CAPE Chemistry Unit 1 May/June 2025 Paper 2, Past Paper Solutions Question 2

Added: 2026-05-15

[00:00:08]Welcome to chem exam explained where the aim is chemistry clarity exam mastery.

[00:00:14]In this video we will be looking at cape chemistry paper 2 unit 1 module 2 2025.

[00:00:21]Let's begin. Question two part a. A mixture of hydrogen gas and iodine gas is sealed in a glass bulb and placed in a thermostat bath. A dynamic equilibrium is established at 450°.

[00:00:39]Part one. Describe the four characteristic features of this dynamic equilibrium. Four examples that we could use out of five are microscopic properties are constant. Microscopic properties are continuous.

[00:00:57]Equilibrium can be established or achieved from any direction. Equilibrium can only be achieved if the system is closed. And according to Latal's principle, if a system is in equilibrium and a change in concentration, temperature or pressure occurs, the system will act in such a way to oppose the change and return to equilibrium. 2.

[00:01:21]A part two. Write a balanced chemical equation to represent the reaction described in two part a above. Here we have hydrogen gas plus iodine gas with your reversible arrow to produce two hydrogen iodide. And this is your equation for two marks. Put in your state symbols and ensure that you have your reversible arrow. 2 a part three.

[00:01:46]It is more convenient to measure the pressure of gases at equilibrium rather than concentrations. Table two provides information on the analysis of the equilibrium mixture of the gases at 450°C.

[00:02:01]So here we have table two and we have our equilibrium pressures or partial pressures. For hydrogen we have 2.5 * 10 4. For iodine we have 1.6 * 10 4. And for hydrogen iodide we have 4.0 * 10 4.

[00:02:18]So we write our partial pressure expression from our equation that was written in part two. So it is the product which is hydrogen iodide raised to the second power over the product which means the multiplication of the reactants and they were not raised to any coefficient cuz their coefficient would have been one each. So we just simply write the equation for KP which is our partial pressure and then we plug in the information. So for hydrogen iodide you plug in 4.0 * 10 4 and you square it you get 1.6 * 10 9. Then you plug in for hydrogen 2.5 * 10 4. For iodine you plug in 1.6 * 10 4 and you multiply and you get 4.0 * 10 8. Then you do your division and your partial pressure ends up to be four. Now there are no units written for the partial pressure here because the partial pressures would cancel out each other.

[00:03:23]Two units above would cancel two units below and so we have no units and that is for three marks. B part one describe each of the following terms. Solubility product. Solubility product is the product of the concentration of each ion in a saturated solution of a sparingly soluble salt raised to the power of their relative concentration.

[00:03:49]Common ion effect. Common ion effect is the decrease in solubility of a dissolved compound by adding a solution of another compound that contains an ion that is common with it.

[00:04:03]B part two calcium phosphate is a water insoluble mineral which is used to make commercial fertilizers. Given that the concentrations of calcium 2 plus ion and phosphate 3 minus ions in equilibrium are 2 * 10 to the 8th they have 8 here but it should have been 2 * 10 the minus 8 and they have 1 * 10 9 it should have been 1 * 10 - 9 but I put in the negative for the 8 and the 9 here with these corrections you are now to determine the solubility product Ksp of calcium phosphate The issue I'm having with this question is that I would have preferred if they asked for the equation first before the calculation because in order to do the calculation, you need the equation to be able to write the expression for Ksp. And so we write the equation first for calcium phosphate to produce three calcium ions and two phosphate ions and they are aquous while the calcium phosphate being sparingly soluble is a solid. Again, you must put in your reversible arrow. So, we pretty much answered part three where they asked us to write a balanc equation for the equilibrium reaction of the formation of calcium and phosphate ions from calcium phosphate. So, that was done already. Now, from that we can write our Ksp expression where we now have the concentration of calcium ions raised to the third power times the concentration of the phosphate ions raised to the second power. Now you must know that these ions are already in equilibrium. So we are not going to multiply um the value we get by three or for phosphate by two because this is already our concentration in equilibrium. So we just simply plug in the value and cube it for calcium and then we plug in the values for phosphate and square it and the and therefore we continue with our calculation. KP is now equal to 8 * 10 - 24 * 1 * 10 -8 and our Ksp value is now 8 * 10 -42 and our unit is moles to the 5th dm -5. How did we get that? We have three calcium ion concentration and two phosphate ion concentrations. So it's moles per dm cub * mole per dm cub * mo per dm cq* mole per dm cub * mo per dm cq five of them.

