MOLE CONCEPT; -006 PRACTICAL VOLUMETRIC ANALYSIS CALCULATIONS (part 2)

Added: 2026-05-12

[00:00:03]Hello students. So, welcome back to the continuation of our last part concerning the practical calculations or the practical theory of volumetric analysis.

[00:00:19]So, proceeding, I said that we're going to proceed with this item.

[00:00:23]More of a back titration, but it's more of a redox analysis.

[00:00:29]So, like I said in the previous lessons, you just have to follow up and then you use the data given to make the necessary equations.

[00:00:38]Since it's the equations that you need to pick the mole ratios to get the number of moles of the respective components that reacted with the standard solution.

[00:00:52]So, in this first item, so this is a continuation of the previous lesson.

[00:00:57]And then we're proceeding to finish up the calculations that are going to be involved in the volumetric analysis of our practical paper.

[00:01:09]So, in this one we're going to look at the analysis of a mixture of ethanedioic acid, which is the same as oxalic acid, and then sodium ethanedioate.

[00:01:21]So, here you have to look for the standard solution.

[00:01:25]Is a standard solution that is going to be titrated with an unstandard solution.

[00:01:30]And for that case, it means that we need to have an equation for the titration procedure. What is taking place in the conical flask?

[00:01:42]So, you're told that 20 cm cubed of a mixture of oxalic acid and sodium oxalate was titrated with 0.1 molar sodium hydroxide using phenolphthalein indicator.

[00:01:57]So, you can know that here our standard solution is a sodium hydroxide simply because its concentration was given for the oxalic acid, we are only given the volume.

[00:02:10]So, since we're only given the volume, it means that you have to find moles of the oxalic acid using the standardized sodium hydroxide.

[00:02:20]So, they proceeded by telling So, how are you going to get the moles of sodium hydroxide that reacted?

[00:02:25]They proceeded by telling you that 10 cm cubed of the base was required to reach end point.

[00:02:33]I said when they talk about end point, they're talking about full reaction or neutralization point.

[00:02:41]So, since our base is a sodium hydroxide and it has an unknown concentration, 1,000 contains 0.1 moles, then what about the volume the the volume that was fully react that fully reacted with oxalic acid?

[00:02:57]How many moles of sodium hydroxide does it contain?

[00:03:01]Now, for us to get the moles of the acid, we are going to compare the equation.

[00:03:06]And it said maybe two moles two moles of sodium hydroxide reacted with one mole of oxalic acid. Then what about the moles of sodium hydroxide we have calculated from this volume?

[00:03:18]And those will be the moles that are present in the 10 cm cubed of the oxalic acid.

[00:03:25]We have already looked at this in details.

[00:03:28]So, proceeding, we said that another fresh 10 cm cubed of the mixture was mixed with 20 cm cubed of two molar sulfuric acid in a conical flask, and then the mixture was heated to 50 to 60°.

[00:03:46]And immediately titrated with potassium manganate seven solution.

[00:03:52]So, here we also to understand that the potassium manganate, what is the effect of the potassium manganate on the mixture of sodium and oxalic acid. Now, remember that when sodium in this mixture, we are going to have sodium oxalate. So, if at all we introduce a potassium permanganate solution, it means that potassium permanganate is going to oxidize these oxalate ions to give us carbon dioxide gas.

[00:04:19]I hope that this is clear. So, this is redox titration or redox calculation combined with the neutralized or the neutralization calculation.

[00:04:34]So, they proceeded by telling you that 25.5 cm³ of potassium permanganate was required to reach the end point.

[00:04:41]Whenever you see end point, it means that you use Since potassium permanganate is a standard solution, 1,000 contains 0.02 Then, what about the volume that fully reacted?

[00:04:53]Then, you compare mole ratio from the equation, and then you get moles of the oxalate ions that were present.

[00:05:01]So, let's look at the theory behind these calculations. So, you are told that in the first titration, one oxalic acid reacts with sodium hydroxide according to this equation. And then, the second titration is oxalate ions from sodium oxalate that react or that get oxidized by acidified manganate seven.

[00:05:26]And since this is a reducing agent and this is an oxidizing agent, that means that you are supposed to write redox equations.

[00:05:35]The two half equations, and then the overall equation. I taught you how to balance the half equations in the previous lesson.

[00:05:46]So, for this case, you are profile are provided with the 12 equations and then the overall equation. So, for the oxalate, it gets oxidized to carbon dioxide and then electrons. For the acidified manganate seven, we have already looked at how we can balance and how we can generate these two half equations.

[00:06:03]So, you have these two half equations.

