Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take hydrochloric acid, very dilute: ten to the minus eight moles per litre. It is a strong acid, fully ionised, so the H plus concentration is ten to the minus eight, and p H, minus log of that, is eight. p H eight? At twenty five degrees Celsius, that is basic! Can adding an acid make water basic? Something is wrong here, right? That exact statement sits in Part C of this chapter, the unit problem bank for Equilibrium. Today: how to use this bank, a few of its problems, and this puzzle, solved. So, let's begin.

0:43How to use this chapter

So, this chapter has three parts. Part A has eight solved problems at JEE Main level, mixing the whole unit, then three more for practice. Okay? The instruction: attempt Part A as a thirty minute test, after Chapters one to six. So sit with a clock, and open the solutions only at the end. Right? Then Part B: six longer problems, at JEE Advanced level, and three for practice. With me? And Part C: concept checks, the ideas that are often tested in statement form, like true or false. Our p H eight puzzle lives there. Clear?

1:21Part A: name the idea before you calculate

Now, the real skill in a mixed test: before you calculate, name the idea. Here is a Part A problem. K c, the K with a small c, is four times ten to the minus four, for N two plus O two giving two N O, nitric oxide. The two half arrows, one each way, mean equilibrium. Okay? Find K c for N O giving half N two plus half O two. Compare the two reactions. What has happened to the first one? It is reversed, and halved, yes! Reversing gives one over K, and halving gives the square root; that is the root sign. So K c is one over the root of four times ten to the minus four, one over zero point zero two: fifty. The double arrow means this leads to, and this rule is from Chapter one. Got it? Next. Zero point one M acetic acid, plus zero point one M sodium acetate; capital M means moles per litre. p K a is four point seven four; p K a means minus log of K a, the acid dissociation constant. We need the p H of this mixture. But first, what kind of solution is this? It is a buffer, correct! Weak acid plus its salt, in equal concentrations, so p H equals p K a: four point seven four. Name the idea first, and Part A becomes fast. Right?

2:43Part C: read every word

Now, Part C, and here is our puzzle, as a statement. True or false: the p H of ten to the minus eight MHC l, hydrochloric acid, is eight. So, true or false? False! An acid cannot give a basic solution. At such a low concentration, water's own H plus matters too; the little plus sign up top means a positive ion. Include it, and at twenty five degrees Celsius, the little circle means degrees, the p H comes out six point nine eight. Still slightly acidic, below seven. Puzzle solved! Okay? One more. True or false: K s p of A g C l, silver chloride, is smaller in zero point one M N a C l, sodium chloride, than in water. K s p, the K with s p written small below it, is the solubility product. True or false? If you said true, careful! It is false. The chloride from N a C l is a common ion, so the solubility is smaller, yes. But K s p itself depends only on temperature. One word changes the answer, so in Part C, read every word. Clear?

3:53Example 1: total pressure over solid NH4HS

Now, Example one, from Part A, okay? Solid N H four H S, ammonium hydrosulphide, breaks up into ammonia, N H three, and H two S, hydrogen sulphide. The small s in brackets means solid, and the small g means gas. K p, the K with a small p, is zero point two five atmosphere squared. Find the total pressure at equilibrium, starting from the solid alone. Pause the video, and try it first. Done? Okay, let's check. Pure solids are left out of K, so K p is just the partial pressure of ammonia times the partial pressure of hydrogen sulphide. That is p with N H three below it, times p with H two S below it. Right? Now, we started from the solid alone. Each time one ammonia molecule forms, what else forms with it? One hydrogen sulphide molecule, yes, exactly! So the two partial pressures are equal; call each one small p. With me? So p squared equals zero point two five, and p is zero point five atmosphere. Got it? Now, careful. The question asks for the total, not one gas. Stopping at zero point five is the trap! Capital P, the total, is zero point five plus zero point five: one point zero atmosphere. Okay?

5:16Example 2: two solids sharing gas B, all pressures

Now a harder one, from Part B, JEE Advanced level. Two solids decompose in the same evacuated vessel, a vessel that starts empty. Solid A B gives gases A and B, with K p one equal to zero point zero four atmosphere squared. Solid C B gives gases C and B, with K p two equal to zero point zero five atmosphere squared. Find all the partial pressures, and the total pressure. Pause the video and try it. Hint: where does gas B come from? Done? Okay, let's check. Solids are left out again, so p A times p B equals zero point zero four, and p C times p B equals zero point zero five. The letter below each p names the gas. With me? Now, in Example one, the two gases had equal pressures. So here too, p A equals p B, and p C equals p B, right? No, no, careful! That is the trap. There is only one gas B in the vessel, and it comes from both solids. Each A comes with one B, and each C comes with one B, so p B equals p A plus p C, the sum of the other two. Clear? So now we have three equations and three unknowns. Is there a quick way in, without heavy algebra? Yes: add the first two! That is the topper's trick. p B times p A, plus p B times p C, is p B times p A plus p C, in round brackets. And that bracket is just p B, so p B squared equals zero point zero four plus zero point zero five, that is zero point zero nine. So p B is zero point three atmosphere. Neat, isn't it? Now divide. p A is zero point zero four over zero point three: zero point one three three. p C is zero point zero five over zero point three: zero point one six seven atmosphere. Check: they add up to zero point three, exactly p B, as they should. Got it? And the total: zero point one three three, plus zero point three, plus zero point one six seven. Zero point six atmosphere. Both K p values hold at the same time, through one shared gas. That is Part B for you: one extra idea, and the problem opens up. Right?

7:33Quick revision: name the idea, shared gas, dilute acid

So, quick revision. Part A: a thirty minute test after Chapters one to six, and name the idea before you calculate. Part B: JEE Advanced level; a gas shared by two solids gets its pressure from both. Part C: read every word, and for a very dilute acid, water's own H plus counts. Now open Part A, start your clock, and begin. All the best!