Buffer Solutions

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take one litre of pure water at twenty five degrees Celsius. Its p H is seven, right? Now add just zero point zero one zero mole of hydrochloric acid. The p H crashes from seven to two. Five whole units! Now add the same acid to one litre of another clear solution, and the p H moves by less than one tenth of a unit! What is in that second bottle? The secret is a buffer solution; even your blood is buffered. Buffer questions come up often in JEE, so let's begin.

0:38What is a buffer?

So, first, the definition. Buffer: a solution that resists a change in p H when you add a little acid or base, or dilute it with water. Resists, okay? The p H changes, but only a little. There are two kinds. An acidic buffer is a weak acid plus its salt with a strong base. For example, acetic acid, C H three COOH, with sodium acetate, C H three COON a. The small numbers below are atom counts, and the slash separates the two partners. Right? And a basic buffer is the mirror image: a weak base plus its salt with a strong acid. Like ammonia, N H three, with ammonium chloride, N H four C l. Clear?

1:26The Henderson–Hasselbalch equation

Now, the one equation you need: the Henderson Hasselbalch equation. For an acidic buffer, 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, with a small a for acid; it is fixed at a given temperature. Square brackets mean concentration, in moles per litre. Okay? For a basic buffer, the same shape, but it gives p O H: p K b, small b for base, plus log of salt over base. Then p H is fourteen minus p O H, at twenty five degrees Celsius; the little circle means degrees. With me? And here is a gift. Salt and acid share the same volume, so the volume cancels in the ratio. Use moles, or millimoles, directly. Got it? Now quick, tell me. If salt and acid are equal, what is the p H? p K a, exactly! The ratio is one, and log of one is zero; the double arrow means, so this gives. So p H equals p K a. Remember this one. Right?

2:38How a buffer fights back

Now, how does a buffer resist? Suppose you add H plus; the small plus up top is its charge. Which of the two partners will eat it up? The salt, yes! Acetate, C H three COO minus, grabs the H plus and becomes acetic acid; the arrow means, turns into. So added H plus turns salt into acid. Okay? And when you add O H minus, the acid steps in. Acetic acid gives its H plus to the O H minus, making acetate and water, H two O, right? So added O H minus turns acid into salt. So, after any addition, just recalculate the ratio of salt to acid. That's it! Now a trap. I dilute a buffer ten times with water. Everything becomes ten times weaker, so the p H must change, right? No, no, careful! Salt and acid are both diluted ten times, so their ratio stays the same. So the p H stays the same, to a good approximation. With me? But the buffer capacity falls: how much acid or base it can absorb before its p H changes much. Clear?

3:53Capacity and the useful range

Now look at this 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 that comes from the salt. Okay? Now see where they cross: right at zero, on the dotted line, where p H equals p K a and salt equals acid, as the caption says. There the buffer has its greatest capacity. Right? Capacity also grows with concentration. Compare a zero point one molar pair with a zero point zero one molar pair; capital M means molar. Same p H, but the first absorbs ten times as much acid or base before its p H changes much; the cross means times. Got it? Now the shaded band, labelled buffer range, from minus one to plus one. Inside it, both forms are present in useful amounts. So the useful range is p K a plus or minus one, the plus minus sign. Outside it, one partner is too scarce to buffer well. With me? Now a topper's trick. You add a strong base to a weak acid, but only enough to neutralise half of it. What is the p H? p K a, yes! Half the acid has become salt, and half is still acid, so salt equals acid. No log needed. And it is a quick way to make a buffer, okay?

5:22Example 1: pH of an acetic acid-sodium acetate buffer

Okay, Example one. Find the p H of a solution that is zero point one zero molar in acetic acid, and zero point two zero molar in sodium acetate. p K a is four point seven four. Pause the video, and try it first. Done? Okay, let's check. Acetic acid with its own salt: an acidic buffer, so Henderson Hasselbalch. p H equals p K a, plus log of salt over acid, which is four point seven four, plus log of zero point two zero over zero point one zero. Right? Before you calculate, think. There is more salt than acid. Will the p H be above p K a, or below it? Above, correct! The ratio is two, and log two is zero point three zero. So p H is four point seven four plus zero point three zero: five point zero four. Very good!

6:23Example 2: HCl added to a buffer vs pure water

Now our puzzle, as a real problem. Zero point zero one zero mole of H C l goes into one litre of a buffer, zero point one zero molar in both acetic acid and sodium acetate; capital L is litre. p K a is four point seven four. Find the new p H, and compare with one litre of pure water. Pause and try. Hint: work in moles. Done? Zero point zero one zero mole of H plus in one litre: that is zero point zero one zero molar, so the p H is two, right? No, no, careful! That is pure water, not a buffer. Here the acetate eats the H plus, so salt turns into acid. Acetate falls from zero point one zero to zero point zero nine, and acetic acid rises from zero point one zero to zero point one one. Okay? Now recalculate. p H equals four point seven four, plus log of zero point zero nine over zero point one one. That log is about minus zero point zero nine. So the new p H is four point six five. Right? Look at that! The p H moved by only zero point zero nine. In pure water, the same acid makes H plus zero point zero one zero molar, and the p H falls from seven to two. Puzzle solved: that second bottle was a buffer, just like this one! Clear?

7:54Quick revision: buffer pH, added H+ or OH-, useful range

So, quick revision. Acidic buffer: p H equals p K a plus log of salt over acid; basic buffer, the same with p O H and p K b. Added H plus turns salt into acid; added O H minus turns acid into salt; recalculate the ratio. Best at p H equals p K a, useful within p K a plus or minus one, and dilution leaves the p H almost unchanged. Now go and try the practice problems on this page. You can do them, I promise. All the best!