[00:06:05]We combine the two equations by making sure that the electrons in both of the equations are the same. So, you multiply the first equation by five and then you add the two equations. You're going to end up with this.

[00:06:19]So, here you needed 12 two electrons to balance up simply because the overall charge this side is -2. And here, three this is zero plus -2 from the two electrons.

[00:06:33]So, so you multiply the second equation by two, which is the coefficient of the electrons in the first equation.

[00:06:40]Then you sum up the two equations to get the overall redox equation.

[00:06:45]I said I've already made a separate topic on balancing redox equations. So, you navigate through the platform and then you try to see how you can work it out on your own.

[00:06:58]So, you're going to see the concentration. So, the task one is to calculate the concentration of the oxalic acid in grams per liter of the mixture. But before you get it in grams per liter, we need to first get it in moles per liter or in the concentration of molar.

[00:07:14]But you don't have the moles of the oxalic acid that reacted, but rather we titrated this oxalic acid with a standard solution of sodium hydroxide.

[00:07:23]So, we start from the standard solution.

[00:07:25]So, since the standard solution of sodium hydroxide was 0.1 molar, that means that 1,000 cm³ of the sodium hydroxide contains 0.1 mol. Then what about the volume that fully reacted? And that volume is the one which is always at the end point.

[00:07:45]And what all that is 10 cm cubed. So, 10 cm cubed contains how many moles? So, it will be 0.1 divided by 1,000 * 10.0.

[00:07:56]And this will give us 0.001 moles of sodium hydroxide. So, these are the moles of sodium hydroxide that fully reacted.

[00:08:04]After getting the moles of the component that fully reacted, you write an equation immediately to write the mole ratios so that you can get also the moles of the other component that reacted. So, the other component in this case was oxalic acid.

[00:08:22]So, you're going to use the mole ratio from the equation to get the moles of oxalic acid that reacted. Remember that you were given the volume of oxalic acid that was used.

[00:08:32]So, it means that volume corresponds to contains those moles. Then, what about 1,000?

[00:08:39]I hope that is clear. So, let's look at that.

[00:08:45]So, from the equation provided from our theory, you can see that two moles of sodium hydroxide reacted with one mole.

[00:08:52]Then, what about the moles of sodium hydroxide that fully reacted that you have calculated for previously?

[00:08:57]So, it will be 1 / 2 * 0.001 moles.

[00:09:02]Putting that in the calculator, you're going to have 0.0005 moles of the oxalic acid.

[00:09:11]So, remember that this oxalic acid we are given the volume which was used. So, that means that the volume which was used is the one which contains these moles.

[00:09:20]Because you want to get concentration.

[00:09:22]And remember that concentration is in 1 L or 1 dm cubed or 1,000 cm cubed. So, 10 cm cubed of the mixture contain 0. 0005 moles of of oxalic acid. then what about 1,000 cm cubed?

[00:09:41]So, it's going to contain 0.0005 / 20 then times 1,000.

[00:09:48]So, you end up with 0.025 moles. But, since it's in 1,000 Since it's in 1,000, it's the same as moles per decimeter cubed or moles per liter or simply molarity.

[00:10:03]But, remember that our task is to get it in terms of grams per liter. That means that since molarity is the same as moles per liter, like in the previous lesson, so here we just need to convert these moles to grams.

[00:10:18]And I said that whenever you're converting moles to grams, always you have to find RFM.

[00:10:23]So, we find RFM of our hydrated oxalic acid, and it's going to be equal to 126. So, here I've not used I've not put a sign of grams simply because I used RFM.

[00:10:36]If you say molecular mass, you have to use the signs of grams.

[00:10:41]So, relation of moles to grams always RFM contains 1 mole or 1 mole of any compound contains its RFM. Then, what about these moles? These 0.0 25 moles. So, it's going to be 126 / 1 then times 0.025.

[00:11:01]And that is going to give us It's going to give us the 0.125 g. So, since the moles of 0.025 were in 1 L, so it's the same as 0.125 g per liter, which is the required concentration as per the task.

[00:11:20]Now, here you are told that you are supposed to find the concentration of the sodium oxalate also in grams per liter. Now, also here I have to remind you You have seen that after reacting oxalic acid with sodium hydroxide, we got sodium oxalate.

[00:11:35]And they told you that in this mixture, we added potassium permanganate. So, it means that potassium permanganate reacted with the oxalate ions or sodium oxalate to give us carbon dioxide gas and manganese ions.

[00:11:50]So, we start with a standard solution as usual.

[00:11:54]So, the standard solution was the manganese seven which was used.

[00:11:58]So, 1,000 since we're given 0.02 molar, it means that 1,000 cm cubed of the manganese seven contains 0.02 moles.

