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*Your ConquerHSC Notes are consistently being revised over the 2019 HSC year to ensure quality.

Week 4 Content

Week 4 Inquiry Question – How does solubility relate to chemical equilibrium?

Learning Objective #1 – Describe and analyse the processes for the dissolution of ionic compounds in water

Learning Objective #2 – Investigate the use of solubility equilibria to remove toxins from food

Learning Objective #3 – Exploring solubility rules

Learning Objective #4 – Predicting formation of precipitate using Ksp value.

Learning Objective #5 –Analysing the composition of compounds formed by mixing ionic solutions

Learning Objective #6 – Deriving the solubility product constant, Ksp, for saturated solutions

Learning Objective #7 – Calculate and determine solubility using Ksp value.

NEW HSC Chemistry Syllabus Video – Solubility Equilibria 

Week 4 Homework Set (Essential for Band 5)

Week 4 Curveball Questions (Moving from Band 5 to Band 6!)

Week 4 Extension Questions 

Solution to Week 4 Questions

Overview Week 4 Inquiry Question

Welcome back to Week 4 of your Year 12 HSC Chemistry Syllabus Notes!

This will be the last week of notes for Module 5! 

Last week, we briefly touched on Ksp alongside other equilibrium constants. In this week’s note we will explore Ksp further. 

We will first look at mechanisms of which ionic compounds dissolve in water and what are the conditions required for the solute to successfully dissolve in water.

Then, we will explore how Aboriginal People use the principle of solubility of toxins in water to remove them from their food before consumption.

Following this, we will explore general solubility rules to allow us to predict if a compound produced will appear as a precipitate or in aqueous state.

After that, we will determine the precipitate that will form when we mix two ionic solutions together. This is done using two methods. One is that we use the solubility rules to determine if the reaction between reactants are capable of forming precipitates using general solubility rules. Also, we compare the reaction quotient (Q) to the Ksp value to predict if a precipitate will form.

Towards the end of this week’s notes, we will go over the mathematical derivation of the Ksp formula and other Keq in general. 

Lastly, we will go through a question to calculate the solubility of an ionic compound. 

Brief recap of ionic compounds from Preliminary HSC Chemistry

Let’s do a quick revision of ionic compounds as this week’s material puts them under the spotlight ~

Ionic compounds are substances that are formed where their atoms are joined via ionic bonds. When one atom completely transfers its valence electron(s) to another atom, an ionic bond is formed between the two. 

The atom that donated its valence electron(s) becomes a charged atom called an ion. 

  • More specifically, it is an cation as it is positively charged by having fewer electron(s) than its neutral atomic state. 
  • The atom that accepted the valence electron(s) is called the anion as it has more electrons than its neutral atomic state. 
  • So, anions are negatively charged atoms (ions). 
  • The magnitude of their charge is proportional on the number of electrons that it has donated or accepted.

The charge of the cation and anion are equal but differ in their signs (positive or negative). This essentially creates electrostatic attraction which is the ionic bond. 

Relationship between ionic compounds and salts

There are many definitions for the term ‘salt’ which differ depending on the level of chemistry you are studying. 

One possible definition for salt is a product (compound) of an acid-base reaction. The formation of a salt is illustrated in diagram below, where the salt is a product of the reaction between a Bronsted-Lowry Acid and a Bronsted-Lowry Base.

  • A Bronsted-Lowry acid is a substance that donates a proton (hydrogen ion) in a chemical reaction
  • A Bronsted-Lowry base is a substance that accepts a proton (hydrogen ion) in a chemical reaction. 

Please do note the difference in definitions between a Bronsted-Lowry Acid/Base and an Arrhenius Acid/Base which we explored in last week’s notes. 

The effect of their differing definitions will have consequences on how you write out the chemical equilibrium reaction for the dissolution of acids and bases. 

Also note that weak Bronsted-Lowry acids and bases will result in the above equation having a two-way arrow to indicate equilibrium. The above diagram shows a single way arrow, depicting a strong acid under Bronsted-Lowry definition where all strong acid molecules (H-X) donates its proton (H+ ion) to the base, “C”.

  • Strong acids and bases do not form equilibrium reactions, their dissociation into ions is depicted as one way arrow only as such above. 

