Lesson video

Hello, my name's Dr.

Warren, and I'm so pleased that you can join me for today's lesson on Ionic Equations: Aqueous Ions Tests.

It's part of the Chemical Analysis unit.

I'm here to teach you the lesson, support you all the way through, especially the tricky parts.

Our learning outcome for today's lesson is, "I can write balanced ionic equations for the identification of cations and anions in aqueous solutions." And here are our keywords.

Ion, a charge particle formed with an atom or a molecule gains or loses electrons.

Ionic equation, an equation that shows only the ions and molecules directly involved in a chemical reaction, excluding spectator ions.

Spectator ion, an ion that remains unchanged during a chemical reaction and does not participate in the formation of the product.

Chemical species, any chemical entity such as atoms, ions, or molecules involved in a chemical process.

False positive, an incorrect test result indicating the presence of a substance or condition that is actually absent.

You may wish to pause the video now and note down these keywords and their meanings so that you can refer to them later on in the lesson.

In today's lesson, we have two learning cycles.

The first one is cation tests, ionic equations, and the second one is anion tests, ionic equations.

So let's get started with our first learning cycle on cation tests and ionic equations.

Ionic equations represent the reactions between ions in a solution.

They focus only on the chemical species that actively participate in the reaction.

They exclude spectator ions which do not participate directly in the reaction.

So before we go any further, let's just check we've got that right.

So ionic equations include all ions present in their reaction, true or false? Well done if you chose false.

So why? Is it A, ionic equations only contain spectator ions? Or B, ionic equations exclude all spectator ions? Well done if you chose B, ionic equations exclude all spectator ions.

They only show the ions that are reacting.

So when we're writing an ionic equation, there is a few steps we can go through, and if you learn these steps, you'll find it easy to do them and get them right each time.

So first of all, let's start by writing the full balanced chemical equation for the reaction.

Two, for compounds in aqueous solutions, we then split the compounds into their constituent ions.

Three, we identify and remove spectator ions, so these are the ones that appear unchanged on both sides of the equation.

And then the remaining ions and compounds form the net ionic equation representing the actual chemical change.

And that is your ionic equation.

So if you're asked to write an ionic equation, it's a really good practise to work through these four steps.

So let's have a look at some examples and we'll start with the reaction between calcium chloride with sodium hydroxide.

So first of all, we're gonna start by writing the full balanced chemical equation for this reaction.

It's also good to include state symbols when you're doing this as well 'cause that will actually help identify the ions as well.

So CaCl2 which is aqueous, plus 2NaOH, which is aqueous, gives CaOH2, which is solid, plus 2NaCl, aqueous.

So we've got our full balance equation.

That is correct.

Next, we want to split the compounds into their constituent ions.

So what we have here is Ca2+, aqueous, plus 2Cl-, so that comes from the calcium chloride, which is why it's 2Cl-, plus 2Na+ plus 2OH- and because we've got the big two in front of the sodium hydroxide, we have two of each ions, that gives us calcium hydroxide, which is solid.

So there's no ions there in aqueous solutions, so it's just the Ca bracket OH bracket 2 plus 2Na+ and 2Cl-, which are both aqueous, which comes from the 2NaCl, sodium chloride.

So we have got our compounds in aqueous solution.

We have got them all our ions.

Right, step three, we are now going to identify and remove the spectator ions.

So we look for things that are unchanged on both sides of the equation and we've put them in pink to make it easier.

So you can see on both sides of the equation, we have chloride ions, two chloride ions, and two sodium ions.

So these are spectator ions.

They are just watching.

It's a bit like when you go to a football match, you are a spectator, you watch the football match.

Well, these ions are watching the chemical reaction.

They are not taking part so we can cross them out.

Then finally, the remaining ions and compounds form the ne ionic equation showing the actual chemical change.

So we can rewrite that.

We've got Ca2+ plus aqueous plus 2OH- aqueous gives Ca bracket OH bracket 2 solid.

That is our ionic equation.

So that's the process that we go through.