[00:12:06]Then, what about the volume of the manganese that was required to reach end point or the volume that fully reacted?

[00:12:16]So, we are told that it is 25.50.

[00:12:19]So, 25.50 volume of manganese seven is going to contain 0.02 / 1,000 then times 25.5 moles.

[00:12:32]>> [snorts] >> So, putting that in the calculator, you're going to end up with 0.00051 moles of manganese seven. So, these are the moles of manganese seven that fully reacted with the oxalate ions. But, you're given the equation in the theory of manganese seven reacting with oxalate ions.

[00:12:52]So, after you get the volume that reached end point, moles of the end point, you use the mole ratio in the equation to get moles of the component that the standard solution reacted with.

[00:13:05]So, from the equation, you can see that two moles of manganese seven reacted with five moles of the oxalate ions.

[00:13:13]Then, what about these moles that you've just calculated?

[00:13:17]So, it will be 5 / 2 * 0.0051 moles of the oxalate ions.

[00:13:26]And this is going to give you 0.001275 moles of the oxalate ions. So, from here these moles that you have got are contained in the volume of the mixture that we used of the oxalate ions. And remember that according to the task or the item, we used 20 cm cubed. So, it means that the 20 cm cubed of the mixture is going to contain these moles of the oxalate ions.

[00:13:57]Then, what about 1,000? Simply because we want grams in 1 L.

[00:14:04]So, after converting it in 1 L, then we convert the moles to grams, just like we have done in the previous task.

[00:14:12]So, since 20 cm cubed contains 0.0012575, then what about those ones in 1 L?

[00:14:22]So, it will be 0.001275 / 20 * 1,000. And therefore, you end up with 0.06375 moles per liter. So, we convert these moles to grams since that's what the task is requiring.

[00:14:42]So, before we do that, the total concentration of the oxalate ions in the mixture is this.

[00:14:48]Therefore, the concentration of the oxalate ion from the sodium oxalate is going to be the same as this one from the mixture, then minus the one that we have initially got.

[00:15:02]Now, you need to first pay attention here properly.

[00:15:05]Here, we have got the concentration of the oxalate ions in the mixture. That means that those both on on sodium oxalate and also those ones from the oxalic acid that didn't react.

[00:15:20]So, what we do here is that for us to get the concentration of the oxide ions only from the sodium oxalate, we get those ones from the mixture and then we subtract those ones from the oxalic acid that we initially got.

[00:15:36]And therefore, you should end up with 0.03875.

[00:15:41]Now, these ones are no longer oxide ions from the mixture, but they are oxide ions from the sodium oxalate. I hope that is very clear.

[00:15:52]So, from there, we get RFM. Since to convert Since to convert moles to grams, we have to find RFM.

[00:15:59]So, here we couldn't find RFM from this point simply because in this concentration was for the oxide ions in the mixture including those of oxalic acid that didn't react.

[00:16:11]So, the subtraction of getting the molarity we got in the first task minus this one that we have got is to get the exact concentration of the oxide ions from the sodium oxalate.

[00:16:24]I hope that is clear.

[00:16:26]So, it means that for us to get to convert this to grams per liter, we get RFM of only the sodium oxalate.

[00:16:35]So, it's going to be 23 * 2 then + 12 * 2 then + 4 * 16 and then you're going to end up with 134.

[00:16:45]So, RFM always contains 1 mole.

[00:16:48]A 1 mole of a substance always weighs the RFM. The number of these moles that you have So, it will be 134 / 1 * 0.03875 and you're going to end up You're going to end up with 5.1925 grams per liter.

[00:17:07]Now, make sure that after writing these examples, you review them out on your own.

[00:17:14]Or, you try out the question out on your own and you see whether you can derive the same answers. Majority of you will be off from this point.

[00:17:29]So, in task three, we are supposed to get a percentage of the oxalic acid in the mixture. So, for this, we shall get them mass in grams per liter of the oxalic acid that we got in part A, then divided by the total concentration of the mixture.

[00:17:46]So, our mixture contains both unreacted oxalic acid and also sodium oxalate.

[00:17:52]So, you have 3.15 / summation of all the grams, 3.515 + 5.1925 * 100. So, this is going to give you 37.76%.

[00:18:06]Then, you also get the percentage of sodium oxalate in the mixture. So, to get sodium oxalate left in the mixture, we simply get 100% minus the percentage of the oxalic acid which is present in the mixture, and that will give us 62.24%.

[00:18:26]Now, that was it for that section. So, you're going to proceed to our final part, which is going to include the double indicator titrations.

