Ions in Solution and Reactions in Solution

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. I take fifty millilitres of zero point two molar calcium chloride, and fifty millilitres of zero point one molar aluminium chloride, and I mix them. Now tell me, what is the concentration of chloride ions in the mixture? Somewhere between zero point one and zero point two, right? Hmm. The correct answer is zero point three five molar. More than both bottles! Wait, how is that possible? Today we will find out. And look, JEE loves this: mix two solutions, or let them react, and find a concentration. So, let's begin.

0:41One formula unit, many ions

So, first, just one salt. Take a strong electrolyte, written here as A sub x, B sub y, okay? Strong means it breaks up completely into its ions in water. And its molarity is c, clear? Now think. One formula unit breaks up. How many ions of the first kind come out? Yes, x of them! And y ions of the second kind. The little subscripts tell you everything. So c moles per litre of salt give x times c moles per litre of that ion. The square brackets on screen mean concentration, in moles per litre, okay? So, the concentration of A equals x times c, and the concentration of B equals y times c, okay? As simple as that. Alright, a quick test. Calcium chloride solution, of molarity c, okay? What is the chloride ion concentration? Just c, same as the salt, right? No, no, careful! This is the JEE trap. Look at the formula on screen: one calcium, and a little two after the chlorine. So every formula unit releases two chloride ions, written C l minus on screen, and the chloride concentration is two times c, okay? And the calcium ion concentration? Just c, got it? And aluminium chloride, with a little three after the chlorine? Three chloride ions per unit, so three times c, right? That's it, the subscript is your multiplier, okay?

2:18Mixing: add millimoles, divide by the total volume

So, now we mix two solutions. First, recall from the molarity lesson: millimoles equal molarity times volume in millilitres. The capital M on screen is the molarity, clear? For one ion, multiply once more, by the number of those ions in one formula unit. That is the subscript we just counted, remember? Now think. Two salts can give the same ion, like chloride from two bottles. Once they are mixed, can a chloride ion remember which bottle it came from? Not at all! Chloride is chloride. So we track the ion, not the salt. So, add up the millimoles of that ion from every solution. One total for the whole mixture. With me? Then divide by the total volume, all the millilitres together, not the volume of one bottle, okay? And a small topper's trick. Millimoles divided by millilitres gives moles per litre directly, so you never need to convert to litres. Very simple.

3:18Reactions in solution: who runs out first?

Now the last idea: what if the two solutions react? Step one, same as before. Convert every reagent to millimoles. Molarities alone cannot be compared, because the volumes may be different, right? Step two: write the balanced equation. It tells you the ratio in which they react. Clear? Step three, find the limiting reagent, the one that runs out first. Easy, isn't it? The one with fewer millimoles runs out first, right? Hmm, not so fast! Suppose one mole of the first reagent needs two moles of the second. Then the second gets used up twice as fast. So divide each amount by its coefficient, and the smaller result is the limiting reagent. Look, this is where many students lose marks. Step four: the limiting reagent is gone. For whatever is left, excess reagent or product, divide its amount by the final volume. Final volume means both solutions added together, okay?

4:21Example 1: chloride concentration on mixing CaCl2 and AlCl3

Now, remember our puzzle from the start? Let's finish it. Fifty millilitres of zero point two molar calcium chloride is mixed with fifty millilitres of zero point one molar aluminium chloride. Find the chloride ion concentration in the mixture. Pause the video, and try it first. Done? Okay, let's check. Equal volumes, so just take the average of zero point two and zero point one: zero point one five molar, right? No, no! That treats each bottle as if every unit gave only one chloride ion. Remember the subscripts! So, chloride from calcium chloride first. Here I will work in litres: fifty millilitres is zero point zero five litres. Zero point zero five, times zero point two, times two chlorides per unit: zero point zero two moles. Now from aluminium chloride. Zero point zero five, times zero point one, times three chlorides per unit: zero point zero one five moles. Add them, because chloride is chloride: zero point zero two plus zero point zero one five, that is zero point zero three five moles. And the total volume? Fifty plus fifty millilitres, so zero point one litres. So the chloride concentration is zero point zero three five divided by zero point one: zero point three five molar. Mystery solved! In chloride, the first bottle was really two times zero point two, that is zero point four molar, and the second was three times zero point one, zero point three molar. So zero point three five sits right between them, doesn't it? Makes perfect sense.

6:09Example 2: excess H2SO4 and Na2SO4 after adding NaOH

Now a harder one, JEE level. Read it with me. One hundred millilitres of zero point two molar sulphuric acid is mixed with one hundred millilitres of zero point three molar sodium hydroxide. Find the molarity of the excess reagent, and of sodium sulphate, in the final solution. Pause and try. Hint: write the balanced equation first. Done? Okay, let's check. First, the balanced equation. One sulphuric acid reacts with two sodium hydroxide, and gives one sodium sulphate and two waters. Note this two; it decides everything. Now the amounts. One hundred millilitres is zero point one litres. Acid: zero point one times zero point two, zero point zero two moles. Base: zero point one times zero point three, zero point zero three moles. Alright, so the acid has fewer moles, zero point zero two against zero point zero three. So the acid runs out first, right? No, no, careful! This is the trap we just saw. Divide by the coefficients. Base: zero point zero three divided by two is zero point zero one five. Acid: zero point zero two divided by one, still zero point zero two. The little less than sign on screen says it: zero point zero one five is smaller, so the base is the limiting reagent. From the equation, zero point zero three moles of base make half as much salt: zero point zero one five moles of sodium sulphate. And it uses up zero point zero one five moles of acid. So the acid left is zero point zero two minus zero point zero one five: zero point zero zero five moles. Got it? Now the final volume. One hundred plus one hundred millilitres: two hundred millilitres, which is zero point two litres. Not zero point one, okay? Now, just divide. Sulphuric acid: zero point zero zero five over zero point two, that is zero point zero two five molar. Sodium sulphate: zero point zero one five over zero point two, zero point zero seven five molar. See? Four steps, and even a reaction problem becomes easy.

8:35Quick revision: ion molarity, mixing and reactions

So, quick revision. For ions, each concentration is the subscript times c, so count the subscripts. For mixing, add the millimoles of the ion from every solution, and divide by the total volume. And for a reaction: millimoles, the limiting reagent using the coefficients, and what is left over the final volume. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!