However, like we said last week, we will formally introduced strong/weak acids and bases in Module 6. We just felt like talking about them here as we were here already and saw some relevance to start building some first impressions on acids & bases. 

Learning Objective #1 - Describe and analyse the processes involved in the dissolution of ionic compounds in water

The dissolution of ionic compounds in water can be separated into three processes:

  • The interaction between solvent molecules (the water)
  • The interaction between solute molecules (the ionic compound)
  • The interaction between the molecules of the solute and solvent

The reason to this is because that, prior to the dissolution process, there is interaction between water molecules via hydrogen bonding which must be broken for ions to dissolve in water. Also, there is electrostatic attraction interactions between the ions of an ionic compound, forming their ionic bonds. These bonds also need to be broken for ions to dissolve in water. 

When the solute is added to the solvent, it is the interaction between the water molecules and the ionic compounds that allows the compound to dissolve in water. 

Water is a polar molecule due to the electronegativity differences between hydrogen and oxygen. Oxygen is more electronegative than hydrogen which results in the oxygen atom ‘pulling/attracting’ more electrons towards oxygen atom and away from two hydrogen atoms. 

This results in the hydrogen atoms being partially positively charged and the oxygen atom being partially negatively charge. Moreover, the lone pair of electrons in water molecules repels the electrons between the electrons in the two O-H bonds, giving water a bent shape which enhances its polarity. Collectively, this result in the polar nature of water molecules.

As ionic compounds comprise of cations and anions, when an ionic compound interacts water, the cations will be attracted to the partially negatively charged oxygen atom of water molecules. Vice versa, the anions will be attracted to the partially positively hydrogen ions of water molecules. This attraction between ions and water molecules is called ion-dipole force. 

The ion-dipole force weakens the strength of the electrostatic attraction and breaks the ionic bonds between the cation and anions holding the ionic compound together. Also, this force also breaks the hydrogen bonds between water. Hydration is the process whereby the ion-dipole force causes ions to be separated and be surrounded by water molecules. This process gives off energy known as hydration energy. 

Hydration energy is the energy that is given off when an ion of an ionic compound are hydrated, or surrounded by water. 

For an ionic compound to successfully dissolve in water, the hydration energy must be greater than the lattice energy of the ionic compound as well as the energy required to break the hydrogen bonding between water molecules. Lattice energy is the energy that was released when the ionic bond was formed between cation and anions via electrostatic attraction to make the ionic compound). 

Also, the hydration energy released due to the formation of the ion-dipole bond or force must be large enough to break the hydrogen bonding between water molecules as well. If not, the ions of the ionic compound will not be hydrated or dissolve. 

Beyond Syllabus Information:

The water molecules surrounding the ions effectively minimises the electrostatic attraction between the cation and anions, disallowing formation the ionic bonds by creating a ‘shield’ of water molecules.

Due to the significantly smaller size of individual ions compared to its undissolved form as an ionic compound, when the ions are hydrated, the ionic bonds are broken and ions separated evenly throughout the water molecules (solvent). This decrease in size from ionic compound crystals to individual ions is what we see as the physical change when dissolving ionic compounds. 

The amount of water molecules that an ion can attract will depend on the ratio between the ion’s charge to its surface area. 

So, an ion with a small surface area but a high charge (e.g. 5+) will attract more molecules than an ion with high surface area but equal charge (e.g. 5+) which we explored in Preliminary HSC Chemistry course. 

  • Recall from earlier, the charge of an ion will depend on how many electrons it accepted or donated. For instance, a neutral atom that donated 5 electrons will have a charge of 5+
  • Vice versa, a neutral atom that accepted 5 electrons will have a charge of -5. 

Establishing equilibrium when dissolving ionic compounds in water

Equilibrium occurs when a solution is saturated by an ionic compound. 

This means that there will not be enough water molecules to be shield every ions away from each other, i.e. when water molecules are attracted to a new ion and, form ion-dipole forces, they will move away and dehydrated another ion which they were previously attracted to. 

Therefore, in an aqueous ionic solution, there is a continuous breaking of ion-dipole bonds, hydrogen bonds and ionic bonds.

Since this process will repeats back-and-forth, an dynamic equilibrium is formed which can be expressed as:

NaCl(s) + H2O(l) <-> Na+ (aq) + Cl (aq) + H2O(l)

  • We could can leave out water as it is a spectator molecule but we are putting it here for illustration purposes.

