2026 WAEC CHEMISTRY PRACTICAL SALT ANALYSIS (final hints)

Hinzugefügt: 2026-05-12

[00:00:09]Good day viewers. This is 300 plus academy where all we do is exams tutoring ensuring that you ace your next and every exams. Yes, W 2026 is upon us.

[00:00:21]Uh we're going to be starting off with uh chemistry uh practical more specifically salt analysis and um for this year uh we're going to be given a satin sample uh C which is going to be a mixture of uh powdery lead Charles 4 and uh starch. Um when you look at uh lead carbonate as you can see on your screen it is powdery and um the starch in itself is powdery. Uh when we look at you know both lead carbonate as you can see and starch both are insoluble in water. Um the lead carbonate is inorganic like I said the appearance is that it is a white powdery uh substance. It is not going to dissolve in water. When you talk about starch that this is H105N N here it can be a very large number. It could be 50 it could be 100 is the organic comp is an organic compound. It belongs to the class poly saccharide.

[00:01:33]uh is also white powdery and insoluble in water. As you can see on your screen when you if we dissolve lead carbonate in water separately uh the result is that the lead carbonates will not dissolve. If we dissolve starch in water separately uh the result is also that uh the starch is not going to dissolve.

[00:01:56]Although if you look at it you may think it is going to dissolve. Now that is the uh amulose part of the starch uh making it looks as though the starch had the soil. Uh as you can see on your screen the result is going to be a colloidal cloudy uh uh solution. That is what you're going to have. But that does not mean that the starch had uh dissolved.

[00:02:22]Um I said here I said starch though insoluble forms a colloidal suspension uh the starch particles is just going to be dispassed through the water but they do not dissolve in water. Starch does not dissolve in water. So that collidal suspension however is small enough to pass through the pores or the holes of the filter paper. So in this uh exam what we had is that once lead carbonate and starch had been mixed together and then dissolved in about 10 cm cube of uh water uh the lead carbonate is going to remain on the filter paper as a residue.

[00:03:06]However, talking about the stash, though it is insoluble, it's going to disperse through the liquid forming a colloidal uh suspension. And this collidal suspension in itself is smaller than an ordinary suspension. However, this colloidal suspension of starch is larger than a true solution. Therefore, it is going to be able to pass through the uh pores or the hole of the filter paper.

[00:03:37]So, summarily, let carbonate here is going to form the uh residue. Whereas the starch which is dispassed through uh the liquid to form a colloid particles or colloid suspension uh is going to pass through the pores of the hose to form uh the fuel trait. More specifically what are the expected uh question for this year's salt analysis?

[00:04:06]Uh the question is going to come this way. It says C is a mixture of an inorganic salt and organic compound.

[00:04:12]Carry out the following exercises on C.

[00:04:16]Record your observations and identify any gas evolved. State the conclusion you draw from the result of each test.

[00:04:24]Now question one says, put C into a beaker and add 10 cm cube of distilled water. Stir the mixture filter. Uh we may be told to you know test the filtrate with litmos paper. Uh keep the residue and the filtrate. So when we are carrying out solubility test on C, what are we going to observe? As you can see on the board, uh if we say C plus 10 cm cube of the still water stirred and filter, uh what we will notice is that um the lead carbonate being insoluble is not going to dissolve. It is going to remain as residue on the filter paper.

[00:05:06]However, when you look at the starch, uh, it is going to disperse through the water. Not that the starch is going to dissolve, it is going to disperse forming colloid particles that are small enough to pass through the pores or the holes of the filter paper. And this colloid particles here, they are larger than a true solution. However, they are smaller than the normal uh suspension we have in uh chemistry. So what we have is that the starch though insoluble is going to be able to pass uh through the uh uh filter paper forming the filtrate as a small colloid particle. So what we have there is that c partially dissolves to give a cloudy colloidal filtrate.

[00:05:56]Emphasis on the word cloudy you can see that on the image on your screen. and then you're going to have a white uh residue on the filter paper. So if that be the case since our filtrate you know is cloudy telling us more like a sample had not dissolved but had actually been dispersed uh through the uh water. We say C contains insoluble salt. That is how uh we are to record our inference. C contains insoluble salt because the starch is not going to dissolve but just disperse to form colloid uh through the water but you know it's going to get through the filter paper and the lead carbonate is not even going to dissolve or disperse through the liquid so it remains on the filter paper as residue.