[00:18:36]So, double indicator titrations is a simple titration technique used when determining the percentage composition by mass of a specific component which is present in a given mixture.

[00:18:50]So, the type of mixtures, like last year, we looked at the complex mixtures.

[00:18:56]So, in the complex mis- mixtures, both the components react with the acid in the same reaction.

[00:19:04]In the complex mixtures, both the components react with the acid in the same reaction. For example, a mixture of sodium hydroxide and sodium carbonate.

[00:19:15]Both the sodium hydroxide and sodium carbonate are going to react with the acid they're being titrated with in an equivalent and simultaneous manner.

[00:19:27]And also mixture of sodium hydrogen carbonate and sodium carbonate. Also, both of these components are going to react with the acid simultaneously.

[00:19:37]So, mixture of sodium hydroxide and sodium hydrogen carbonate are also going to do the same. So, let's go to an example of that. So, for example, this we're going to look at the analysis of a mixture of sodium hydroxide and sodium carbonate by the continuous method. This one is the easy to understand method simply because we use first principles which enables the student to attain a lot of scores.

[00:20:08]So, going through the question, here the question is saying that 25 cm³ of a solution containing a mixture of sodium hydroxide and sodium carbonate required a standard solution of hydrochloric acid of 0.1 molar 23.50 cm³ to react using the phenolphthalein indicator. Now, I told you in any calculations, we start with a statement.

[00:20:39]You start with a standard component which has been given. A thousand contain 0.1 molar. Then what about these ones which were required for the for the what? For the reaction. How many moles does it contain? So, to get moles of the components, sodium hydroxide and sodium carbonate, you react the acid with with both of these components. And then you use the given volume to to determine the number of moles of each of the components that reacted with the standard component, which is our hydrochloric acid.

[00:21:16]So, this was the volume which was used to reach the end point.

[00:21:20]And then they told you that the same solution was titrated further using the methyl orange indicator, and then it required 10.7 10.7 cm³ of a 0.1 molar hydrochloric acid also to reach end point. Also, for this case, you doing the methyl orange indicator, you also first start by interpreting this statement. You also budget for the respective number of moles of the components, sodium hydroxide and sodium carbonate that are present. So, let's look at the tasks.

[00:21:58]Yes, so in task one, we are supposed to find the concentration of sodium hydroxide present in grams per liter.

[00:22:05]We also do the same for sodium carbonate present in grams per liter. Now, before I start the solution, I'm first going to provide a few theory that you need to understand about the double indicator titrations.

[00:22:20]So, one, the reaction of sodium hydroxide occurs in a single step, while that of the carbonate occurs in two steps.

[00:22:30]So, first, the carbonate half of it is neutralized by the phenolphthalein indicator.

[00:22:37]And then secondly, when using another indicator, another half of it is also neutralized.

[00:22:43]So, for sodium hydroxide, it's one step, but for sodium carbonate, it's it occurs in two steps.

[00:22:50]So, with phenolphthalein indicator, sodium hydroxide is going to be completely neutralized, whereas sodium carbonate, for it, it gets halfway neutralized when we use the phenolphthalein indicator. So, it means that the other half of sodium carbonate will be neutralized using another indicator of methyl orange.

[00:23:13]So, the step where we are going to use phenolphthalein indicator, we are going to use a before that. So, it's going to be V.

[00:23:21]We are going to denote this with VP when you're using the phenolphthalein indicator. And then when you're using the methyl indicator, we are going to denote that as VM.

[00:23:32]So, write the equations for those neutralizations. So, sodium hydroxide is one step. It gets fully neutralized.

[00:23:42]So, we let we are going to let sodium hydroxide reacts with hydrochloric acid fully in one step to give us the sodium chloride and then the water molecule.

[00:23:53]And then for sodium carbonate, half of it is neutralized simply because when it reacts with hydrochloric acid, we are going to get sodium hydrogen carbonate and then sodium chloride. But remember that sodium hydrogen carbonate is also reactive with the hydrochloric acid.

[00:24:09]So, when you use when you transfer or when you go to another indicator, the methyl orange indicator, the sodium hydrogen carbonate which is formed is going to be converted to sodium chloride and then the carbon dioxide gas. That's when the complete sodium carbonate is fully neutralized.

[00:24:27]Simply because it's half neutralized in the first step simply because the product it forms is also reactive with the excess hydrochloric acid that is going to be added. So, if at all we use another indicator, its product also gets further reduced to the stable product of sodium chloride and then carbon dioxide gas.

[00:24:47]So, when you use another indicator, we are going to denote that reaction with VM showing that reaction with methyl orange as the indicator.

[00:24:57]VP will indicated this is the reaction with phenolphthalein indicator.

