Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Two students find the p H of point zero five molar sulphuric acid. One writes one point three. The other writes one. So who is right? Hold that thought. Today, Part A of our equilibrium problem bank: short questions at JEE Main level, and we solve seven of its eight here. Questions like these come up often, and each one hides a small trap. So, let's begin.

0:31Tool 1: rewrite the equation, rewrite K

So, first tool: change the equation, and K changes too. Problem one. For N two plus O two, giving two N O, K c is four times ten to the minus four. The two half arrows mean equilibrium, and K c, capital K with a small c below, is the equilibrium constant in concentrations, fixed at a given temperature. Okay? Now we want K c prime, the K with a little dash: the new constant, for N O giving half N two plus half O two. It is the reverse reaction, so just flip K: one over four times ten to the minus four, right? Not so fast! Two changes here, not one. Reverse a reaction, and K prime is one over K, okay? Multiply every coefficient by n, and K prime is K to the power n. Here the coefficients are also halved, so n is one half: a square root. Clear? So K c prime is one over the square root of four times ten to the minus four; see the square root sign. That root is two times ten to the minus two, and one over that is fifty. Flip, and take the root. Got it?

1:38Tool 2: strong acids, count every H+

Second tool: strong acids. p H equals minus log of the H plus concentration, with log to the base ten. The square brackets mean concentration in moles per litre, and the small plus up top is a positive charge. Now, a strong acid ionises completely, so the H plus comes straight from the acid. With me? Problem two. Point zero zero one molar H C l, hydrochloric acid, is diluted ten times; capital M means molar. What is the new p H? Four, yes, exactly! Point zero zero one is ten to the minus three. Ten times the volume, one tenth the concentration: ten to the minus four molar. So the p H is four. Okay? Now our puzzle: problem three. Point zero five molar sulphuric acid, H two S O four, with complete ionisation; the small two and four count the atoms. Minus log of point zero five is about one point three; the wavy equals means approximately. So p H one point three, right? No, no, careful! Each H two S O four gives two H plus ions. So the H plus concentration is two times point zero five: point one molar. And minus log of point one is one. Puzzle solved! The second student is right; the first forgot one proton. Clear?

2:58Tool 3: a buffer, then a solubility product

Third tool: ready-made results. First, the Henderson equation for an acidic buffer: a weak acid with its salt. p H equals p K a, plus log of the salt concentration over the acid concentration, where p K a is minus log of K a, the acid dissociation constant. Right? Problem four. Take a solution point one molar in both acetic acid and sodium acetate, with p K a four point seven four. What is the p H? Do you even need a calculator? No calculator at all! Equal concentrations, so the ratio is one, and log one is zero. So p H equals p K a: four point seven four. Clear? Next, problem five: silver chromate, A g two C r O four, a sparingly soluble salt. In its saturated solution it gives two A g plus ions and one chromate ion, charge two minus; s and a q mean solid and aqueous. Its molar solubility, small s, is ten to the minus four molar. So K s p, the solubility product, with s p written small below, is just s squared, right? Oh no, careful! Look at the two in front of A g plus. Silver is two s, and chromate is s. So K s p is two s, squared, times s: four s cubed. Four times ten to the minus twelve. Got it?

4:20Example 1: total pressure over solid NH4HS

Now, example one: problem eight. Solid ammonium hydrosulphide, N H four H S, breaks up into ammonia and hydrogen sulphide gases; s in brackets is solid, and g is gas. K p is 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. K p uses partial pressures, the small p with each gas written below it. And the solid? Left out: a pure solid never appears in the equilibrium constant. So K p is just the ammonia pressure times the hydrogen sulphide pressure. Clear? Now think. We started from the solid alone. How do the two gas pressures compare? Equal, exactly! Each N H four H S that breaks up gives one ammonia and one hydrogen sulphide. So both pressures are the same; call each one small p. With me? Then p squared is point two five, so p is point five atmosphere. So the answer is point five atmosphere, right? No, no, wait! Point five is each gas; the question asks for the total. Total pressure, capital P, is point five plus point five: one point zero atmosphere. Read to the end! Got it?

5:44Example 2: degree of dissociation of PCl5 when Kp = P/3

Now the harder one, example two: problem six. Phosphorus pentachloride, P C l five, gives P C l three and chlorine, C l two. At some total pressure, capital P, K p equals P over three. Find alpha, the Greek letter on screen: the degree of dissociation. Pause and try. Hint: start from one mole. Done? Okay, let's check. Start from one mole of P C l five; m o l is short for mole. Alpha of it breaks up, so one minus alpha is left, with alpha of each product. Total: one plus alpha. Okay? Partial pressure is mole fraction times capital P, so for P C l five, one minus alpha over one plus alpha, times P; for each product, alpha over one plus alpha, times P as well. With me? Put these into K p: products multiplied, over P C l five. One P and one factor of one plus alpha cancel, so K p equals alpha squared P, over one minus alpha squared. Here P is the total pressure at equilibrium, not at the start. Right? Now, a shortcut many students remember: alpha is about the square root of K p over P, as on screen. Can we use it here? Careful! That shortcut holds only when alpha is small. Here it gives the square root of one third, about point five eight. Not small at all! So it fails, and we solve exactly. Clear? So set alpha squared P, over one minus alpha squared, equal to P over three. Look, capital P cancels! The double arrow means it follows: three alpha squared equals one minus alpha squared. Okay? So four alpha squared equals one, and alpha is one half: point five. Half of the P C l five has dissociated, not point five eight. Got it?

7:40Quick revision: K rules, strong acid pH, buffer, Ksp, alpha

So, quick revision. Reverse a reaction: one over K, okay? Multiply the coefficients by n: K to the n, so halving gives a square root. Count every H plus. An equal buffer has p H equal to p K a, and silver chromate has K s p equal to four s cubed. Solids stay out of K, the alpha shortcut needs small alpha, and read to the end. Now try the practice problems on this page. You can do it. All the best!