[00:06:48]Now on the uh colloidal filtrate uh we may be told to test with litmos paper.

[00:06:54]Of course when we are testing a solution with litmos paper either blue litmos or red litmos it is to test if uh that is acidic or alkaline or neutral and in this case there is no effect on litmos.

[00:07:10]So we say the filtrate is neutral.

[00:07:14]Moving on to question two. It says to about 2 cm cq of the filtrate. You can see that on your screen.

[00:07:24]Put that into a boiling tube. Eat until it just boils and allow to uh cool. So it means that we the colloidal filtrate.

[00:07:32]Now we want to eat. So when we eat the colloidal filtrate our result is going to be our observation is going to be a cy paste. Our starch that had once formed colloidal particles all through the water now is going to look pastelike. So a cy paste or gel or colloidal solution is formed and that observation in itself is not definitive.

[00:07:58]Uh so we cannot actually infer anything from that. So uh question two Roman figure 2 it says add a few drops of iodine solution to the mixture from two Roman figure one.

[00:08:13]So from the mixture that we have from two Roman figure one if we add iodine solution that's our potassium iodide solution if we get a blue black or that coloration of course this is very synonymous with uh testing for starch.

[00:08:30]So you say starch is present. And then question two, Roma figure 3 says we should eat the mixture we have from two roma figure two to boiling and then allow to cool. So the mixture we have from two roma figure two. If we eat it to boiling and then allow it to cool, what is going what are we going to uh notice? You see the blue black color that we have here is going to disappear when we eat. However, if we allow the set of the mixture to cool, the blue black color or the dark coloration is going to reappear on cooling and that tells us that starch is truly uh present. So, that ends question two. So, moving on to question three. Uh we were told to work on the lectarbonate residue. Question says uh to about half of the residue add about 5 cm cube of dilute HNO3 in drops divide the resulting solution uh into three equal portions. Okay. So if you look at it the residue plus dilute HNO3. No. Uh the moment we add dilute HNO3 to the residue effvescence occurs and a colorless odorless gas which turns lime water milky and turns moist blue litmos red.

[00:09:55]Of course that's the confirmatory uh test for carbon foxide. So once we have colorless gas you know with a fessence you know we know that is carbon foxide evolved or librated and the fact that it turns lime water that's the confirmatory test we say Charles carbonate ion is present or confirmed. So that ends question 3 A. Uh question 3B uh Roman figure 1. B Roman figure 1. It says to the first portion remember uh in question 3A we were told to divide uh the resulting solution which we've uh uh to which we've added HNO3 into three equal portions. So B1 to the first portion add ammonia solution in drops and then in excess. So to the first portion plus ammonia in drops we're going to have a white chalky precipitate white chalky precipitate not gelatinous precipitate if you write gelatinous precipitate your mark is gone it is white chalk like precipitate and when we had ammonia in essence the precipitate here is insoluble in SS ammonia so that tells us lead is present B Roman figure 2 says to the second portion add dilute hydrochloric acid and it goes further to say eat the resulting mixture and allow to so to the second portion of what we have in 3A we are told to add dilute hydrochloric acid when we have dilute hydrochloric acid it is still white chalky precipitate uh then the resulting mixture heated and later cooled you see the white precipitate ate when you eat the white precipitate will dissolve but when you allow the mixture to cool the white precipitate is going to reappear that is confirmatory for lead PB2+ confirmed B roman figure three it says uh to another portion of the residue solution you know the third portion now um we should add iodine uh solution so you see the residue solution if we add iodine solution as potassium iodide solution. A yellow if a yellow precipitate is formed that is also confirmatory for lead 2.

[00:12:23]When we use the iodine solution it doesn't necessarily mean it has to be about the starch. We should always expect blue black. No remember that the residue is lead carbonate. So if in 3A the residue plus the NO3 that gives us a sort of solution. the first portion, the second portion, the third portion. If we add iodine solution, that is going to be confirmatory for lead two in. And with that, we have come to the end of the probable questions that you can have on uh salt analysis for 2026's y chemistry practical. From myself and the entire team, it's bye for now.