KCSE 2026 Chemistry paper 3 predictions

Added: 2026-06-04

[00:00:01]By the end of this lesson, you will be able to differentiate the graphs in energy changes in this one and the other one which contains the X, this one. You can see this one contains an an X in the middle. This is break. So, you will be able to know the difference and how the graphs look like. Stay tuned to the end.

[00:00:23]Name is teacher David.

[00:00:25]So, let's look at this paper.

[00:00:27]You are provided with solution J.

[00:00:32]You are provided with solution J, copper two ions.

[00:00:38]Uh solution K, 0.1 molar sodium thiosulfate. I want to explain [music] 0.1 capital M. So, 0.1 capital M means number of moles in a liter because this one is molarity.

[00:00:56]And what is molarity? Molarity is the number of moles in a liter. A liter is equal to 1,000 cm cubed. So, it means that the 0.1 is the number of moles which are available in 1,000 cm cubed and that makes a liter. When you understand that, this thing becomes very simple.

[00:01:17]Now, aqueous sodium iodide solution L, solution N, sodium hydroxide and then it's touch solution M.

[00:01:25]So, in practical, you need to understand what you are provided for with what you are provided with and the procedure.

[00:01:33]Which reaction, which solution is reacting with which solution? You need to understand that before. So, when I read the procedure, I am not wasting time. Some people think I'm wasting time.

[00:01:45]You need to read the process so that you can understand because this is these are the steps, the things you are supposed to do.

[00:01:51]Using a pipette, uh fill or place 25 cm cubed of solution J into a 250 volumetric flask. So, we are putting J 25 because when you pipette you actually pipette 25.

[00:02:10]So, we are putting 25 of J into a 250 ml volumetric flask.

[00:02:17]Then we are adding up with distilled water up to the mark. So, it will form J2.

[00:02:25]J2. So, what is the difference between J and J2?

[00:02:30]J2 if it was like J which solution was J? J was copper two ions. So, J2 is copper two ions. The difference is that this one is diluted. J2 is diluted more diluted than J.

[00:02:46]Now, retain solution J for the use in procedure two. Let's go to part B.

[00:02:52]Place solution K in a burette. Solution K is the sodium thiosulfate.

[00:03:01]Using a clean pipette, place 25 cm cube of solution J in a 250 conical flask. Add 10 cm cube of potassium iodide solution L. Shake well, then add 2 cm of starch indicator solution M. Titrate until the blue black color appears and continue titrating until the blue black color just disappear.

[00:03:30]So, you follow those instructions and you will get this is what you get as the final reading.

[00:03:38]19.2 Here is 19.1 and here is 19.0.

[00:03:46]Here and then we have the initial always just put one decimal place. Assuming you started each experiment you were feeling the burette.

[00:03:57]So, when you subtract this is is called arithmetic.

[00:04:01]Uh that is 19.1.

[00:04:04]19.0.

[00:04:05]Then the next thing is to calculate the average.

[00:04:09]So, calculate the average of solution K.

[00:04:11]Solution K that is the thiosulfate. So, you simply add all of them that's 19.2 + 19.1 + 19.0.

[00:04:23]Then you divide by three. You write the answer direct as 19.

[00:04:29]1 cm cube.

[00:04:34]Uh they say that the uh the volume should be in 2 cm³ so you can simply write 19.10 cm³ to two decimal places.

[00:04:48]Calculate the number of moles. Calculate the number of moles of the sulfate.

[00:04:53]Remember this is the volume of potassium of potassium sodium thiosulfate used. K is the sodium thiosulfate used. So, we have used the 14 we have used it 19.1 of K which is sodium thiosulfate. What are the number of moles? You need to go back to what you were provided. We were told that the K is sodium thiosulfate is 0.0.1 molar. That that means and I have explained that in the up there. So, we can get number of moles of this one is simply we know number of moles in 1,000 so 0.1 is equal moles.

[00:05:32]This one is found in 1 L which is 1,000 cm³.

[00:05:37]We will simply ask 19 the one that was used cm³ will contain which volume uh which moles, sorry. That is now 19.1.

[00:05:50]You times 0.1, then you divide and 1,000, and that gives you the number of moles.

[00:06:02]Which is 0.0 0.0 0 0.00191 moles.

[00:06:19]So, that is the number of moles of K used.