Having explored dynamic equilibrium in the first week, we know that the rate of the forward reaction is equal to the rate of the reverse reaction. 

Hence the rate at which the water molecules surround an ion is equal to the rate at which water molecules are removed from the ion as they attracted to another ion (dehydrating the ion.). 

Learning Objective #2 - Investigate the use of solubility equilbria by Aboriginal and Torres Strait Islander Peoples when removing toxicity from foods

Macrozamia is a type of cycad plant as per learning objective. 

Let’s explore how Aboriginal People and Torres Strait Islander People are able to remove toxins from macrozamia seeds so they can consume it harmlessly. 

Cycads such as Macrozamia have a large, high density seed that is surrounded by a layer of flesh called sarcotesta. The pigments in the sarcotesta gave the cycads’ seeds their colourful colours. 

Prior to the colonisation of Australia by the European in the white settlement, there were Dutch explorers that consumed the seeds of Macrozamia plants. It was documented by James Drummond, a botanist who was a early settler in Australia, that eating the fruit without prior processing resulted in many symptoms such as vomiting and unconsciousness. 

The Aboriginal people are able to use a variety of food processing methods to detoxify Macrozamia. According to sources from Grey and Moore, the aboriginal people used anaerobic fermentation techniques to process and detoxify the seeds of Macrozamia.

The detoxification process involved cutting the sarcotesta of the seed and submerging the food directly in shallow lakes for leaching. There are toxic substances such as cycasin and macrozamin in macrozamia’s sarcotesta. The slicing of the sacrotesta increased the surface area whereby toxins can be leached out. Sometimes the cycad was grounded down to further increase the surface area for leaching. 

The washing process was done delicately to prevent toxic substances of the fruit to pollute nearby water channels which was used as potable water. 

Following this, the seeds were buried in holes and concealed the hole from sunlight using a blanket of leaves. The fermentation process for the fruits was approximately two weeks until the sarcotesta became mouldy. The holes had a depth of approximately a woman’s arm length (50 to 60 centimetres deep) with a diameter of 30 centimetres. 

Overall, these deep holes provided an environment that was deprived of oxygen (anaerobic), absent of sunlight and at a temperature suitable for fermentation to occur. Moreover, this minimised the risk of other organisms such as bugs and marsupials with the capacity to burrow from eating the seeds. 

As time was forwarded to the late nineteenth century, bags were used submerge seeds in salt water which was usually attached to trees in the nearby lakes. A rope was used to close off the bag’s opening to prevent the water and endosperm of the food (see diagram below) from escaping.

It is important to note that the Noongar People, aboriginal people from southwest Australia, did not eat the endosperm of the Macrozamia. Instead, they only ate the sarcotesta (outermost flesh section). 

On the other hand, Aboriginal people from the eastern board of Australia took a different approach in consuming the Macrozamia. 

Rather than discarding the endosperm, they consumed it. However, unlike the Noongar People, they did not consume the sarcotesta. It is still not clear to why this is the case. 

Below is a diagram of the fruit, Macrozamia.


There are toxic substances such as cycasin and macrozamin in the sarcotesta and endosperm. Aboriginal people removed most of such toxins from by leaching them out in water of shallow lakes (‘soak pools’). 

Cycasin is soluble in water so it can be dissolved leached out from the fruit and into the water. It is important to note that since the leaching was opened in an open system (flowing water in the shallow lakes), the solubility equilibrium, Cycasin(s) <-> Cycasin(aq), was never allowed to be reached to reach equilibrium. However, while the solubility equilibrium could exist in nature, it was never reached equilibrium. Either way, the removal of cycasin and macrozamin through leaching was successful as it is soluble in water.

Other methods that were used included leaving the fruits out to age where sunlight help breakdown the chemical structure of the toxin, effectively reducing the toxins in the food.

Sometimes the fruits and its seeds were roasted prior to leaching which also helps breakdown the toxins’ chemical structure and reduce their levels. 

The overall fermentation process allowed Aboriginal People to make the taste of the sarcotesta more potent, elevate texture, increase nutritional value (Vitamin A and D) and facilitate the remove the endosperm from the sarcotesta for consumption (Aboriginal people on eastern Australia) or disposal (Noongar People).

