapour. Pause the video, and try it first. Done? Okay, let's check. Five moles in all, so x A is one over five, zero point two, and x B is zero point eight. Okay? The solution is ideal, so Raoult's law holds for both: each partial pressure is mole fraction times pure vapour pressure. p A is zero point two times two hundred and fifty: fifty. p B is zero point eight times fifty: forty. With me? The total pressure is their sum: fifty plus forty, ninety millimetres of mercury. Got it? Now the vapour. The liquid is twenty percent A, so the vapour is twenty percent A too, right? No, no, careful! Never use liquid mole fractions for the vapour. y A is the share of the pressure that comes from A: p A over P, fifty over ninety, zero point five five six! One fifth of the liquid, but more than half of the vapour, because A is far more volatile. Clear?

5:59Example 2: Ka of a weak acid from its boiling point

Now the harder one: our opening puzzle. One molal aqueous solution of a weak acid, H A, boils at one hundred point five seven two degrees Celsius; the small circle means degrees. Find K a, capital K with a small a, the acid dissociation constant, taking molality as roughly molarity. Pause and try. Hint: find i first. Done? Okay, let's check. Water boils at one hundred degrees, so delta T b is zero point five seven two. Okay? Now think. If H A did not ionise at all, what would delta T b be? Just K b times m, yes, exactly! Zero point five two for water, times one: zero point five two, as in our puzzle. Right? So i is observed over expected: zero point five seven two over zero point five two, one point one. More than one, so the acid has ionised a little. Got it? H A gives H plus and A minus; the half arrows mean equilibrium, and the signs on top are charges. Two ions, so i equals one plus alpha; the Greek letter alpha is the degree of dissociation. The double arrow means so: alpha is zero point one. With me? Now K a. Alpha is small, so K a is just m alpha squared, zero point zero one, right? Oh no, careful! K a equals m alpha squared, over one minus alpha; with alpha at ten percent, that one minus alpha matters. One times zero point zero one, over zero point nine. Clear? So K a is about zero point zero one one: one point one times ten to the power minus two. Puzzle solved! That tiny extra rise gave us i, then alpha, then K a. Very good!

7:59Quick revision: molality, Henry's law, i and Ka

So, quick revision. Molality is moles of solute per kilogram of solvent, never of solution. In Henry's law, p equals K H times x, and match the units of p and K H first. Compare solutions by i times m; in the vapour, y A is p A over P; and for a weak acid H A, which gives two ions, alpha is i minus one. Now try all six problems of Part A on this page as a thirty minute test; the one on molar mass and osmotic pressure is waiting for you. Then the four practice problems. All the best!