Okay, I'm gonna do one more example and then you're gonna have a go.

So for this example, we have got iron chloride reacting with sodium hydroxide to form iron hydroxide plus sodium chloride.

So you've been given the chemical equation this time, let's split it up into the ions.

So Fe2+ plus 2Cl- plus 2Na+ plus 2OH-, iron hydroxide, that's solid, so we leave that as it is, plus 2Na+ plus 2Cl-.

Next, we want to identify those spectator ions and remove them.

So we've drawn the pink line through the chlorides and the sodium ions and then we can rewrite the equation.

So Fe2+ plus 2OH- gives Fe bracket OH bracket 2.

Okay, your turn.

So write the ionic equation for the following reaction, which is copper sulphate plus sodium hydroxide gives copper 2 hydroxide plus the sodium sulphate.

So remember the steps, pause the video and have a go at this ionic equation.

Okay, so on our next line, what you should have done is sorted it and rewritten it, giving the ions in aqueous solution, so that's Cu2+, plus SO4 2- plus 2Na+ plus 2OH-, then that gives us our carbohydroxide, we leave the same as it's a solid, and then we have 2Na+ plus SO4 2-.

So the next thing is identify those spectator ions and cross them out.

So our spectator iron this time are the sulphate ions and the two sodium ions on both sides of the equations.

We cross them out and we can rewrite the ionic equation, which is Cu2+ plus 2OH- gives Cu bracket OH bracket 2, which is of course solid.

So very well done if you got that correct.

The law of conservation of mass states that matter cannot be created or destroyed.

So just as a number of each type of atom needs to be balanced in a symbol equation, so does the charge.

And that's quite important.

When you come to look at your ionic equations, you'll see that the charge is also balanced.

So for example here, calcium 2+ plus 2OH- gives calcium hydroxide and you can see the 2+ charges and the 2- charges cancel each other out, so you get Ca bracket OH bracket 2, calcium hydroxide.

The reactants cancel each other out to be neutral and the product is a neutral compound with no charges and that's really important.

Quick check when you are writing these ionic equations just to make sure you've got everything right.

So quick check for understanding.

What is the total charge of the reactants in the equation? Al3+ plus 3OH- gives Al bracket OH bracket 3.

So the total charge, is it 3-? Is it 1-? Is is 0? Or 3+? We're looking at the reactants.

Well done if you chose 0.

We have three pluses from the aluminium and three OH minuses, so three minuses from the hydroxide, so that makes a total charge of zero.

So that is the correct answer.

That brings us to Task A.

We've got three questions here.

One, explain why spectator ions are not included in the ionic equation.

Question two is about the reaction between aqueous aluminium nitrate and sodium hydroxide to form a precipitate of aluminium hydroxide.

So the first thing we'd like you to do, A, is to write a full balanced equation for this reaction.

Don't forget to include the state symbols.

And part B, identify the spectator ions and write the ionic equation for the reaction.

Then in question three, you're given an ionic equation for the formation of calcium hydroxide.

Explain how the charges are balanced.

So have a go at these questions and pause the video while you do that and then when you are ready, we'll look at the answers together.

So question one, explain why spectator ions are not included in the ionic equation.

This is because they do not undergo any chemical change during the reaction.

They remain in the same state before and after the reaction.

They simply spectate and they do not directly participate in the formation of the products.

That is why they're called spectator ions.

So very well done if you got that correct.

Question two.

So first of all, we need our full balanced chemical equation.

And remember, all the information has been given to you in the question.

You just gotta get the formula right.

So aluminium nitrate, so that's Al bracket NO3 bracket 3 aqueous plus 3NaOH aqueous gives aluminium Al bracket OH bracket 3, that's your aluminium hydroxide, that's the solid, plus 3NaNO3 aqueous.

So very well done if you got that equation correct.

It's really important to get the formula right and then balance out all the atoms. So we need to identify the spectator ions.

So our spectator ions are 3Na+ and 3NO3-.

So our sodiums and nitrates are spectator ions.