[00:06:24]Calculated the the question is telling that the the root of the equation is here, calculated the concentration of in moles per liter. Concentration in moles per liter means molarity.

[00:06:36]Moles per liter, it is moles in 1,000, and moles per 1,000 is mo- molarity of copper two ion in solution J, given that the number of moles of copper two ions in 25 of solution J are the same as the number of moles of sodium thiosulfate used. That is very important. You have been given that the number of moles We have been given a very important information here.

[00:07:00]Given that the number of moles of copper two ions in 25 solution J2 meaning the ratio was 1 to 1 are the same. So, that is okay. So, it means that moles of 25 cm³ is a J 2 is equal to the same as this, 0.00191 moles.

[00:07:34]Because it is the same.

[00:07:36]Then, now we can get the number of moles in 250. Remember, we diluted solution J here, solution J into J2, and the volume for J J2 was 250 cm cubed. So, let's get let us get the number of moles in 250.

[00:07:58]So, moles if it this is in 25, so we can ask 25 250 cm cubed of J2 will contain how many moles?

[00:08:09]So, that one it is 250 * this divided by this, you will get the number that is 250 * 0.00191.

[00:08:20]You divide by 25.

[00:08:24]And that one simply gives you 0.

[00:08:29]0191.

[00:08:32]In a similar 0.1 This is the number of moles in 250.

[00:08:38]Now, there is comes now a very something that you need to know a very important point here.

[00:08:45]That now we have the number of moles of J2 in 250. You need to know that the number of moles in 250 solution J2 will be equal to the number of moles of J in 25, the one that was pipetted and put into J into the conical flask. This 25 of J will contain equal number of moles in 250 cm cubed solution J.

[00:09:12]Very important.

[00:09:14]Remember, we are asked to calculate the more number of moles of J. Now, we we know the number of moles in 250 of solution J2. This one you need to know that number of moles moles in 250 cm cubed J2 will be equal to number of moles of 25 cm cubed solution J.

[00:09:43]Now, when you know that, this maths is dead. This one is over. Therefore, if that is true, then we will ask now 25 cm cubed of J contains this number of moles, 0.0191 moles. What about uh in moles per liter is 1,000. So, we ask 1,000 cm cubed of J will contain how many moles? That means it is 1,000 * 0.0191 you divide by 25 and that gives you the number of moles and that is 0.764 moles. Capital M stands for molarity and the molarity is the number of moles in a liter.

[00:10:35]Very important.

[00:10:38]Let's now go to procedure two.

[00:10:42]The graph.

[00:10:48]The graph. Using a clean burette, measure place 25 5 cm of solution N into each test tube, six test tubes.

[00:10:59]Uh using 100 ml measuring cylinder, place 20 cm cubed of solution J in 100 ml plastic beaker. Measure the temperature of solution J and record in the table. To solution J in the beaker, add the sodium hydroxide solution from one of the test tube. Stir the mixture with a thermometer and record the maximum temperature reached.

[00:11:27]Continuously with step D, uh continue with step D immediately. Add the sodium hydroxide solution N from the another test tube to the mixture obtained in C above. Stir and record the mixture uh the minimum the maximum temperature reached. Then put two, go record it there. Continued adding the solution of hydroxide solution N from each of the test tube, stirring the mixture and record the mix the mix the record the max the recording the maximum temperature each test tube here.

[00:12:03]So, here you see we have the volume of sodium hydroxide we started with zero, then five, then like that.

[00:12:12]Because we had placed here at the first here we had placed 5 cm solution in each test tube.

[00:12:19]So, we start uh we fill the table now. Let me give you the values of the table.

[00:12:32]Actually, the table is here.

[00:12:37]And now, this is where I have the problem.

[00:12:53]This is not movable.

[00:13:05]when solution N has not been added that is a zero, then mark the maximum temperature was 22.5.

[00:13:12]After adding five of solution N, the temperature rose to 24.5.

[00:13:20]Then 10, 26.5 continuously. I have like him.

[00:13:26]So, remember uh the temperature you have to wonder small places, you feed the table with if it is 25, like you see it is 27, you put 27.0.

[00:13:37]27.0, so 20.5.

[00:13:40]22.5.

[00:13:42]23.0.

