Dilution and Mixing

Transcript

0:00Introduction

Hello everyone! First, one small puzzle before we start. In the lab, you have two bottles of hydrochloric acid. One is ten molar, and the other is two molar, okay? Your teacher says: make exactly one litre of five molar acid. Hmm. Take half a litre from each bottle? Sounds fair, right? But will that really give five molar? By the end of this video, you will find the exact answer in under a minute. And you know, dilution and mixing sit inside so many JEE questions on solutions. Once you see the one idea behind them, these are sure marks. So, let's begin.

0:39Dilution: the solute does not change

So, first, dilution. Diluting simply means adding water to a solution, okay? Think of a glass of lemonade that is too strong. What do you do? You pour in more water. Now tell me, what changes, and what does not change? Yes, exactly! The volume goes up, but the amount of solute stays the same. Nobody added or removed any solute, did they? We only added water. So the moles of solute are unchanged on dilution. That is the whole idea. Now, molarity times volume in litres gives you the moles of solute, okay? Before dilution, the moles are M one V one. After dilution, they are M two V two. And the moles did not change, so M one V one equals M two V two: moles before equal moles after, as you see on screen. Any volume unit works here, as long as both sides use the same one. Write this down. Clear? Quick check. If you double the volume by adding water, what happens to the molarity? It becomes half, exactly! Molarity times volume must stay the same, so if the volume goes up two times, the molarity comes down two times. But look, here is the JEE trap. V two is the final volume of the solution, not the volume of water you add, okay? Questions love to ask: how much water must be added? Then you must subtract: final volume minus starting volume. We will catch this trap in the first example. And the same formula works the other way too. If you evaporate water, the volume falls, the moles of solute still stay the same, so the molarity rises. Same equation, M one V one equals M two V two. As simple as that.

2:28Mixing two solutions of the same solute

Alright, now mixing. You take two solutions of the same solute. One has molarity M one and volume V one, the other has M two and V two, and you pour them together. What adds up when we mix them? The moles of solute, yes! The first solution brings M one V one moles, and the second brings M two V two moles. So the total moles are M one V one plus M two V two. And the volumes also add. In these questions, we take the volumes as additive, okay? So the total volume is V one plus V two. Now, molarity is just moles divided by volume. So M mix equals M one V one plus M two V two, divided by V one plus V two. Total moles on top, total volume at the bottom. Got it? Now a quick question. Many students just add the two molarities and divide by two. Is that right? Only when the two volumes are equal! Otherwise, no. The solution with the bigger volume brings more moles, so it pulls the answer towards its own molarity. So always use the full formula. No shortcuts here, clear?

3:40Mixing by mass: the same idea

Alright, but what if the concentrations are given as mass percent, w by w, instead of molarity? Will the same thinking work? Yes, of course! Look at the formula on screen. Here, m i is the mass of each solution, and w i percent is its mass percent. The big sigma sign just means: add over all the solutions, okay? On top, each mass times its mass percent: this counts the solute each solution brings. At the bottom, the total mass of solution. That gives the w by w percent of the mixture. So again a weighted average, but weighted by mass this time. Got it? Now let me share a trick that toppers use. Look at all three ideas: dilution, mixing by molarity, and mixing by mass. What are they all really doing? Counting the solute, and dividing by the total, yes! So in any dilution or mixing question, don't hunt for a formula. Just ask: how much solute in total, and how much solution in total? Total solute divided by total solution, the line on screen. That's it, the answer follows.

4:51Example 1: water to add to take 0.5 M HCl to 0.2 M

Let's take our first example. How much water must be added to two hundred millilitres of zero point five molar HCL, to make it zero point two molar? Pause the video, and try it first. Done? Okay, let's check. Moles before equal moles after. M one is zero point five, V one is two hundred millilitres, and M two is zero point two. So, zero point five times two hundred equals zero point two times V two. Notice, I kept both volumes in millilitres, the same unit on both sides. Zero point five times two hundred is one hundred. Divide by zero point two, and V two is five hundred millilitres. So the answer is five hundred millilitres, right? No, no, careful! Read the question again. It asks how much water must be added. Five hundred is the final volume, V two. Remember the trap? We already had two hundred millilitres in the flask. So the water to add is five hundred minus two hundred: three hundred millilitres. Honestly, that one subtraction decides your marks. Clear?

6:03Example 2: mixing 10 M and 2 M HCl to make 5 M

Now the harder one, and look, it is our puzzle from the start. What volumes of ten molar and two molar HCL must be mixed, to make one litre of five molar acid? Pause the video and try it. Hint: call one of the volumes x, okay? Done? Okay, let's check. First, our guess from the start: half a litre of each bottle. Will that give five molar? Count the moles. Half a litre of ten molar gives five moles. Half a litre of two molar gives one mole. Six moles in one litre: six molar, not five! Too concentrated. So we need less of the ten molar acid, and more of the two molar one, right? So, let x litres be the ten molar acid. The total must be one litre, so the two molar acid is one minus x litres, okay? Now, the moles of solute add up. Ten x from the first, plus two times one minus x from the second. That must equal five molar times one litre, which is five moles. Open the bracket: ten x, plus two, minus two x, equals five. So eight x equals three, and x is zero point three seven five litres. So, three hundred seventy five millilitres of the ten molar acid, and the rest of the litre, six hundred twenty five millilitres, of the two molar acid. Puzzle solved! And toppers always check. Ten times zero point three seven five is three point seven five moles. Two times zero point six two five is one point two five moles. Total, five moles in one litre: five molar, absolutely correct. Got it?

7:47Quick revision: M1V1 = M2V2 and molarity of a mixture

So, quick revision. Dilution: M one V one equals M two V two, and V two is the final volume, not the water added. Mixing: total moles by total volume, not a simple average, unless the volumes are equal. And in every question, just count the total solute, and divide by the total solution. Now go and try the three practice problems on this page. You will find them easy, I promise. All the best!