Another food source that was consumed by Aboriginal People and Torres Strait Islanders People (specifically the Tiwi People living from the Tiwi islands) in Northern Territory was bitter yam. 

The process was similar to the detoxification process of cycads where they were roasted using earth ovens and leached prior to consumption. This process of leaching removed toxic oxalates from the yam.

Learning Objective #4 - Derive equilibrium expressions for saturated solutions in terms of Ksp and calculate the solubility of an ionic substance from its Ksp value

Notice in the learning objective, the term ‘saturated solutions’ is used instead of unsaturated solutions.

This is because for an equilibrium to be established in the event of the dissolution of an ionic substance, the ionic compound must saturate the solution. This is because unsaturated solutions do not have enough ions dissolved to form trace quantities of precipitate and hence cannot form an equilibrium between the ionic compound and its ions. 

Recall the concept of reaction quotient (Q) that we touched on last week. It was used to compare to the Keq value. By comparing the Q and Keq value, it is possible to determine which way the equilibrium position will shift in order to make Q equal to Keq, which is happens at equilibrium. 

Recall that there many types of Keq including Ka, Kb, Kc, Kp and Ksp

Let’s take a closer look into the relationship between Qsp and Ksp here.

The generic formula for the dissolution of an ionic compound such as Calcium Fluoride is: 

CaF2 (s) <-> Ca2+(aq) + 2F  (aq)

As we have talked about in last week’s note, to calculate the reaction quotient, Q, you just take the initial or current concentrations of the calcium ions and fluoride ions and multiply them together (after taking into account mole ratio). There is no fractions so we don’t need to divide by the reactants because we exclude solids and pure liquids from equilibrium expressions where the reasons have already been discussed in the prior weeks’ notes.

There are questions for you to practice at the end of this week’s notes. You can check solutions after you do em’.

If Q is greater than Ksp, it would mean that the equilibrium must shift to the left so that Q equals to Ksp (Please refer to previous week’s note if you are unfamiliar with the reaction quotient and Keq relationship). This would mean that more CaF2 solid, or precipitate, will form. The precipitate will stop forming when the value of Q is equal to Ksp. At this point, the solution will be saturated, i.e. at equilibrium. Solutions with Q greater than Ksp are called supersaturated solutions – precipitate will form as long as Q is greater than Ksp or until Qsp = Ksp.

If Q is equal to Ksp, it would mean that the dissolved ionic compound is at equilibrium and the solution is completely saturated. In this situation, addition of more solute (ionic compound) will not dissolved. Solutions with Q equal to Ksp are called saturated solutions. These solution contains the maximum amount of ions, in terms of concentration, that can be dissolved or exist in solution. In this situation of saturated solutions, the rate of precipitation is equal to rate of dissolution so no precipitate will form and no more solid will dissolve. This is because as precipitate are formed, they are immediately dissolved, i.e. no precipitate is formed as there is no shift in equilibrium position as system is at equilibrium. 

NOTE: For HSC purposes, trace (very small quantities) quantities of precipitates are formed when Qsp = Ksp. This is because the concentration of ions is just enough to make a saturated solution, allowing trace quantities of precipitates to be formed. However, if the question only says ‘precipitate’ and NOT ‘trace precipitate’ then no precipitate will form when Qsp = Ksp.

If Qsp is less than Ksp, it means that the equilibrium will shift to the right so that Q equals to Ksp. When Q is less than Ksp, it means that the solution is unsaturated and no precipitate will form from the aqueous ions. In this situation, if you add more of the ionic compound, more of it will dissolve to form its ions. So, the concentration of the ionic compound will not increase as no precipitate (ionic compound) will form. This will continue to happen as you add more ionic compound and dissolution will only stop you have added enough so that Q is equal to Ksp. Solutions with Q less than Ksp are called unsaturated solutions and precipitate will not form as long as Q is less than Ksp

Learning Objective #4 - Conduct an investigation to determine solubility rules, and predict and analyse the composition of substances when two ionic solutions are mixed

Exploring Solubility Rules


Analysing the composition of compounds formed by mixing ionic solutions

Suppose we have the chemical reaction between potassium chloride and silver nitrate. 

We can express the reactants as follows: 

KCl(aq) + AgNO3 (aq) –> 

How can we determine the products?