[00:13:45]Yes, temperature should be recorded that way. Now, we have this one. Then you are told on the grid provided, plot a graph of temperature uh against the volume of solution sodium hydroxide solution in.

[00:14:01]Now, we will have our temperature uh that is on the x on the y-axis. Then the word that comes from after the word against is on the x-axis. So, we'll have x-axis sodium hydroxide solution used.

[00:14:17]So, I have drawn the table. So, let us follow the table slowly and to see how I have drawn it. So, I started because I have 22. I broke the people at the I broke the graph and started at 22. Then it means 22.5 is here.

[00:14:35]Then when temperature when volume is zero, that is when it is zero here. When it is five, let me reduce this one so that we can have the values.

[00:14:49]When temperature is zero here. Then when it is five, five, that is 20.

[00:14:58]24.5. When it is 10, 10, 26.5 like that.

[00:15:05]Then we have 27, 27.

[00:15:08]We have 27, then we have 27, then we have 26 and that one. So, this graph, you see the the the as we are going this is the last digit or a point I had here, then followed by this and then this one. So, this curve this curve requires you to draw a line a straight line that takes more points. So, this one up to here and this one up to there. Some points you are going to leave them because this is a heat or a heat change that is here energy changes. You must note the graph that we have in energy changes. So, this one requires you now after here you extrapolate from this point and extrapolate from this point here. Where the two extrapolated lines meet, that gives us the maximum temperature is so that is why we have this temperature here. Now, you read it there. That one gives you the maximum temperature and our minimum temperature was 22.5. Now, we are going to get uh delta T. Delta T, that is how it is obtained.

[00:16:17]Now, I want to differentiate between this graph and this one.

[00:16:23]When we have a graph temperature then time, then we have this break here. There is this break.

[00:16:31]Always remember this graph is going to be very different. It is important for you to note that this is when we have this break X here that is at the point at the time when time is 3 minutes, nothing is going to happen here.

[00:16:45]Therefore, that X requires you to note that this is the break and that is going to to be our reference point.

[00:16:54]It's going to be our reference point.

[00:16:55]So, when you draw this graph, it should be like this. So, like now we have from here we have 14 13 13. So, I started 14 13 13. Then 7 30 30. So, I did that 7 30 30 like that. Then join the lines the points which gives you the where we join many many points. So, like here you join this this and this one gives you the then up to this point here. Then also draw a line a complete from here up to here.

[00:17:31]Up to here like I drew here.

[00:17:38]Then from this point you extrapolate to the reference point. From this point you extrapolate to the where X was. This is X3 X it was here. That is our reference point. So, at this point this is where we have the maximum temperature and this point it is where we have the minimum temperature. When you subtract now you will have delta H.

[00:18:03]Temperature change. So, these two graphs you must not confuse any of them and that is how they should look like. This is for exothermic reaction. If it is endothermic it will be like this but the opposite it ain't number If it is endothermic for this one is exothermic. This one.

[00:18:24]I hope that one is clear. So, we have our graph. Now, we are going to use our graph to answer the questions that follow.

[00:18:35]Use your graph to determine the volume the volume or the volume the volume of sodium hydroxide solution N that reacted completely with 20 cm cube of solution J. 20 So, this is the maximum temperature here. That is meaning this is where the reaction ended. That was the final the maximum that they after the reaction ended it is the point. So, come down and read the volume here. So, that one gives you 13 cm cube. So, the maximum uh uh when temperature the the the volume of sodium hydroxide is 13 cm cubed.

[00:19:22]Determine the temperature change. So, temperature change the temperature to change. So, from here you read at this point, then you subtract this point.

[00:19:31]This is the minimum and this is the maximum.

[00:19:37]So, that one gives us maximum minus minimum, which is 27.7 subtract 20 22.5, and that is 5.2 degrees Celsius.

[00:19:54]Calculate the enthalpy change of the reaction per mole of copper two ions.

[00:20:00]So, the enthalpy change of the reaction per mole per mole. In other words, they're asking us to calculate molar enthalpy change. Molar enthalpy change.

[00:20:10]Molar enthalpy of copper two ions. That is very important.

[00:20:15]Let me use this white board here.

[00:20:19]Now, you need to remember things something here. For you before we calculate the molar enthalpy, we need to calculate heat change. Heat change is equal to this molar the the volume this molar the mass of the solutions the all of the mixture times specific heat capacity times temperature change.