And the ionic equation is 2Al3+ plus 6OH- minus gives 2Al bracket OH bracket 3 in the solid state.

So very well done if you got that right.

And remember that you might not just give the answer straight away.

It's good practise to go through and just draw all the ions separately and then cross out the spectator ions like we did before.

So excellent work if you got that right.

Let's move on to question three.

So we've got our ionic equation for the formation of calcium hydroxide.

Explain how the charges are balanced.

Right, so the charges are balanced by ensuring that the total charge on the reactant side is equal to the total charge on the product side.

So that's the first thing.

So here, the calcium 2+ ion has a 2+ charge, and the 2OH- ions each have a 1- charge giving a total of 2- charge.

The charge cancel each other out, resulting in a neutral product of Ca bracket OH bracket 2, which has no charge.

So very well done if you got that right as well.

Excellent work.

So that brings us to the end of our first learning cycle.

We're now going to move on to our second learning cycle and look at anion tests and the ionic equations they produce.

An anion test involves reactions with specific reactants to identify the presence of anions like carbonate, sulphate, and halide ions.

So we've got a variety of different tests to look at this time.

They can also be represented using ionic equations.

So we'll start with an example of sodium carbonate reacting with dilute acid.

So just as we did with the anion test, we're gonna start by writing the full balanced equation for the reaction.

Sodium carbonate plus two hydrochloric acids gives water plus carbon dioxide plus two sodium chlorides.

Again, put your state symbols there and it makes it easier to identify the ions in aqueous solution because the second thing we're gonna do for compounds in aqueous solution, we're gonna split them into their constituent ions.

So that results with 2Na+ plus CO2-, that comes from the sodium carbonate.

Then from a hydrochloric acid, we have 2H+ and 2Cl- ions.

On the other side of the equation are water and carbon dioxide are molecules in the liquid and gas state.

So we don't include them, but we do have 2Na+ and 2Cl- ions from the sodium chloride.

So once we've got all of our ions, we now need to, as before, identify the spectator ions and remove them.

So we're looking for things that are unchanged on both sides of the equation.

And what we find is we've shown them in pink again, we've got sodium plus ions and chloride minus ions on both sides of the equation.

We cancel them out, so just draw a line through them.

And then finally, the rest gives us our ionic equation.

So we have CO3 2-, the carbonate ion, plus the hydrogen ion, 2H+ plus, gives water plus carbon dioxide gas.

So although it may seem a little bit more complicated than the hydroxide ones that we used for the anion tests, if we work methodically through the same method, we will always get to the right answer.

So we've got a couple more examples to work through.

This time, we're going to look at the sulphate test, barium chloride plus iron sulphate gives barium sulphate plus iron chloride.

You've been given the symbol equation, so all you need to do now is identify the ions in the aqueous solution, so split it up into the ions.

So we've got Ba2+ plus 2Cl-, that comes from the barium chloride, plus Fe2+ and SO4 2- minus from the iron sulphate.

We leave the barium sulphate alone because it's in the solid state.

And then for the iron chloride, we've got Fe2+ and 2Cl-.

So the next thing is to identify the spectator ions and cross them out.

So we have done that and we can then write the ionic equation.

So that's Ba2+ plus SO4 2- gives BaSO4.

Right, there's an example for you to have a go at now.

This time, it's silver nitrate, so it's like the test for the halide.

So silver nitrate plus sodium chloride gives silver chloride plus silver nitrate.

So pause the video while you have a go at writing this ionic equation.

Let's have a look at the answer.

So splitting it into the ions Ag+ plus NO3- plus Na+ plus Cl-, that gives silver chloride, which is a solid, and Na+ and NO3-.

So if we identify those spectator ions, we will find that we have got nitrates and sodium ions on both sides of the equation, so let's cross those out, and our ionic equation is Ag+ plus Cl- gives AgCl.

So very well done if you've got that right.

Just remember the steps, work through those and you'll get it right every time.

When testing for anions, it's crucial to use acids that do not introduce additional interfering ions, which could cause false positives.

And the reason we use acids often is to remove impurities.