First, split up each compound into its constituent ions.

K+(aq) + Cl (aq) + Ag+(aq) + NO3(aq) -> 

Secondly, pair up cations with anions which gives you the products.

K+(aq) + Cl(aq) + Ag+(aq) + NO3 (aq) -> AgCl + KNO3

Done! We matched silver ions with chloride ions and potassium ions with nitrate ions to give us two new compounds as our products.

According to the general solubility rules table, which we examined previously, silver chloride is formed as precipitate and potassium nitrate is soluble. 

This means we can write the overall chemical reaction as the following:

KCl (aq) + AgNO3 (aq) -> AgCl(s) + KNO3 (aq)

Please note that if the products are ALL soluble, then no reaction will occur. 

For instance, suppose a reaction involving magnesium chloride and sodium acetate.

MgCl2 (aq) + 2NaC2H3O2 (aq) -> Mg(C2H3O2)2 (aq) + 2NaCl (aq)

From the general solubility rules, both magnesium acetate and sodium chloride are soluble products. Hence, no chemical reaction will occur. You need to form either a liquid, solid or gas to drive the reaction. 

Predicting if a precipitate will form when two ionic solutions are mixed

Let’s suppose you mix silver nitrate with sodium chloride. Will a precipitate form?

You are given the following information:

  • 20.0 ml of silver nitrate at a concentration of 0.010M.
  • 20.0 ml of sodium chloride at a concentration of 0.010 M.
  • The Ksp for the formation of silver chloride is silver chloride is 1.9 x 10-10.

Answer to Question

Let’s first write out the chemical equation for the reaction as below.

AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)

AgNO3 and NaCl are ionic solutions dissolved in water, therefore, we assign them with aqueous states (aq).

So, the relevant solubility equilibria here can be expressed as: AgCl(s) <-> Ag+ (aq) + Cl (aq)

How will you determine if silver chloride precipitate will form? Well, we need to compare the reaction quotient value with Ksp value!

We are given Ksp value. We need to find reaction quotient value. Since we are dealing with concentration values here for the dissociation of ionic compounds, we need to substitute concentration values into the Q (reaction quotient) equation to find out the reaction quotient value. 

As mentioned in prior week’s notes, we always substitute the initial values (e.g. initial concentration) of the ions into the reaction quotient equation.

First, let’s calculate the initial concentration of silver and chloride ions. 


Since the concentration of silver and chloride ions are greater than at equilibrium, the equilibrium position will shift to the left, forming silver chloride precipitate in attempt to establish equilibrium by lowering the Qsp value so that it is equal to Ksp (i.e. 1.9 x 10-10)

Recall that Qsp = Ksp at equilibrium which is what the system is trying to achieve or move towards.

Learning Objective #6 - Derive equilibrium expressions for saturated solutions in terms of Ksp and calculate the solubility of an ionic substance from its Ksp value

Deriving the equilibrium constant

Suppose we have the chemical reaction involving the reaction of substance A and B to form substance C.

A + B <-> C

According to the Rate Law,

Rate of forward reaction = Kforward x [A] x [B]

  • Kforward is the rate constant for the forward reaction

Rate of reverse reaction = Kreverse x [C]

  • Kreverse is the rate constant for the reverse reaction

As mentioned in Week 1, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.

So, Kforward x [A] [B] = Kreverse x [C]

If we divide Kreverse from both sides, we get:  Kforward / Kreverse = [C] / [A] [B]

As we have equated the rate for both the forward and reverse reactions to mirror an equilibrium reaction, we can re-write the above expression as:

Keq = [C] / [A] [B]; where Keq is the equilibrium constant for the reaction between A and B to form C.

Since the equilibrium constant is derived from rate law, only factors that affect the rate law will affect the value of the equilibrium constant. 

The following equation is known as the Arrhenius equation which you have seen in your Year 11 Chemistry course where K is the rate constant for a reaction (forward or reverse). 


From the Arrhenius equation, we can see that the only variable that affects Kforward and Kreverse is temperature. Hence, only temperature will affect the rate constant, K, and thus the equilibrium constant.

Other factors such as changes in concentration of reactants, for example, will only put the system out of equilibrium but will NOT affect the equilibrium constant. We have already touched on why in previous homework questions. But let’s do recap anyways.

