Acid–Base Titrations and Indicators

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. When hydrochloric acid is titrated with sodium hydroxide, the equivalence point is at p H seven. But methyl orange changes colour between p H three point one and four point four, and phenolphthalein between eight point three and ten. Neither changes at seven, and yet either one gives the correct end point. How is that possible? Today we learn how indicators work and how to choose one, and then we solve this puzzle. This comes up often in JEE practical chemistry, so let's begin.

0:36What an indicator really is

So, what is an indicator? It is a weak organic acid, or a weak organic base, whose ionised and un-ionised forms have different colours. Two forms, two colours, that is the whole idea. Okay? For an acid indicator we write H I n, where the I n stands for indicator. It ionises into H plus and I n minus, with the charges written up top. The two half arrows, pointing both ways, mean this is an equilibrium. Right? Now, phenolphthalein is one such weak acid, H P h, and as the question on screen says, its anion is pink. So tell me, in an acid, which form will we see? Yes, we see the un-ionised form. The acid supplies plenty of H plus, a common ion, which suppresses the ionisation. So only colourless H P h is present; the pink anion, P h minus, appears in alkali. Clear? And the colour changes over a p H of about p K I n, plus or minus one. That is capital K with a small I n below it; the lowercase p in front means minus log, and the wavy equals sign means approximately.

1:50The two indicators you must know

Got it? Now, two indicators to know by heart. Methyl orange is red in acid and yellow in alkali; the arrow means changes to. Its range is p H three point one to four point four. On the graph, ignore the curves for now: this range is the lower shaded band. With me? Phenolphthalein is colourless in acid and pink in alkali, with a range of p H eight point three to ten point zero, the upper band. So methyl orange changes on the acid side of seven, and phenolphthalein on the alkaline side. Okay?

2:24Choose by the steep part of the curve

So which one do we use? The graph shows p H against the volume of zero point one molar sodium hydroxide added. The solid curve, labelled H C l, is for twenty five millilitres of zero point one molar hydrochloric acid; the dotted line labelled equivalence marks the equivalence point. The black dot on the dashed curve, labelled p H equals p K a, we will not need today. There the curve shoots straight up, and the indicator must change colour within this steep part. Right? Now look at the jump of the solid curve. Which shaded band does it cross? Yes, exactly, it crosses both bands. For a strong acid against a strong base, the jump runs from about three to about eleven. So either indicator works here. Clear? Now the dashed curve, labelled C H three COOH, which is acetic acid, a weak acid, with the same strong base. The salt formed is hydrolysed, so equivalence lies above seven; that sign on screen is greater than. Its steep part sits only in the upper band, so we use phenolphthalein. Got it? Now here is a classic trap. Suppose a student titrates acetic acid with sodium hydroxide, but uses methyl orange by mistake. Will the titre, the volume of alkali used, come out too high or too low? If you said too low, you are absolutely right. Follow the dashed curve: it crosses the lower band early. Methyl orange turns yellow near p H four point four, when only part of the acid has been neutralised, so the student stops too soon. Don't fall for this one, okay?

4:04The other two pairs, and a topper's trick

Now flip it: a strong acid against a weak base, like hydrochloric acid with ammonia. The salt is hydrolysed again, but now equivalence lies below seven; this sign is less than. So we need methyl orange, which changes colour on the acid side. Right? And now both weak: acetic acid against ammonia. Which indicator would you pick? That was a trick question, because no indicator is suitable. With a weak acid and a weak base, the p H changes only gradually near equivalence, with no sharp jump, so no indicator changes colour sharply with one drop. With me? So here is a topper's trick: ask who the weak partner is. With a weak acid, equivalence is above seven, so phenolphthalein; with a weak base, it is below seven, so methyl orange. If both are strong, either works, and if both are weak, none does. Okay?

5:00Example 1: indicator for oxalic acid vs NaOH, colour change

Alright, example one. Which indicator is used for titrating oxalic acid against sodium hydroxide, and what is the colour change when the alkali is in the burette? Pause the video, and try it first. Done? Okay, let's check. Oxalic acid is a weak acid, and sodium hydroxide, N a O H on screen, is a strong base. The weak partner is the acid, so the double arrow, meaning therefore, leads to phenolphthalein. Right? Now the colour change. The acid is in the flask, and the alkali comes from the burette. So, colourless to pink, or pink to colourless? In the flask we have acid, and phenolphthalein is colourless in acid. At the end point, the first extra drop of alkali makes the solution just alkaline, and it turns pale pink. Got it? So the answer is phenolphthalein, colourless to pale pink. Always check which liquid is in the burette.

6:02Example 2: pH at 24.9 mL and 25.1 mL of NaOH added

Okay? Now the harder one, which solves our puzzle. Twenty five point zero millilitres of zero point one molar hydrochloric acid is titrated with zero point one molar sodium hydroxide; capital M means molar. Find the p H when twenty four point nine, and when twenty five point one millilitres of alkali have been added. Pause the video and try it first, and count millimoles. Done? Okay, let's check. Millilitres times molarity gives millimoles, the m mol on screen. Acid taken: twenty five point zero times zero point one, which is two point five millimoles. At twenty four point nine millilitres, the base added is two point four nine millimoles. With me? So the acid left over is zero point zero one millimoles, in about fifty millilitres. Zero point zero one over fifty gives two times ten to the power minus four molar; the square brackets around H plus mean its concentration. Okay? Now p H is minus log of that, which is four minus log two. Log two is about zero point three, so the p H is about three point seven, still acidic. Right? Now add just zero point two millilitres more, up to twenty five point one. Where do you think the p H will go? Now the alkali is in excess, again by zero point zero one millimoles, in about fifty millilitres. So the hydroxide ion concentration, O H minus in square brackets, is two times ten to the power minus four molar. Clear? So p O H is about three point seven. Since p H plus p O H is fourteen, the p H is fourteen minus three point seven, which gives ten point three. Got it? And our puzzle is solved. Only zero point two millilitres of alkali, just a few drops, carry the p H from three point seven to ten point three, across the ranges of both indicators. So either one gives the correct end point, even though neither changes at seven. Beautiful, isn't it?

8:06Quick revision: pKIn plus minus 1, choosing the indicator

So, quick revision. An indicator, H I n, changes colour over a p H of about p K I n, plus or minus one. Choose one that changes colour on the steep part of the p H curve. Weak acid, phenolphthalein; weak base, methyl orange; both weak, no indicator. Now try the practice problems on this page. All the best!