[00:20:37]Now, remember we had solution N solution J 20 plus solution N we want we determined from the graph as 13. That gives us 13 33 cm cubed mixture of the solution.

[00:20:54]Then you need to know that we have been given the density here that 1.0 cm it means that 1 cm cubed is equal to 1 g.

[00:21:09]So, we will ask 33 centimeters cubed would it be how many grams? So, you can see it's going to be 33 g. So, this one is the same as 33 g.

[00:21:22]33 centimeters cubed is the same as 33 g because you've been given the density.

[00:21:27]So, here we would have 33 mass over 1,000 to convert to kilograms times specific heat capacity 4.2 as provided here here. Then, we multiply with times temperature change, which was 5.2.

[00:21:43]This one gives us 0.0 7072 kJ.

[00:21:51]Now, the next thing we need to get is the number of moles of copper two ions that produced this one.

[00:21:57]Remember, copper two ions equal to solution J, that's one thing which you must put in mind.

[00:22:07]So, let's go back.

[00:22:09]How many number of moles of J do we have in a liter? In J, we have 0.764 in 1 liter. Now, we can get it the number of moles of J that was used.

[00:22:20]Remember, J we used 20. Why did I tell you that it is a very important to understand the the procedure. Here we say we were told here in this procedure two that uh in procedure two, using this one, uh using a 100 ml place 20 cm of solution J. So, copper two ions we have used only 20.

[00:22:45]That's the way it is important for you to understand the procedure to make to put the information provided in the procedure into your mind.

[00:22:53]So, here we have 20 that one we have used. But, we know that 1,000 centimeters cubed of J contains 0.7 64 moles, the one we calculated in a liter. So, now we will define out 20 cm cube of J will contain how many moles.

[00:23:14]That is 20 * 0.764 you divided by 1,000 and that gives us 0.015 28 moles.

[00:23:28]So, that is the number of moles of J that was used in this reaction. Now, the next thing you need to understand now this in 20 this number of moles in a 20 are the one that produced this one.

[00:23:41]So, 0.0152 8 moles of J, which is copper two ions, is was able to produce 0.07072 kJ. So, we will only ask one mole will produce because one mole if that is what we call uh one mole will produce how That's 1 * 0.07072 / 0.01528 and that one gives us 47.12 kJ per mole.

[00:24:21]But now we need to ensure we remember this reaction was exothermic because temperature was increasing from 22 to 20 something. Therefore, the answer The correct answer should be -47.12 kJ per mole because the reaction is exothermic. The reaction is exothermic.

[00:25:10]>> You are provided with solid P.

[00:25:14]Cut out the tests below and write your observations and inferences in the space provided. Describe the appearance of the substance P. Substance P was a white crystalline. So, if it is white or if it is green, you say a green crystalline substance. For this one, it was white.

[00:25:37]White crystalline.

[00:25:40]Crystal-like, okay?

[00:25:44]Crystals.

[00:25:53]Put about 1/3 of the substance in a dry test tube. Heat it strongly. So, when you heat a substance, uh you expect uh some water of crystal test for water of crystallization, that's the first thing.

[00:26:09]So, what you will see if you heat a substance, then you expect a colorless liquid which condenses on the uh colorless liquid. Don't say water.

[00:26:22]So, the first observation is that a colorless a colorless liquid a colorless liquid condenses.

[00:26:39]On the cooler parts of the test tube. On the cooler parts of the test tube.

[00:26:53]Then, that is the major thing because if we were told to test with the red litmus paper, we can we could make red or blue.

[00:27:03]We could make more observation, but for this case we just make that one observation there.

[00:27:10]Then if it is that one, then you say that this one is an hydrated substances.

[00:27:16]hydrated Just to say substance because we don't know whether it is a salt or what.

[00:27:26]Or you can say substance P contains water of crystallization. This is the other way instead of using the first this one say contains water of crystallization.

[00:27:47]Contains water of crystallization. Don't use both of them. Just use one. Either the first one or the second one.

[00:27:56]Place the remaining amount of substance P in a boiling tube. Add what distilled water. Retain the mixture for D below.

[00:28:03]So here, if the space is provided and you are asked to add water, you need to tell us whether the solid dissolved or it did not dissolve. For in this case, the solid dissolved to form a colorless solution.

