Neutralisation and Ionisation

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Neutralise hydrochloric acid with sodium hydroxide. Now use sulphuric acid instead. Different acid, right? But for every mole of water formed, the heat that comes out is exactly the same: fifty seven point one kilojoules. Now try acetic acid, and less heat comes out. Try hydrofluoric acid, and more heat comes out! So why is the number fixed for some acids, and not for others? Interesting, isn't it? Today we crack this, and learn to find the heat used to ionise a weak acid. Questions on this come up often in JEE, so let's begin.

0:43Enthalpy of neutralisation

So, first, the definition. When an acid is neutralised by a base, heat comes out. But per mole of what? Per mole of acid, or per mole of water formed? Per mole of water, yes! The enthalpy of neutralisation is the enthalpy change when one mole of water is formed, in dilute solution. On screen, the little triangle is the capital Greek letter delta, meaning change in, and H is enthalpy. The small word neut, written low on the right, stands for neutralisation. Okay? Now, back to our puzzle. Any strong acid, with any strong base. Why should the heat be the same every time? Because in dilute solution, a strong acid and a strong base are fully ionised. So whatever the acid and the base, the only real change is this one: H plus, aqueous, plus O H minus, aqueous, gives H two O, liquid. The little plus and minus up top are the charges on the ions, aq in brackets means aqueous, dissolved in water, l means liquid, and the arrow means gives. Same reaction, same heat. Clear? And that heat is fixed: delta H neut equals minus fifty seven point one kilojoules per mole. The k J mol to the power minus one means kilojoules per mole. In kilocalories, the k cal on screen, it is minus thirteen point seven. Minus sign, because heat is given out: neutralisation is exothermic. Got it?

2:17Weak acid or weak base: less heat

Now, acetic acid is a weak acid, only partly ionised in water. So, when we neutralise it with sodium hydroxide, will the heat be more than fifty seven point one, or less? Less, exactly! Before H plus can meet O H minus, the weak acid must first be ionised, and that takes in some heat. So with a weak acid, or a weak base, some heat is used to ionise it. Okay? So less heat comes out than fifty seven point one kilojoules per mole of water. Part of it was spent on ionisation, right? So, a neat formula. Delta ion H, the enthalpy of ionisation, with the small word ion written low after the delta, equals delta H neut for the weak acid, minus, minus fifty seven point one. The word weak in brackets just tells you which value to put in. Measured value, minus the strong acid value. With me? Now tell me. Weak acid with a weak base, or weak acid with a strong base. Which one gives less heat? Weak acid with a weak base, correct! Both the acid and the base must be ionised, so the heat is smaller in magnitude still. Clear?

3:31Two special cases: HF and water

Now, the twist in our puzzle. Hydrofluoric acid, H F, is also a weak acid. So with sodium hydroxide, it gives less heat than hydrochloric acid, right? No, no, careful! H F is the exception. The fluoride ion, F minus, has a large hydration enthalpy: a lot of heat comes out as water molecules surround it. So for H F, the value is more negative than minus fifty seven point one. More heat, not less. Okay? So, let's arrange them. The magnitude of the enthalpy of neutralisation with sodium hydroxide, for hydrochloric acid, for acetic acid, and for hydrofluoric acid. Biggest first. H F first, then hydrochloric acid, then acetic acid. The sign between them is greater than. H F gives more than fifty seven point one, hydrochloric acid exactly fifty seven point one, and acetic acid less. And there, our puzzle is fully solved! Right? Last one. What about the reverse of neutralisation: water splitting into H plus and O H minus? What is its delta H? Plus fifty seven point one kilojoules per mole! Reverse the reaction, and the sign flips. So the enthalpy of ionisation of water is plus fifty seven point one: heat must be taken in. Got it?

4:55Example 1: ionisation enthalpy of a weak acid HA

Now, Example one. The neutralisation of a weak acid, H A, by sodium hydroxide releases forty nine point eight kilojoules per mole of water. Find the enthalpy of ionisation of H A, okay? Pause the video, and try it. Done? Okay, let's check. First, the sign. The question says releases. So is delta H neut plus forty nine point eight, or minus forty nine point eight? Minus, correct! Released means heat given out, so delta H neut is minus forty nine point eight kilojoules per mole. Right? Now our formula. Sodium hydroxide is a strong base, so only the acid needs to be ionised. Delta ion H equals delta H neut for the weak acid, minus, minus fifty seven point one. So, minus forty nine point eight, minus, minus fifty seven point one. Careful with the double minus. Okay? Minus forty nine point eight plus fifty seven point one: plus seven point three kilojoules per mole. Positive, so ionisation takes heat in, exactly as it should. See it another way: fifty seven point one should have come out, only forty nine point eight did, and the missing seven point three went into ionising H A, right? So if you get a negative answer here, check your signs. Perfect!

6:21Example 2: heat released when H2SO4 is mixed with NaOH

Now a harder one. One hundred millilitres of zero point two molar sulphuric acid, H two S O four, is mixed with one hundred millilitres of zero point three molar sodium hydroxide. Capital M means molar, moles per litre, and m L means millilitres. Find the heat released. Pause and try. Hint: count the ions, not the molecules. Done? Okay, let's check. One hundred millilitres is zero point one litre. So moles of sulphuric acid: zero point one times zero point two, that is zero point zero two. So zero point zero two moles of H plus, right? No, no, wait! Sulphuric acid has two hydrogens, and each molecule gives two H plus ions. So H plus is two times zero point zero two: zero point zero four moles. Clear? Now the base. Sodium hydroxide gives one O H minus each, so zero point one times zero point three: zero point zero three moles of O H minus. Okay? Now think. Zero point zero four moles of H plus, and zero point zero three moles of O H minus. How many moles of water can form? Zero point zero three, yes! The O H minus runs out first, and the extra zero point zero one mole of H plus has nothing left to react with. So only zero point zero three moles of water form. With me? Strong acid with strong base, in dilute solution, so every mole of water gives fifty seven point one kilojoules. Heat released is zero point zero three times fifty seven point one: one point seven one kilojoules. And notice: we used the moles of water formed, not the moles of acid. That's it!

8:15Quick revision: 57.1 kJ, weak acids, HF and water

So, quick revision. One: any strong acid with any strong base, in dilute solution, gives minus fifty seven point one kilojoules per mole of water. Two: with a weak acid or base, less heat comes out, and delta ion H equals delta H neut, minus, minus fifty seven point one. Three: H F gives more heat, and the ionisation of water takes in plus fifty seven point one kilojoules per mole. Now go and try the practice problems on this page. All the best!