[00:25:03]So sodium >> [clears throat] >> hydrogen carbonate is neutralized by the excess hydrochloric acid according to the following reaction. We get sodium chloride, carbon dioxide together with the water motion. So now let's start the calculations.

[00:25:19]So you're going to note this down properly.

[00:25:22]So since half of the sodium carbonate is going to be neutralized in the first titration using the phenolphthalein indicator, then the other half is going to be neutralized using the methyl orange indicator. So it means that the volume of the acid required to completely neutralize to completely neutralize sodium carbonate is going to be twice the volume that is used to completely neutralize it with the methyl orange indicator.

[00:25:59]So that's why we made the conclusion that therefore the volume of the acid needed for complete neutralization of sodium carbonate is twice VM that when we are using methyl orange indicator simply because we have In other words, to see to easily understand it is that we have to titrate twice using the same volume that we use in methyl indicator methyl orange as the indicator. We have to use it twice, the volume of the acid to completely neutralize sodium carbonate.

[00:26:31]So therefore, the volume of the acid for complete neutralization of sodium carbonate is going to be given by 2 VM the volume that we use when we are using the methyl orange indicator.

[00:26:44]And the volume that we got using the methyl orange indicator was 7.7. So volume of the acid for complete neutralization of sodium carbonate is going to be 21.4 0 cm cubed.

[00:26:57]Now, remember that this volume is important when with the standard solution. So, you get the standard solution which was acid and then you get the moles that fully reacted in the complete neutralization with this volume.

[00:27:12]And then later, you get the concentrations.

[00:27:15]Then the volume required for the Remember that sodium hydroxide for it is completely neutralized straight away in one equation. So, the volume needed volume of the acid needed for complete neutralization of sodium hydroxide, we get it by the volume that we used for phenolphthalein indicator minus the volume that we used in the methyl orange indicator. So, it's going to be 23.50 then minus 10.7.

[00:27:41]And that of therefore, the volume of the acid needed for the complete neutralization of sodium hydroxide is going to end up with 12.80 cm cubed. So, what you need to memorize here is that sodium carbonate for it, [clears throat] it gets completely neutralized in two steps.

[00:28:02]And for us to get the volume for its complete neutralization, it's twice the volume that we we use when you're using the methyl orange indicator.

[00:28:12]Then for the volume of the acid required for complete neutralization of sodium hydroxide, which is one step, one step fully neutralized, we get volume used when you're using phenolphthalein indicator minus the volume used when you're using the methyl orange indicator. And this is going to give us 12.80 cm cubed.

[00:28:33]So, now we can proceed with the task since we have got the information that we need. The other calculations are going to remain the same.

[00:28:41]So, first we we we calculate for the moles of hydrochloric acid that reacted.

[00:28:48]So, remember that hydrochloric acid was a standard solution. We are told that it is 0.1 molar.

[00:28:54]And the acid is reacting with both sodium hydroxide and also sodium carbonate.

[00:29:02]So, I'm going to come here and select from the standard solution 0.1 molar. It means that 1,000 cm cubed of hydrochloric acid contains 0.1 mol.

[00:29:12]Then, what about the volume?

[00:29:15]What about the volume it used in the complete neutralization of the sodium hydroxide, which is 12.8.

[00:29:24]So, 12.8 moles of the acid contained these moles of hydrochloric acid.

[00:29:32]But, remember that you're interested in the moles of sodium hydroxide. So, since you have got the moles from the complete reaction or from the end point of the acid with sodium hydroxide, then we can get moles of sodium hydroxide using mole ratios from the equation.

[00:29:52]So, by that you shall write an equation of sodium hydroxide reacting with hydrochloric acid to give us sodium chloride and then the water molecule.

[00:30:02]So, from the equation you can see that 1 mol of hydrochloric acid, since this is the one which is known or standardized, reacted with 1 mol of sodium hydroxide.

[00:30:12]Then, what about the moles that you have calculated for the acid that completely reacted with sodium hydroxide? So, it's going to be 0.00128 moles of hydrochloric acid reacted with 1 over 1 * 0.00128 moles of sodium hydroxide. And you're going to remain with the same same number of moles of sodium hydroxide that reacted. But, remember that you're not interested in the moles, we're interested in the concentration.

[00:30:39]So, you convert these moles to moles per liter.

[00:30:43]So, which volume of sodium hydroxide did we contain? Remember that the mixture that we used was was for sodium hydroxide and then sodium carbonate.

[00:30:55]So, 25 cm³ of the mixture contained these moles of sodium hydroxide.

[00:31:01]So, what was the concentration of sodium hydroxide in the mixture?