So if we think about an example of an inappropriate acid, so if we want to remove impurities and we add sulfuric acid, this will introduce some sulphate ions.

If barium ions are present, this could lead to the formation of barium sulphate falsely indicating the presence of sulphate ions and that would not give us the correct answer.

So that's an example of using an acid and introducing a false positive.

So what would be more appropriate when testing for carbonate ions, for example, to use hydrochloric acid because chloride ions are spectator ions and they do not interfere with the reaction.

So that's something that we need to bury in mind when we're carrying out cation tests.

So that brings us to Task B and we've got three questions here for you to have a go at.

So one, explain why hydrochloric acid is used in the sulphate test and not sulfuric acid.

Two, we have a question here about the reaction between aqueous potassium carbonate and hydrochloric acid.

First of all, we want you to write the full balanced chemical equation for the reaction, and then secondly, identify the spectator ions and write the ionic equation for the reaction.

Question three is about the reaction between aqueous lithium chloride and silver nitrate.

Again, we'd like you to A, write the ionic equation for this reaction, and B, explain the role of nitric acid in this reaction and why it's important to use this specific acid.

Pause the video while you have a go at these questions, and then when you're ready, we'll work through the answers together.

Let's have a look at the answer to question one.

Explain why dilute hydrochloric acid is used in the sulphate test and not sulfuric acid.

Well, let's have a look at what the sulphate test actually is.

So barium chloride plus iron sulphate gives barium sulphate plus iron chloride.

Dilute hydrochloric acid is used in the sulphate tests to acidify the solution and remove any impurities such as carbonate ions, which could give a false positive result by precipitating as barium carbonate.

Sulfuric acid contains sulphate ions which would interfere with the test and lead to false positives.

The chloride ions from hydrochloric acid are spectator ions and do not interfere with the formation of the barium sulphate precipitate, so Ba2+ plus SO4 2- gives BaSO4.

So that is why hydrochloric acid is used in the sulphate test.

Very well done if you got that correct.

Let's have a look at the answer to question two.

First of all, we need to write the full balanced equation for this reaction.

So we have potassium carbonate, which is K2CO3 aqueous plus hydrochloric acid, which is 2HCl aqueous, gives potassium chloride, which is 2KCl aqueous, plus CO2 carbon dioxide gas plus water H2O liquid.

So it's important to make sure you have all of those formulae right and that it is balanced.

So well done if you've got that correct.

We now need to go on and identify the spectator ions and write the ionic equation for this reaction.

Well, the spectator ions are the ones that don't take part, so that's the 2K+ and the 2Cl- ions.

The ionic equation is CO3 2- aqueous plus 2H+ aqueous gives CO2 gas plus H2O liquid.

So very well done if you've got that correct, including all of the state symbols.

Excellent work.

Right, now for question three part A, we need to write the ionic equation for this reaction.

Well, before we write the ionic equation, it's probably easiest to write out the full chemical equation, and that is lithium chloride aqueous plus silver nitrate aqueous goes to silver chloride solid plus lithium nitrate aqueous.

So make sure you have got all of those formulae correct.

Then we need to identify and get rid of the spectator ions.

So we're going to lose our lithium plus ions and our nitrate NO3- ions, and that leaves our ionic equation as Ag+ aqueous plus Cl- aqueous goes to AgCl solid.

So very well done if you got that correct.

And B, explain the role of nitric acid in this reaction and why it's important to use this acid.

So nitric acid is added to remove any impurities that could react with a silver nitrate and form insoluble precipitates leading to false positives.

Nitrate ions are spectator ions and do not interfere with the detection of halide ions.

So again, very, very well done if you got that correct.

That's excellent work.

Now we just need to summarise our key learning points.

An ionic equation represents the reaction between positive and negative ions.

The electric charge on each side of an ionic equation adds up to zero.

Solid, liquid, gas, and aq in ionic equations identify the state of each substance as solid, liquid, gas, or in solution.

I hope that you have enjoyed today's lesson and I look forward to learning with you again very soon.