[00:06:37]Then you multiply the moles five times.

[00:06:39]You get moles to the fifth power. And of course dm now is dm -3 * dnus 3 is dus 6 * dm -3 is dus 9 * dus 3 dmus 12 * dm3 d mm -15. And that's how we got our unit moles to the 5th dm to the minus15. Part C. A saturated solution of silver chloride is filtered and the residue washed with dilute hydrochloric acid instead of water. Comment on the solubility of silver chloride in water compared with the solubility of silver chloride in dilute hydrochloric acid and justify the difference if any.

[00:07:26]The solubility of silver chloride is decreased in dilute hydrochloric acid than in water. The hydrochloric acid has the common ion CL minus ions and this would decrease the solubility of silver chloride forming a white precipitate. So this is dealing with the common ion effect. Part D. Write the steps in an experimental procedure for the determination of the solubility product constant of berium hydroxide.

[00:07:54]First you'll prepare a saturated solution of berium hydroxide solution.

[00:07:59]You will then filter the solution to remove the excess solid. After that you'll titrate about 25 or 20 cm cube of the berium hydroxide saturated solution with standard hydrochloric acid. And this is an equation showing the reaction between the berium hydroxide and hydrochloric acid to produce a salt and water. After that you would then determine the concentration of berium hydroxide solution from the result. The berium hydroxide would produce berium ions and hydroxide ions. After that you can use that equation to calculate the Ksp using Ksp is equal to the concentration of berium ions times the concentration of hydroxide ions squared.

[00:08:47]Part E. Buffer solutions are used to provide suitable media for a variety of activities ranging from industrial to essential life processes. E part one define the term buffer solution. A buffer solution is a solution that resists changes in pH when small amounts of acid or alkali is added. Part two.

[00:09:14]explain how the molecular structure of amino acids relates to their function as buffer solutions in human blood. Well, we must first talk about amino acid. And as you can see that amino acid consists of two parts, two groups. Here we have the acid CO and here we have NH2 which is the amine. So the first thing we say is that the amino acid contain the caroxilic acid group and the amine group. In solution the amino acid exists as a dipole ion called zuta ion. So here we see where the acid donated the proton to the base forming the conjugate base co minus and the base accepted the proton to form the conjugate acid NH3+.

[00:10:01]So now this is our dipole ion which is our zwitta ion. So this dipole ion will now interact with things in the blood.

[00:10:11]So if you add acid to our blood then the H+ ions are neutralized by the conjugate base. And so if you look at the equation you'll see that this part is a part that will counteract the increase in the H+ ions in the blood forming this section which forms back the acid that was already there leaving our pH constant.

[00:10:34]If alkali enters the blood, the added O minus ions are neutralized by the conjugate acid which is the NH3+ group.

[00:10:43]And as you can see here, the this group is what counteracts the O minus ions.

[00:10:50]And so this should be minus. And this then the conjugate acid will donate a proton to O forming water. And this again will produce something that's already present in our blood which maintains a constant pH. Part three state one industry in which buffer solutions are used and that is easy.

[00:11:11]That's our pharmaceutical industry and we can think of things like eye drops that but when you put that in your eyes it maintains a balance or or a constant pH so that there there would be no change in pH to damage your eyes. the lotions you put on your skin, um the diaper rash that baby has, you use a powder that will maintain a constant pH and prevent diaper rash. And so the pharmaceutical industry is the answer for this question. This ends the question for module 2, unit one, paper 2, 2025. Thank you.