[00:31:07]Since the given volume that we used uh in the question knowing the item was 25, then what about 1,000?

[00:31:14]So, 2 base 0.00128 / 25, then times 1,000.

[00:31:20]So, this is going to give us 0.0 512 moles per liter or moles per decimeters cubed or what you can simply rate as molarity.

[00:31:33]But, remember that the item or the task wanted it in grams per liter.

[00:31:38]So, to convert these moles to grams, we use RFM. So, RFM of sodium hydroxide calculated for it is going to give us 40. So, 1 mole weighs 40 g. Then, what about these moles that you have just calculated? Therefore, the concentration of sodium hydroxide in our mixture is going to be 2.048 g per liter.

[00:32:05]I hope this was very clear.

[00:32:07]Sodium hydroxide calculated in the titration of sodium hydroxide, you only have to consider the volume of the acid that fully reacted with sodium hydroxide. After that, you get the number of moles of the acid present in that volume.

[00:32:22]Since the acid was already standardized, 0.1 molar.

[00:32:26]So, after getting the number of moles of the standard component, which was HCl, that reacted, you write an equation to get the number of moles of sodium hydroxide that you had.

[00:32:40]So, from there you convert those moles to concentration.

[00:32:43]But, before you have to understand that those moles were contained in which volume, which you are given 25 cm³ of the mixture. Then, what about 1,000?

[00:32:54]So, by that you have got moles in 1 L.

[00:32:58]But, since it was it was required in grams per liter, so we had to convert these moles to grams. And you already looked at this in the previous lessons.

[00:33:07]So, I hope you have understood this up to this point.

[00:33:13]Now, for us to approach task number two, which was getting the concentration of sodium hydrogen carbonate, we also have for the same procedures. We get the volume of the acid that fully reacted with sodium carbonate. Remember that I've told you it's in two steps.

[00:33:29]The first step, it's having initial readings, and then the second step, using the methyl orange indicator, is also a couple of initial readings. So, to get the total volume or the actual volume of the acid that reacted that completely neutralized it, you get twice the volume used when using the methyl orange indicator, which you got as 21.40.

[00:33:50]So, since HCl was a standard solution, 1,000 cm³ of HCl contains 1 0.1 mol.

[00:33:57]Then, what about the total volume of the acid that was required for the complete neutralization of sodium carbonate?

[00:34:05]Contained how many moles? So, these are the moles of HCl that reacted with that fully reacted with sodium carbonate. So, for that we write the overall equation that sodium carbonate reacts with HCl to give us an end result of sodium chloride, water, and then carbon dioxide. Now, how do you How do you derive this equation?

[00:34:30]The other equation The other two equations that we wrote are the are the step-by-step equations. So, we write the overall equation when we are getting the final product of sodium chloride.

[00:34:42]I hope that is clear.

[00:34:44]The first equation, you first get sodium hydrogen carbonate using phenolphthalein. But when you go to methyl orange, the sodium hydrogen carbonate gets oxidized to sodium chloride or reduced to sodium chloride.

[00:34:57]So, you write the overall equation instead of writing those two equations simply because the volume that you considered was for the complete neutralization. And this is the equation related with the complete neutralization of sodium carbonate.

[00:35:10]I hope that is clear.

[00:35:12]So, from the equation, two moles of the acid reacted with one mole of sodium carbonate. So, many moles of sodium carbonate reacted fully.

[00:35:22]So, from the moles of hydrochloric acid that completely reacted with sodium carbonate, we can get the moles of sodium carbonate that reacted with the acid. So, it's going to be a half the moles of hydrochloric acid that reacted. And we're going to end up with 0.00107 moles of sodium carbonate.

[00:35:42]Now, these moles are present in which volume?

[00:35:45]Remember that the mixture of sodium carbonate and sodium hydroxide that we used was 25 cm cubed.

[00:35:51]25 cm cubed of that mixture contained these moles of sodium carbonate. Then what about 1,000?

[00:36:00]So, 1,000 cm cubed of the mixture is going to contain 0.00107 / 25 then times 1,000. So, you end up with a total concentration in moles per liter of 0.0428 moles per decimeter cubed or molar or moles per liter. You use the signs that you are well conversant with for molarity.

[00:36:32]And so, after that, we convert that concentration moles per liter in terms of grams per liter. So, get RFM 1 mol Where is the RFM? Denominator by the moles that you've got. So, you shall end up with a concentration grams per liter to be 4.5368, which was the required concentration as per the basic requirements.

[00:36:55]So, that was an example of the indicators when we have used a monobasic acid.

[00:37:05]And we have used the formula that is well conversant with most of the students. So, we're going to use another separate formula called the separating method. But for this case, we are going to be using a dibasic acid.

