Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. I have one hundred millilitres of a solution with a p H of four point seven four. Now I add one millimole of sodium hydroxide, a strong base. Strong base, so the p H should shoot up, right? It goes to four point eight three. Less than one tenth of a unit! So where did that strong base go? Hold on to that question. Today: Set two of our mixed problems on equilibrium, the topics its eight problems test, and two solved problems. The second one is this exact puzzle. Mixed questions like these come up often in JEE Main. So, let's begin.

0:43Set 2: eight problems, four topics

So, Set two has eight solved problems from four topics. First, equilibrium constants: K p, with a small p for pressures, and K from delta G nought. The little triangle means change, and the little circle means standard. So that is the standard Gibbs energy change. Three problems. Okay? Second, acids and bases: spotting Lewis acids, the electron pair acceptors, and the p H after mixing an acid with a base. Two problems. Right? Third, salt hydrolysis and buffers: two problems. Clear? And fourth, the solubility product, K s p, with a small s p below: one problem. Today's two examples come from these last two lines: K s p first, then a buffer. With me?

1:33Solubility product and the common ion

Now, the solubility product. Solid silver chloride, A g C l, with the s in brackets for solid, sits in its saturated solution. It gives silver ions, A g plus, and chloride ions, C l minus; the sign up top is the charge, and a q means dissolved in water. The two half arrows mean equilibrium. Okay? So K s p equals A g plus times C l minus; square brackets mean concentration, in moles per litre. Let small s be the molar solubility. In pure water, each ion is s, so K s p equals s squared. Right? Now think. I add silver ions from outside, from silver nitrate, A g N O three, the three written small below. Will more silver chloride dissolve now, or less? Less, exactly! The extra silver ions push the equilibrium back to the left. That is Le Chatelier's principle. This is the common ion effect. For a one to one salt like silver chloride, if the common ion is already c molar, capital M for molar, then s is approximately K s p over c; the wavy equals sign means approximately. This works when s is much smaller than c, the double less than sign. Got it? And one trap to avoid. The solubility falls, but K s p itself does not change. It depends only on temperature. Clear?

3:03Buffers: the Henderson–Hasselbalch equation

Now, buffers. An acidic buffer is a weak acid with its salt, like acetic acid with sodium acetate. For it, we use the Henderson Hasselbalch equation: p H equals p K a, plus log of salt over acid. p K a is minus log of K a, the acid ionisation constant, and it is fixed at a given temperature. Okay? Now look at the graph. Along the bottom is p H minus p K a; up the side, the fraction of the acid in each form. The dark blue curve is H A, the acid itself, and the orange curve is A minus, the anion from the salt. Where they cross, salt equals acid. So what is the p H there? p K a, exactly! The ratio is one, and log of one is zero. And the shaded band, from minus one to plus one, is the buffer range: p K a plus or minus one, the plus minus sign. With me? Here is a shortcut. Salt and acid share the same volume, so the volume cancels in the ratio. So just use millimoles: molarity times volume in millilitres. Right? Now you add O H minus, the hydroxide ion. Which partner reacts with it, the acid or the salt? The acid, correct! Acetic acid gives its H plus to the O H minus, and becomes acetate. So added O H minus turns acid into salt. Then recalculate the ratio. Clear?

4:33Example 1: solubility of AgCl in 0.1 M AgNO3

Now, Example one, an easy problem from Set two. Find the solubility of silver chloride in zero point one molar silver nitrate. K s p is one point eight times ten to the minus ten. Pause the video, and try it first. Done? Okay, let's check. K s p equals s squared, so s is just the square root of K s p, right? No, no, careful! That is pure water. Here, silver nitrate already gives zero point one molar silver ions. So A g plus is zero point one plus s, but C l minus is only s. And s is tiny, so zero point one plus s is about zero point one. With me? So K s p equals zero point one times s; the cross means times. And that equals one point eight times ten to the minus ten. Okay? Divide by zero point one. s equals one point eight times ten to the minus ten, over zero point one: one point eight times ten to the minus nine molar. Far smaller than zero point one, so our approximation was fine. Very good! Now compare. In pure water, s is the square root of K s p, about one point three times ten to the minus five molar. So in pure water, silver chloride dissolves more than seven thousand times as much. That is the common ion effect, and K s p never changed. Got it?

6:06Example 2: buffer pH after adding 1 mmol NaOH

And now, Example two, a medium one, and it is our puzzle from the start. One millimole of sodium hydroxide is added to one hundred millilitres of a buffer, zero point one molar in both acetic acid and sodium acetate. p K a is four point seven four. Find the p H, okay? Pause and try. Hint: work in millimoles. Done? Okay, let's check. Zero point one molar times one hundred millilitres: ten millimoles of acid, and ten millimoles of salt. Salt equals acid, so at first the p H equals p K a, four point seven four. Got it? Now one millimole of O H minus goes in. So the salt becomes eleven millimoles, and the acid stays at ten, right? No, no, wait! This is the trap. Each hydroxide ion uses up one acetic acid molecule to make one acetate. The arrow on screen means turns into. So the acid falls from ten to nine, and the salt rises from ten to eleven millimoles. Okay? Now Henderson Hasselbalch. p H equals four point seven four, plus log of eleven over nine. Clear? Log of eleven over nine is about zero point zero eight seven. So the p H is about four point eight three. Puzzle solved! The acid swallowed the strong base, so the ratio moved only from one to eleven over nine. That is why the p H rose by less than one tenth. Perfect!

7:40Quick revision: s = Ksp/c and buffer pH in millimoles

So, quick revision. Common ion, for a one to one salt like silver chloride: s is approximately K s p over c, and K s p itself depends only on temperature. Buffer: p H equals p K a plus log of salt over acid, with the ratio in millimoles. And added O H minus takes the acid down and the salt up, by the same amount. Now go and try the three practice problems on this page. You can do them, I promise. All the best!