Mixed Problems: Set 3 (Concentration, Equivalents and Titrations)

Transcript

0:00Introduction

Hello everyone! Now, one small puzzle before we start. You mix two hundred and fifty millilitres of zero point two molar sodium hydroxide with seven hundred and fifty millilitres of zero point four molar sodium hydroxide. Same solute, two strengths. Zero point two and zero point four: the average is zero point three. So the mixture is zero point three molar, right? Ah, that is exactly what most students write, and it is wrong! Today, in our mixed problem class, Set three, we will find the right answer. Concentration, equivalents and titrations: these are JEE Main favourites, and three simple tools solve almost all of them. So, let's begin.

0:45Tool 1: count millimoles

So, tool number one, and it is the most useful tool in this whole chapter: count millimoles. Molarity, capital M, into the volume in millilitres, gives the millimoles directly. The m m o l you see on screen is just short for millimoles, okay? No litres, no conversions. Clear? Now dilution. When you add water, the volume changes, and the molarity changes. But tell me, what stays the same? The millimoles of solute, yes! Nobody added or removed any solute, right? So M one V one equals M two V two. Quick one. What volume of six molar sulphuric acid do we need to make five hundred millilitres of zero point three molar acid? Try it with M one V one equals M two V two. Six into V equals zero point three into five hundred, that is one hundred and fifty. So V is twenty five millilitres. The double arrow you see on screen just means, so, okay? That's it, one line. Alright, now ions. Mix one hundred millilitres of zero point one molar sodium chloride with one hundred millilitres of zero point one molar magnesium chloride. What is the chloride ion concentration? Both are zero point one molar, so the chloride is zero point one molar, right? No, no, careful! Count the chloride ions in magnesium chloride. Magnesium chloride gives two chloride ions per formula unit, you see? So it gives zero point one, into one hundred, into two: twenty millimoles of chloride. Second mistake: the volume. After mixing, it is two hundred millilitres, not one hundred. So the chloride is ten plus twenty, thirty millimoles, in two hundred millilitres: zero point one five molar. Count the millimoles of each ion, add them, and divide by the total volume. Got it? Sodium chloride gives just ten millimoles. Now don't fall into the second trap: zero point one plus zero point two, zero point three molar? No! After mixing, the volume is two hundred millilitres, not one hundred. So the chloride is ten plus twenty, thirty millimoles, in two hundred millilitres: zero point one five molar. Count the millimoles of each ion, add them, and divide by the total volume. With me?

3:12Tool 2: mass percent and density to molarity

Tool number two. Here is a twenty percent sodium hydroxide solution, by mass, with density one point two grams per millilitre. The w by w on screen means weight by weight, and small d is the density, okay? Find its molarity. Many students do this. Twenty percent means twenty grams in every hundred millilitres. So two hundred grams per litre, divided by forty, five molar, right? No, no, careful! Twenty percent by mass means twenty grams in a hundred grams of solution, not in a hundred millilitres. So here is the trick: take a convenient amount. For molarity, take one litre of solution. One thousand millilitres into one point two: one thousand two hundred grams of solution. That is exactly what the density is for. Twenty percent of one thousand two hundred grams is two hundred and forty grams of sodium hydroxide. Divide by forty, the molar mass of sodium hydroxide: six moles, in one litre. So, six molar. Now the topper's shortcut. Molarity equals ten, into the mass percent, into the density, divided by the molar mass of the solute, written M sub B on screen. Ten into twenty into one point two, by forty: six again. Exactly the same. Try one in your head. Concentrated nitric acid is sixty three percent by mass, with density one point four. Got it? Ten into sixty three into one point four, by sixty three, the molar mass of nitric acid. The sixty threes cancel! Fourteen molar. Perfect!

4:52Tool 3: equivalents and normality

Tool number three: equivalents. Normality, capital N, equals molarity into the n-factor. And milliequivalents, the m e q on screen, equal normality into the volume in millilitres. Same idea as millimoles, right? Now find the normality of zero point one five molar aluminium sulphate. There is no redox here, so the n-factor is the total charge on the cations. Aluminium is plus three, so the n-factor is three, right? No, no, careful! There are two aluminium ions in one formula unit, see? Two into three: the total charge is six, so the n-factor is six. Normality is zero point one five into six: zero point nine normal. And why do we love equivalents? Because substances always react in equal numbers of equivalents. So in a titration, N one V one equals N two V two. Clear? Last, back titration. When a solid will not titrate directly, we add a measured excess of acid, and titrate what is left. So the milliequivalents of the sample equal the milliequivalents added, minus the milliequivalents left. Like paying with a big note and getting change back.

6:07Example 1: molarity on mixing 0.2 M and 0.4 M NaOH

Now, remember our opening puzzle? Let's finish it. Example one. Two hundred and fifty millilitres of zero point two molar sodium hydroxide is mixed with seven hundred and fifty millilitres of zero point four molar sodium hydroxide. Find the molarity of the mixture, as x into ten to the power minus two molar. Pause the video, and try it first. Done? Okay, let's check. Tool one: count millimoles. Two fifty into zero point two: fifty millimoles. Seven fifty into zero point four: three hundred millimoles. Total: three hundred and fifty millimoles, in one thousand millilitres of solution. So M mix, the molarity of the mixture, is three fifty by one thousand: zero point three five molar, and x is thirty five. Not zero point three! There you go, puzzle solved: three times as much of the stronger solution went in, so the answer leans towards zero point four. The simple average works only when the volumes are equal. Got it?

7:15Example 2: purity of MgO by back titration

Now a harder one, JEE level. Zero point three zero grams of impure magnesium oxide is dissolved in fifty millilitres of zero point five normal hydrochloric acid. The excess acid needs twenty five millilitres of zero point five normal sodium hydroxide. Find the percentage purity of the magnesium oxide, to the nearest integer. Pause and try. Hint: added minus left. Done? Okay, let's check. First, the acid added: zero point five into fifty, twenty five milliequivalents. That is our big note. Now the change. The leftover acid needed twenty five millilitres of zero point five normal base: twelve point five milliequivalents. Equal milliequivalents react, so twelve point five milliequivalents of acid were left. So the magnesium oxide used twenty five minus twelve point five: twelve point five milliequivalents of acid. With me? Now convert this to mass. Many students multiply by forty, the molar mass of magnesium oxide, right? No, no, careful! That gives five hundred milligrams, more than the whole sample of three hundred milligrams. That is impossible! Milliequivalents always go with the equivalent mass, the capital E on screen. Here E is twenty: forty, divided by an n-factor of two. Twelve point five into twenty: two hundred and fifty milligrams of pure magnesium oxide. Two hundred and fifty milligrams is zero point two five grams. Purity: zero point two five, out of zero point three zero, into hundred: eighty three point three percent. So the answer is eighty three. See how all three tools worked together?

9:02Quick revision: millimoles, density shortcut, normality

So, quick revision. One: count the millimoles, add them, and divide by the total volume. Never average molarities blindly. Two: from percent and density, take one litre of solution, or use ten into percent into density, by molar mass. Three: normality is molarity into the n-factor, equal milliequivalents react, and back titration is added minus left. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!