[00:37:21]Monobasic acids are those ones which have one hydrogen atom in them.

[00:37:29]Dibasic acids are those ones which are having two hydrogen atoms in them, such as sulfuric acid.

[00:37:35]So, let's look at the sulfuric acid using the separate method.

[00:37:40]So, that whichever method that you're being taught at your school, it matches with your understanding.

[00:37:47]So, by this, we're going to look at the analysis of a mixture between sodium hydroxide and sodium carbonate by the separate method. By instead of HCl, we're going to use a dibasic acid of sulfuric acid. So, the concept remains the same, and the calculations are going to remain the same.

[00:38:04]It's just that they have a different variance in the explanations.

[00:38:09]So, still here, 25 cm³ of a mixture of sodium hydroxide and sodium carbonate were used. And they required 18.70 of a 0.05 molar sulfuric acid using phenolphthalein indicator.

[00:38:26]And then another 25 cm³ of this mixture sodium hydroxide and sodium carbonate required 23.50 cm³ of the standard solution sulfuric acid of 0.05 molar by the same round using the methyl orange indicator.

[00:38:44]So the task requirements are to find the concentrations of sodium hydroxide in grams per liter and then later we find the concentration of sodium carbonate in grams per liter.

[00:38:56]I've already told that this one reacts straight away. But this one reacts in steps.

[00:39:03]I hope you have already understood that in the previous example. So let's first look at this one in details.

[00:39:12]So here the theory is going to remain the same.

[00:39:16]Sodium hydroxide reacts in one step but sodium carbonate reacts in two steps when subjected to the phenolphthalein indicator simply because of the phenolphthalein indicator it really forms sodium hydrogen carbonate which further undergoes a reaction with excess acid.

[00:39:34]So sodium hydroxide reacts with sulfuric acid to form a salt and then water and then sodium carbonate reacts with sulfuric acid when subjected to phenolphthalein indicator to form sodium sulfate and sodium hydrogen carbonate.

[00:39:47]But when subjected to the methyl orange indicator our volume is going to be the same as sodium hydroxide is fully neutralized.

[00:39:57]Both of them are fully neutralized when they are subjected to the methyl orange indicator.

[00:40:03]So sodium hydroxide will give us the salt and then water. Similarly, sodium carbonate while using the methyl orange indicator is going to give us the salt, water and then the carbon dioxide gas.

[00:40:16]So therefore the volume of the acid that is going to be used for the complete neutralization of sodium carbonate this time round is not going to be twice VM.

[00:40:27]Like the one that we looked at with a monobasic acid solution of sulfuric acid.

[00:40:32]This time round is going to be twice the difference between the volume using the methyl orange indicator and then that using the phenolphthalein indicator.

[00:40:43]So it's going to be 23.50 - 18.70 and then you multiply it with two and then you end up with 9.60 cm³.

[00:40:54]It's about knowing how to get the volumes for the complete neutralization of each component present in the mixture.

[00:41:03]Then the other calculations are going to remain in the same procedure as just like we have looked at them in the previous lessons. But the main key here is to understand how the volume is going to be derived.

[00:41:14]With the dibasic acid, this is the formula for the sodium carbonate. For the acid required for complete neutralization of the sodium carbonate and then with the monobasic acid is simply twice VM.

[00:41:27]So you have to understand the difference between the two.

[00:41:34]It's also here when you're dealing with a monobasic acid the volume that is required to completely neutralize the sodium hydroxide is going to be twice the volume when you're using phenolphthalein indicator minus the volume when you have used the methyl orange indicator.

[00:41:50]These are the only issues that you need to work on. The other calculations are going to follow the same or similar workings that you have already looked at in the previous lessons. The only constants where students make mistakes from are the volumes required to neutralize to fully neutralize volume of the acid required to fully neutralize both of the components in the given mixture. That's the main issue.

[00:42:17]So, you have to understand between the monobasic acid, what are the two formulas for getting sodium carbonate and sodium hydroxide? And then when dealing with a dibasic acid, what are the respective two formulas? Those are the only things that you need to master.

[00:42:33]The workings are going to remain in the similar pattern.

[00:42:37]So, after getting the volumes for the neutralization, or is the acid given is standard?

[00:42:43]So, if we start with sodium carbonate or sodium hydroxide, we're going to start by standardizing our acid. 1,000 cm cubed contains 0.1 mol. Then what about those moles that are required to the volume that of the acid that reacted completely with sodium hydroxide, which was 13.9? So, it's going to be 0.05 / 1,000 * 13.9. And it's going to give us 0.000695 moles of sulfuric acid. So, since these are the moles that fully reacted with sodium hydroxide, we have to make an equation for that.