[00:28:28]So if it forms a colorless solution, then it means that if the colored ions are absent, the copper copper sulfate that is the copper the ion two and the ion three are absent.

[00:28:49]So, we use this one absent because there are so many ions could be present if it forms a white uh a colorless solution.

[00:28:57]So, we use this one to say that. Use two of a centimeter about 2 cm³ of portion of the mixture.

[00:29:08]And and this video, add two drops of aqueous barium nitrate to the mixture.

[00:29:13]So, we are adding uh barium. So, when we whenever we are adding barium, we expect a white precipitate because barium may form a white precipitate nitrate will form a white because we know this one if there's a carbonate, if there is a sulfate, it will form a white precipitate.

[00:29:29]So, what was the observation is that a white precipitate is formed.

[00:29:39]So, let me caution you about using this PPT.

[00:29:43]PPT, if you use this one in paper one, if you write PPT in paper one, if you write PPT in paper two, you will get zero.

[00:29:51]But, in paper three, it is allowed to use PPT.

[00:29:55]So, mark that. Never use PPT in your writing short form in paper one or a paper two. Don't write any short form in paper one or a paper two chemistry. But, paper three, it allows.

[00:30:09]So, if you a white precipitate is formed, then it means uh barium forms a white precipitate with the carbonate so that it is barium carbonate.

[00:30:19]I it forms with the sulfate. So, we start with the sulfate.

[00:30:25]The sulfite and the carbonate. These are the three ions which forms a white precipitate with you say that these one are are present.

[00:30:37]These are the ones that forms a white precipitate with barium nitrate.

[00:30:44]Add five drops of dilute nitric acid to the mixture.

[00:30:49]So, this mixture here we have been asked to add Uh so, if we are adding dilute nitric acid, so we expect that if it was the sulfate present, the sulfate, the first one, a white precipitate will remain, will persist. But if it is the second and the third present, the white precipitate will will dissolve to form a colorless solution when you add an acid. So, an acid is added. Let's see the observation.

[00:31:21]The first observation is that there was no effervescence.

[00:31:30]No effer- effervescence.

[00:31:39]So, if there is no effervescence, it means that the copper two sulfate Uh there's no effervescence. Another thing is that the white precipitate persist.

[00:32:04]The white precipitate plus persist does not dissolve. So, if that is the case, then then it means the sulfate are the one that are present.

[00:32:20]So, what happens if there was the sulfite and the carbonate? You receive effervescence. When acid acid is added, you receive bubbles. So, effervescence will occur.

[00:32:31]And what happens again? The white precipitate will dissolve to form a colorless solution if this one is what?

[00:32:38]Present.

[00:32:41]Add to the mixture aqueous sodium hydroxide drop wise until in excess. So, we are adding to the same mixture. Then a white precipitate is formed insoluble in excess. White PPT which is dissolved soluble soluble in excess.

[00:33:03]No, insoluble.

[00:33:18]A white precipitate which is insoluble in excess.

[00:33:32]A white precipitate is formed which is insoluble in excess.

[00:33:38]So, if I wait we are adding sodium hydroxide So, for sodium hydroxide we and there is a white precipitate so insoluble in excess, the ions which we suspect to be present are barium we expect calcium and then magnesium two ions.

[00:33:58]So now we eliminate this and this one. So, it is magnesium two which is present.

[00:34:14]I know the question would be, why did I cancel my barium and my calcium and they said it is magnesium ions. Why did I cancel this one? We have said when you add the sodium hydroxide drop wise until in excess then a white precipitate is formed and insoluble and in this insoluble in excess, That these three ions, barium, calcium, magnesium are expected to be there. But now I have cancelled this barium, and why did I do that? Remember we have said here what is present is sodium the sulfate. For the anion, it is the sulfate. [music] And we said it is soluble in it is soluble. It dissolves to form a colorless solution. So, if we if we say it was barium sulfate, barium sulfate barium sulfate barium sulfate is insoluble. It is a solid. Calcium sulfate is an insoluble solid. It is insoluble. It doesn't dissolve in water. So, that's why barium I had moved them because barium and calcium forms a white precipitate, and it cannot it cannot be a soluble salt.

[00:35:22]So, you need to understand form two solubility of salts, and that is what makes this qualitative analysis very simple.