Past JEE Questions: Titrations

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Some sodium hydroxide mixed with sodium carbonate is titrated with hydrochloric acid. The first end point needs seventeen point five millilitres of acid, and the second needs only one point five millilitres more. So which number tells us the carbonate, the big one or the small one? We will settle it today, with two past JEE Main questions on titrations, a kind that comes up often. So, let's begin.

0:29Every titration question: count millimoles

So, here is the first tool: molarity, in moles per litre, times the volume in millilitres, the m L on screen, gives millimoles; the cross means times. Okay? Now, suppose capital A and capital B react in the ratio small a to small b. If we know the millimoles of A, how do we get those of B? We divide each side by its own coefficient; the arrow in the equation means gives. So molarity times volume of A, over small a, equals molarity times volume of B, over small b, and the little letters below M and V name the solution. Right? Normality, capital N, is molarity times the n factor, which for an acid is the number of H plus ions one molecule gives. Then normality times volume is the same on both sides. With me? And millimoles times molar mass gives the mass in milligrams, the m g on screen, which we need for percentages. Clear?

1:29Choosing the indicator

Next, the indicator. This figure shows the p H as zero point one molar sodium hydroxide is added: the solid curve for hydrochloric acid, the dashed one for acetic acid. The dotted line at twenty five millilitres marks the equivalence point, and the black dot, where p H equals p K a, we will not need today. The shaded bands are where the indicators change colour. Okay? Methyl orange is red in acid and yellow in alkali, and it changes in the lower band, from three point one to four point four. Right? Phenolphthalein is colourless in acid and pink in alkali, and it changes in the upper band, from eight point three to ten. Got it? Now, the solid curve shoots straight up near twenty five millilitres, and the indicator must change colour on this steep part. So which band does this jump cross? Yes, exactly, it crosses both, from about three to about eleven. So for a strong acid with a strong base, either indicator works. Clear? For acetic acid, a weak acid, the equivalence point lies above seven, and its steep part, from about seven to about eleven, crosses only the upper band. So a weak acid with a strong base needs phenolphthalein. With me? And for a strong acid with a weak base, the equivalence point lies below seven, so we use methyl orange. Okay?

2:54Carbonate: two end points

Now the idea behind our puzzle: sodium carbonate takes acid in two stages. In stage one, carbonate, C O three with a two minus charge up top, takes one H plus and gives bicarbonate, HCO three minus. Phenolphthalein turns colourless at the end of this stage. With me? In stage two, bicarbonate takes a second H plus and gives water and carbon dioxide, and methyl orange changes only here. Right? So V one, a capital V with a small one below, is the volume to the phenolphthalein end point, and V two is the further volume, from the first end point to the second. Clear? Now take sodium hydroxide, N a O H, with sodium carbonate. In stage two, which ion is still taking acid? Only the bicarbonate made from the carbonate, because the hydroxide was used up in stage one. So carbonate corresponds to V two; the three-line equals sign means corresponds to. And hydroxide corresponds to V one minus V two. Got it?

3:58Example 1: molarity of H2SO4 from a Ca(OH)2 titration

Alright, example one, from JEE Main twenty twenty three. Twenty millilitres of calcium hydroxide solution is consumed by ten millilitres of sulphuric acid, H two S O four. Also, twenty millilitres of zero point five molar hydrochloric acid with phenolphthalein, titrated with the same calcium hydroxide, turns pink when the burette reading changes from twenty five point five to thirty five point five millilitres. Find the molarity of the sulphuric acid. Pause the video and try it first. Done? Okay, let's check. The burette reads thirty five point five at the end, so thirty five point five millilitres of calcium hydroxide was run in, right? That is the trap, so be careful here: thirty five point five is only the final reading. The calcium hydroxide run in, written with O H in brackets and a small two, is the change: thirty five point five minus twenty five point five, which gives ten point zero millilitres. Okay? The flask holds twenty times zero point five, which gives ten millimoles of hydrochloric acid, the m mol on screen. So how much calcium hydroxide does that need? Each unit has two O H groups, so it takes two H plus, and the ratio is one to two; the colon means to. So it needs five millimoles, in ten point zero millilitres, and five over ten gives zero point five molar, the capital M on screen. With me? Now the first sentence: twenty millilitres of it holds twenty times zero point five, which is ten millimoles. Sulphuric acid gives two H plus and calcium hydroxide takes two, so they react one to one. Right? So ten millimoles of sulphuric acid sit in ten millilitres, which gives one molar. Read the burette as a change, and count the H plus ions. Got it?

5:56Example 2: percent Na2CO3 in NaOH mixture, double indicator

Now a harder one, from JEE Main twenty twenty one: our opening puzzle. Zero point four grams of a mixture of sodium hydroxide, sodium carbonate and inert impurities is titrated with N over ten hydrochloric acid using phenolphthalein, needing seventeen point five millilitres. Then, with methyl orange, it needs a further one point five millilitres. Find the percentage by mass of sodium carbonate, to the nearest integer. Pause the video and try it first. Here is a hint: start with the second end point. Done? Okay, let's check. Capital N over ten means a normality of zero point one, and hydrochloric acid gives one H plus, so its n factor is one and its molarity is zero point one. Okay? Now, our puzzle. Which volume measures the carbonate, seventeen point five or one point five? Yes, exactly, it is the small one. V one is seventeen point five millilitres, V two, the further volume, is one point five, and carbonate corresponds to V two. Right? In stage two, each bicarbonate ion takes one H plus and gives carbon dioxide. One point five times zero point one gives zero point one five millimoles of acid. With me? Now, carbonate takes two H plus in all, one in each stage. So should we double this, to zero point three millimoles of carbonate? That is the trap, so be careful here: doubling counts the acid for both stages, not the carbonate itself. Each bicarbonate in stage two came from one carbonate ion, so there are zero point one five millimoles of sodium carbonate. Multiplying by its molar mass, one hundred and six, gives fifteen point nine milligrams. Clear? Zero point four grams is four hundred milligrams. Fifteen point nine over four hundred, times one hundred, gives three point nine seven five; the wavy equals sign means approximately, so it is four percent to the nearest integer. So the small volume was the carbonate, only about four percent of the sample. Got it?

8:09Quick revision: millimoles, indicator choice, V1 and V2

So, quick revision. Millimoles are molarity times volume, and each side is divided by its own coefficient. Weak acid with strong base needs phenolphthalein; strong acid with weak base, methyl orange. For sodium hydroxide with carbonate, carbonate corresponds to V two, and hydroxide to V one minus V two. Now go and try the practice problems on this page. All the best!