2026 WAEC Chemistry Practical TITRATION End point | CALCULATIONS | volumetric analysis

Hinzugefügt: 2026-05-12

[00:00:00]Yeah, good day viewers. Welcome to my channel. This is Radical Science. If today is your first time of visiting my channel, all I want you to do is to hit the subscription button and also turn the subscription bell on so that any time post in this channel, you can be able to get notification to watch our videos. So on today's edition, I'll be taking you on chemistry um titration which is on the volutric analysis for this year 2026.

[00:00:24]So here is my setup. Now I'm going to take the the instruction. I'm taking the instruction so that you will know what we are doing. Please I want you to pay attention to this video because everything I'm going to tell you in this video you will definitely see it as far as work is concerned including the end point for this USM titration. Here they said 150 cm cube of a solution tetra oxopate six in a fl or bottle level a. This should all be the same containing 6.0 cm of concentrated tetra oxos 6 per dm cube of solution. Now B says 150 cm cube of sodium hydroxide solution in a fl or bottle label bn. This should all be the same containing 4.6 4.0 g of sodium hydroxide per dm cq. So what this simply mean is that we are going to be measuring this um tetra oxosate 6 acid. We are going to be measuring um 5.6 cm cube of this. Okay. And here is what we'll be using to measure it. We'll be using this to measure it. So I just want you to pay attention to this so that you don't miss any step at all. Okay. And here is our sodium hydroxide. This is our sodium hydroxide. So we are going to be measuring you measuring it using our scale using our weighing scale. And we are instructed to measure this 4.0 g. So all I want you to do is just to pay attention to this. So all we're going to do is we are going to measure it and fill it down into this place.

[00:02:14]Remember this is this is 1,00 m and 1,00 mil is equal to 1 dm cube. So you are going to be dissolving you're going to be dissolving um 5.60 cm of this tetra of zosoped 6 into this 1,00 m. Okay that is it. I will give you the procedure as well as the titration and the end point.

[00:02:39]Just pay attention to it as we proceed.

[00:02:42]If you have any question, all I want you to do is to drop your comment in the comment section and I promise you that as soon as I get to it, I will give you your reply instantly.

[00:03:15]That's why >> [snorts] [snorts] >> Okay, let's take a look at the calculations. you must see in this year's W chemistry practical 2026. Stay tuned and remember don't forget to like and share this video. If you look at the board here, A is 0.125 moles DMQ of what surface is which is H2 SO4. Now B is a solution containing X gmQ of what? Sodium hydroxide which we know is Na O. Now put A into into the boret and titrate it against 25.00 cmq portion of B using metic orange as indicator. Record the volume of your pipet. Titrate your readings and calculate the average volume of what? A use the equation for the reaction involved in this titration. Now if you look at this equation, you see that here that tyros reacted with what? sodium hydroxide to give you sodium trox and also your water. Now if you look at the question that they give you see b says from your result and information provided calculate the amount of what surface 6 in the average volume of what a and also concentration of b in mole dq and the value of what x given that your h is what 1, o is 16 and what na is what 23. First question that we we will answer [snorts] is the amount of what titos of in the average volume of what acid. Now to calculate that remember that we are given that the molar con of a which is that H2SO4 is okay. Now the con of A is what? 0.125 mo DMQ. Well the question say find the amount of what the surfaces in the average volume of what as of A. Here remember we have that our volume of A is equals 12.00 cm cube. That means for you to get the amount of a in the average volume.

[00:14:38]It will be the volume times what?

[00:14:45]The molar con over 1,000. Here everything has been listed. You now have therefore the volume is 1200 * 0.15 moles over this is 1,0005 moles.

[00:15:15]Now on the second question they said concentration of B in mo DMQ and to find the concentration of B in DMQ we already have our concentration of O A. So now we now say con of B in DM cub. We use a formula C A VA over CP VB equals N A / N B.

[00:15:57]Now we we have that our CA was given to be 0.125 mole dm cubed. Our VA was given to be 12.00 cm cq. Our VB was given to be 25.00 cmq. Our cb was not given. Then our NA which is the ratio of A or and the ratio of B which is NB this is the ratio of A and the ratio of B that means A which is for the acid and that of what base which is gotten from what the equation for the reaction. Now if you look at that you see that you have 1 is to 2 that is the ratio that means we are now going to use these parameters to solve for CB which is what the concentration of B.

[00:17:00]All right. First from here now you make what? CB the subject of the formula. So which you have CB equals you have CA VA over what? CB VB which is equal to N A over what? N B. First of all you cross multiply. Now you have that CB VB N A equals C A VA and what B?

[00:17:32]Now you asked to make CB the subject of the formula. So you divide both side by what? VB N A over VB N A that means this cancels this. So you have CB equals CA A VA NB over VB N A which is equals which is because [snorts] our CA is 0.125 times our VA is 12.0 * our NB is 2 over our VB is 25.00* our NA is 1 which will give us So we have 0.115 2 moles gm.

[00:18:43]So this is that for that. Then the next question said you should find x. Now what that x simply means that you should find the mass con of what x that is the mass concentration of that sodium hydroxide. So now for you to find the mass concentration you use the formula that is what you do. So now they say for you to find X which is the mass con of that B which is sodium hydroxide. So you have the formula con of B equals to mass con of B over what mass? Now from there what you asked to find is mass con that means we have to find the mass of B and then multiply it by what the mass con of what B. So now you have mar mass of B in brackets ND A equals you >> that's X >> okay X the value of X given H1 >> okay >> oxygen 16 sodium 23.

[00:19:57]>> Okay, you have that sodium is 23 plus oxygen 16 + 1 which will give us 40 that is what the mass of B. Therefore, Therefore, we now have the mass of B to be this. Therefore, for you to get the mass of now, you now say mass equals to equals mass over mass. So, you cross multiply you have that mass of equals M time mass that is therefore the molar con was gotten when you use when you found the molar concentration of B in that's what we call the mar. So now what we got there was m there was you have 0.1152 mole dm cub * 40.

[00:21:24]So therefore must come into 4.608 g d.

[00:21:38]So that is it about what the calculations involved.