Acid–Base Titrations

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. We add sodium hydroxide to twenty five millilitres of hydrochloric acid, both zero point one molar. After twelve and a half millilitres, half the acid is gone, but the p H has risen by less than half a unit. Then, at one moment, one extra millilitre pushes the p H up by more than four units! Why does the p H crawl first, and then jump? Today we find out. Questions on titration curves come up often in JEE, so let's begin.

0:36Reading a titration curve

So, here is a titration curve. We add a strong base, zero point one molar sodium hydroxide, from a burette; the capital M means molar, moles per litre. p H goes up the side; base added goes along the bottom. The key point is equivalence, where millimoles of base equal millimoles of acid; the short form on screen means millimoles. Here, that is at twenty five millilitres, the dotted line marked equivalence. Okay? The solid curve is H C l, a strong acid; the dashed one is C H three COOH, acetic acid, a weak acid. The small three is a subscript. The shaded bands show where two indicators, phenolphthalein and methyl orange, change colour. With me?

1:23Start, and before equivalence

At the start, no base has gone in, so the p H is that of the acid alone. Zero point one molar H C l gives p H one. Acetic acid is weak, so the dashed curve starts higher, near three. Okay? Now add some base, not enough to reach equivalence. For H C l, the H plus left over, the excess, decides the p H; the small plus sign up top is a positive charge. Right? In acetic acid, part of the acid has become acetate, its salt; the rest is still acid. Weak acid plus its own salt: what is that? Yes, a buffer! So we use Henderson's equation: p H equals p K a, plus log of salt over acid. p K a is minus log of K a, the acid constant; square brackets mean concentration. That is why the dashed curve is so flat here: the buffer resists change, okay? Now, half-way: half the acid is neutralised, so salt equals acid. What is the p H? Correct! Log of one is zero, so p H equals p K a: the black dot on the dashed curve, at twelve and a half millilitres. Topper's trick: read p K a off the curve at half-way. Only for a weak acid, okay?

2:43At equivalence, and after it

Now, equivalence. All the acid is just neutralised, so the p H is seven, always, right? No, no, careful! Only for strong acid plus strong base. H C l and sodium hydroxide leave sodium chloride, which does not hydrolyse. And this seven needs twenty five degrees Celsius, where K w is ten to the minus fourteen; the small circle means degrees. Okay? With a weak acid, the flask holds sodium acetate at equivalence, and acetate hydrolyses, leaving O H minus. So the p H is greater than seven, the sign on screen. Got it? The mirror case: strong acid plus weak base, like H C l with ammonia. The ammonium ion hydrolyses, so the p H is less than seven. Clear? After equivalence, no acid is left: the excess O H minus, the hydroxide ion, decides. See, after twenty five millilitres, the two curves join. With me? Now, which curve has the longer vertical jump, the solid one or the dashed one? The solid one, H C l! For acetic acid, the buffer holds the p H higher before equivalence, and hydrolysis puts equivalence near eight point seven, so its jump is shorter. The rule: the steeper and longer the jump, the easier the end point, where the indicator changes colour. The double arrow means so. Right?

4:11Example 1: pH during HCl vs NaOH titration

Now, Example one, our opening puzzle. Twenty five millilitres of zero point one zero molar H C l is titrated with zero point one zero molar sodium hydroxide. Find the p H after zero, twelve and a half, twenty five, and twenty six millilitres of base. Pause the video and try it. Work in millimoles. Done? Okay, let's check. Acid: twenty five times zero point one zero, two point five millimoles. At zero millilitres, H plus is zero point one zero molar, so p H one point zero zero. Okay? Twelve and a half millilitres of base leave one point two five millimoles of H plus. Now divide by twenty five millilitres, right? Oh no, careful! The base brought its own water. The total volume is thirty seven point five millilitres. So H plus is zero point zero three three three molar, and p H is one point four eight. Half the acid gone, and the p H barely moved: the slow crawl from our puzzle! At twenty five millilitres, equivalence. Strong acid, strong base, so p H seven point zero zero. Now one more millilitre. How far can the p H move? That millilitre brings zero point one millimoles of O H minus, with no acid left to use it up. In fifty one millilitres, that is one point nine six times ten to the minus three molar. Okay? So p O H, minus log of O H minus, is two point seven one, and p H is fourteen minus that: eleven point two nine. Over four units, with one millilitre! Puzzle solved. That's it!

6:01Example 2: pH during acetic acid vs NaOH titration

Now a harder one, JEE level. Twenty five millilitres of zero point one zero molar acetic acid, p K a four point seven four, is titrated with zero point one zero molar sodium hydroxide. Find the p H after zero, twelve and a half, twenty five, and thirty millilitres. Pause and try. Hint: ask what is in the flask. Done? Okay, let's check. At zero, only the weak acid: p H equals half of, p K a minus log c, with c the acid concentration. Log of zero point one is minus one, so half of, four point seven four plus one: two point eight seven. This shortcut needs ionisation below about five percent, true here. Okay? At twelve and a half millilitres, exactly half the acid is neutralised. What is in the flask? Yes, exactly! Half acid, half acetate: a buffer. So p H equals p K a, four point seven four. No calculation! Clear? At twenty five millilitres, equivalence. So p H seven, right? No, no, wait! Weak acid, strong base: the flask holds sodium acetate, which hydrolyses. Two point five millimoles in fifty millilitres: c equals zero point zero five molar. Got it? Use the hydrolysis formula, at twenty five degrees Celsius: p H equals seven, plus half of, p K a plus log c. Log of zero point zero five is minus one point three zero, so the bracket is three point four four. Seven plus half of that: eight point seven two. Basic, as we said! At thirty millilitres, five millilitres past equivalence: zero point five millimoles of extra O H minus. Divide by the total, fifty five millilitres, not five! That is nine point one times ten to the minus three molar. Right? Excess strong base decides now. p H equals fourteen minus p O H, that is, fourteen plus log of O H minus: eleven point nine six. See how fast that was?

8:20Quick revision: pH before, at and after equivalence

So, quick revision. Before equivalence: the excess H plus for a strong acid, or a buffer for a weak acid. Half-way for a weak acid, p H equals p K a. At equivalence, p H seven only for strong plus strong; after it, the excess O H minus decides. Now go and try the practice problems on this page. All the best!