Equivalent Mass and the n-Factor

Transcript

0:00Introduction

Hello everyone! So, one small puzzle before we start. Take sodium carbonate, and titrate it with hydrochloric acid. One student finds its equivalent mass is one hundred and six. Another student, same sodium carbonate, same acid, gets fifty three. Same compound, two different equivalent masses? Hmm, one of them must be wrong, right? Or maybe, both are right! By the end of this video, you will know exactly how. And look, this is not a small topic. In JEE, almost every normality and titration question stands on this one idea: equivalent mass and the n factor. So, let's begin.

0:45Equivalent mass: molar mass by the n-factor

So, what is equivalent mass? Take the molar mass, capital M, and divide it by a number called the n factor. E equals M, divided by the n factor, okay? Write this down. This one line is the heart of today's lesson. Now think. The molar mass of a compound is fixed, right? So if the equivalent mass of the same compound changes, what must be changing? Yes, exactly! The n factor. So the whole game today is one thing: find the n factor correctly. Clear? Once you have E, the number of equivalents is easy. Number of equivalents equals small m, divided by E, okay? Or, the same thing in another form: moles, times the n factor. Here small m is the mass you take, and small n is the number of moles, okay? So one mole of something with n factor two gives two equivalents. See, this line will power every titration question in the coming lessons.

1:48Acids and bases: count what is replaced

Now, let's find the n factor, one type at a time. First, acids. For an acid, the n factor is the number of replaceable H plus ions per molecule, actually used in the reaction. This is called the basicity of the acid, okay? So HCL gives one H plus, and its n factor is one. Sulphuric acid, H two S O four, gives two, so its n factor is two. And H three P O four, when it is fully neutralised, gives three. Simple counting, right? Now bases. Same idea, but we count the replaceable O H minus ions. This is the acidity of the base. Sodium hydroxide has one, so n factor one. Calcium hydroxide has two, so n factor two. And look, acids have basicity, bases have acidity. The names are crossed, so don't mix them up!

2:39JEE trap: which H atoms really count?

Alright, now a question for you. Look at H three P O three. Three hydrogen atoms in the formula. So the n factor is three, right? No, no, careful! This is the JEE trap. In H three P O three, one hydrogen is bonded directly to the phosphorus atom. That one is not replaceable. So the acid is dibasic, and the n factor is two, not three. Next, H three P O two. This one is monobasic, so its n factor is just one, okay? And boric acid, H three B O three. It also shows three hydrogens, but it is a monobasic Lewis acid, so the n factor is one. So: two, one, one. Never three! Remember these, they come in JEE again and again. Now one more twist. For H three P O four, we said three. But is it always three? Not always! When H three P O four is converted into sodium dihydrogen phosphate, N a H two P O four, only one H plus is replaced. So here the n factor is one. See, the n factor depends on the reaction, not just the formula. So here is a topper's habit: look at the product first, then count. Got it?

3:58Salts and redox: total charge, or electrons

Next, salts in a non redox reaction, where no oxidation state changes. Here the n factor is the total positive charge in one formula unit, or the total negative charge. Both give the same number. Sodium carbonate has two sodium ions, each plus one. Total positive charge two, so n factor two, okay? Aluminium sulphate has two aluminium ions, each plus three. Two times three: n factor six. Check from the negative side: three sulphate ions, each minus two, again six. Exactly the same. And in a redox reaction, the n factor is the number of electrons gained or lost per formula unit, okay? Let's try one. Potassium dichromate, K two C r two O seven, going to chromium three plus. Each chromium goes from plus six to plus three: three electrons gained. So the n factor is three, right? No, no, careful! Count the chromium atoms in one formula unit. Two, isn't it? So the n factor is two times three: six. Not three. Six! Students lose easy marks here, so always multiply by the number of atoms that change.

5:12Equivalent mass of an element

One more case: the equivalent mass of an element. It is the mass of the element that combines with, or displaces, one point zero zero eight grams of hydrogen, or eight grams of oxygen, or thirty five point five grams of chlorine. For hydrogen gas, you can use the volume too: eleven point two litres of H two at STP, okay? And the link with atomic mass: atomic mass equals equivalent mass times valency. So for a divalent metal, the atomic mass is twice the equivalent mass. Really simple.

5:45Example 1: equivalent mass of H2SO4, Ca(OH)2, Na2CO3

Right, on to Example one. Find the equivalent masses of sulphuric acid, calcium hydroxide and sodium carbonate, in non redox reactions. Pause the video, and try it first. Done? Okay, let's check. Sulphuric acid has two replaceable H plus, so n factor two. Molar mass ninety eight, divided by two: forty nine. Calcium hydroxide has two O H minus, so n factor two again. Seventy four by two: thirty seven. Now sodium carbonate. Only one carbonate ion in it, so the n factor is one, right? No, wait! We don't just count the ions. For a salt in a non redox reaction, which rule do we use? Total positive charge, yes! Two sodium ions, each plus one, so n factor two. One hundred and six by two: fifty three. Notice that, fifty three. That is one of the numbers from our puzzle, remember? Keep it in mind.

6:49Example 2: Na2CO3 to phenolphthalein and methyl orange

Now a harder one, JEE style. And this is our puzzle from the start! Sodium carbonate is titrated with hydrochloric acid. In part A, only up to sodium hydrogen carbonate: that is the phenolphthalein end point. In part B, all the way to carbon dioxide: the methyl orange end point. Find the equivalent mass in each case. Pause and try. Hint: look at the product. Done? Okay, let's check. Part A, first. The carbonate ion, C O three two minus, takes one H plus and becomes hydrogen carbonate, HCO three minus. One H plus per formula unit, so n factor one. E equals one hundred and six by one: one hundred and six. The full molar mass! Now part B, okay? Same sodium carbonate, same acid. But now we go all the way to carbon dioxide. So what changes? Yes! Now each carbonate takes two H plus, and gives carbon dioxide and water. So n factor two, and E equals one hundred and six by two: fifty three. And there it is, puzzle solved! One hundred and six, and fifty three: both students were right. Same compound, but a different reaction, so a different n factor. So in a titration, always check which end point the question is talking about.

8:13Quick revision: E = M / n-factor and how to find n

So, quick revision. Equivalent mass is molar mass divided by the n factor, and the number of equivalents is mass by equivalent mass, or moles times the n factor. The n factor is the H plus used for an acid, the O H minus for a base, the total charge for a salt, and the electrons for a redox reaction. And always remember: the reaction decides the n factor, so look at the product first. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!