[00:43:19]So, the equation for the reaction between sodium hydroxide and sulfuric acid is 2:1. So, 1 mol of sulfuric acid reacted with 2 moles of sodium hydroxide. Then what about these moles that you've calculated?

[00:43:31]So, it's going to be 2 / 1 * 0.000695 moles of sodium hydroxide.

[00:43:39]So, you end up with a volume with a number of moles of sodium hydroxide that fully reacted to be 0.00139 moles of sodium hydroxide.

[00:43:49]So, now we get the volume. So, we use the same volume like in the previous task 25.

[00:43:54]So, 25 cm cubed of our mixture contain these moles of sodium hydroxide. Then what about 1,000? So, from there we convert that concentration to grams per liter.

[00:44:06]Which you have already looked I said that the main issue is about understanding the formulas that you're going to be using when given a dibasic acid and then a monobasic acid.

[00:44:17]That's the only trick in double indicators.

[00:44:20]The others remain the same.

[00:44:26]So, my titrate, as I've said, was 25 cm cubed of the mixture between sodium hydroxide and sodium carbonate. So, 25 cm cubed of the mixture contained this moles of sodium hydroxide. Then, what about 1,000? So, it'll be 0.00139 moles divided by 25 times moles 1,000. So, since it's moles in 1,000, we can give a conclusion of our concentration to be 0.0556 moles per liter.

[00:44:54]But, the task is in need of concentration in grams per liter. That means that we have to convert these moles to grams. Since they're already present in a liter, we just convert moles to grams.

[00:45:06]To do that, we have to find RFM. So, 1 mole contains RFM of sodium hydroxide, which is 40 g. What about the moles that you've got above?

[00:45:17]So, it's 40 divided by 1 times 0.0556 g. So, putting that in the calculator, we're going to end up with 2.224 g per liter of sodium hydroxide. And that was the required concentration by the task.

[00:45:34]So, we're going to follow the same procedure since we have already got the volume of the acid that fully neutralized or that fully reacted to the end point with sodium carbonate. So, we approach task task two, and then we do the same or the similar working.

[00:45:55]So, we shall follow the same procedures that we have already looked at. This time, we use the acid.

[00:46:01]Find the moles present in the acid that reacted fully with sodium carbonate.

[00:46:06]Write an equation.

[00:46:08]Find the moles of sodium carbonate that reacted with the acid.

[00:46:13]Then, which volume contains these moles?

[00:46:17]It was 25.

[00:46:19]So, you convert these moles to concentration.

[00:46:23]25 cm³ contain those moles. Then, what about 1,000?

[00:46:27]We follow the same procedures.

[00:46:30]So, you're going to get 0.0192 moles per liter. So, you convert these moles to grams per liter.

[00:46:37]For us to get the required concentration, that means that you have to get the RFM of sodium carbonate.

[00:46:46]The RFM of sodium carbonate is 106.

[00:46:50]So, 1 mole of sodium carbonate weighs 32. Then, what about the moles that you have got? And that's the concentration, which is going to be 2.03 52 g per liter.

[00:47:00]As simple as that.

[00:47:03]I see the only issue that you have to master in double indicators is how to get the volume of the acid that fully reacted with the components present in the mixture.

[00:47:15]We use different formulas when you are given a monobasic acid like HCl, and also when you are given a dibasic acid like sulfuric acid. So, that's what you should note in your summary.

[00:47:29]So, this was it for the topic of mole concept.

[00:47:33]And these are going to be the questions below for the practice.

[00:47:37]I'll be tagging or I'll be leaving a WhatsApp group link in this description for the form six members who may need some guidance and facilitations before they go back to school on terms of scenario approach and answering approach to exhaust enough marks required for us to pass well and excel properly in our chemistry papers.

[00:48:04]So, you'll be pausing the video and then you take down the questions that are going to test your understanding. Although I provided a few.

[00:48:15]>> [clears throat] >> Now, this has marked the end. This has marked the end of our mole concept topic.

[00:49:55]Like I've said, I'm going to leave a WhatsApp group link below the description in this video.

[00:50:05]So, you tap more more down the video and then you'll be directed to a WhatsApp group link that you should join.

[00:50:17]We'll see where we can get time and then we'll hold some facilitations concerning or covering the basic principles that you're going to look at while approaching our final and external examinations.

[00:50:33]And also for any inquiries that you may have that I may listen to.

[00:50:39]So, for this we're going to end here. We meet in the next topic or in the next lesson.

[00:50:46]I remain Mr. Olugbenga Essu.