According to Le Chatelier’s principle, if there are changes in the concentration of reactants participating in the reaction, the equilibrium position will shift itself to restore equilibrium. This means that the reaction quotient value will change as the system counteracts and minimises the disturbance. 

However, throughout this process in restoring equilibrium, there will be NO effect on the ratio between products and reactants at equilibrium, i.e. no effect to the equilibrium constant Keq. This is because the ratio, Keq, is same before AND after the equilibrium has been restored, hence the term equilibrium constant.

Derivation of Ksp

The derivation of Ksp, a type of Keq, is similar.

Suppose we have the dissolution of silver chloride in water. 

We can express it as the following:

AgCl(s) + H2O(l) <-> Ag+(aq) + Cl (aq) + H2O(l)

The rate of the forward reaction can be expressed as the following according to the Rate Law.

Rateforward = Kforward x [AgCl] [H2O]

Ratereverse = Kreverse x [Ag+] x [Cl]

So, Kforward x [AgCl] = Kreverse x [Ag+] x [Cl] [H2O]

Kforward/Kreverse = [Ag+] [Cl] [H2O] / [AgCl] [H2O]

As we equated the rate of reaction for both forward and reverse reactions which occurs when equilibrium has been established, we can re-write the ratio of the rate constants,  Kforward and Kreverse, as Keq, i.e. the equilibrium constant. 

Specifically, since we are dealing with the dissolution of solids here, we can re-write Keq as Ksp.

So, Ksp = [Ag+] [Cl] [H2O] / [AgCl] [H2O]

Notice that we can cancel water from the numerator and denominator. This is because water is acting as a pure solvent here for the dissolution of solid silver chloride. This means that it appears on both sides of the reaction. This means that water as a pure solvent will not affect the Ksp value and so we do not include it in equilibrium constant expression. 

But what happens if water is not a pure solvent? There is a homework question on this at the end of this week on this scenario. Give it a shot and check solutions to see if you are right!

Secondly, solids such as solid AgCl in this question should not be included in the equilibrium constant expressions. This is because their concentrations are constant which we have explained why in last week’s notes. 

Hence, we can re-write the solubility product constant as Ksp = [Ag+] [Cl] when given the situation of solid silver chloride (ionic compound) being dissolved dissolved in a beaker of water. 

Calculating the solubility of a compound – ICE Table 

You made a saturated solution of silver chloride by dissolving 0.3 grams of silver chloride in 50 ml of water. Using this information, answer question 1 and question 2.

Question 1: Calculate the molar solubility of silver chloride (i.e. the concentration of silver chloride in solution)

Question 2: Calculate the molar solubility product constant for silver chloride. 

NOTE to Question 1: Molar solubility is the number of moles of the sparingly soluble compound that can dissolve in a litre of water before the solution becomes saturated e.g. reaches a state of equilibrium where Qsp = Ksp.

NOTE to Question 2: It is important to note that solubility product constant (Ksp) is sometimes referred to as molar solubility product constant. This is different to molar solubility! Keyword to look out for is ‘Constant’

Answers to Question One and Question Two


To reinforce your understanding of molar solubility, let’s try the following question. 

Question 3: Suppose you dissolve a compound with the formula, A2B3 in a beaker of water and the ions do not react with water. Calculate the molar solubility of A3B2. You are given that the Ksp = 1.1 x 10-19


Week 4 Homework Problem Set

Week 4 Homework Question #1: Describe the processes used by Aboriginal People to remove toxins from one of their named food.

Week 4 Homework Question #2: Explain how the reaction quotient can be used to determine how an equilibrium position will shift

Week 4 Homework Question #3: Explain when you will use Le Chatelier’s principle and reaction quotient to determine how an equilibrium position will shift, i.e. what information you need for you to use each one. Also identify which of the two can only be used when an equilibrium reaction is at equilibrium. 

Week 4 Homework Question #4: Describe the processes involved in the dissolution of ionic compounds in water

Week 4 Homework Question #5: Suppose you have a reaction between lead chloride and sodium sulfate. The formation of lead sulfate has a solubility product constant of 1.77 x 10-10. Assume that the volume of the solution will not change when lead sulfate was added into the sodium sulfate solution. How much lead (II) chloride solid you need to add in order to form the precipitate from 0.001M of sodium sulfate. Note: Always assume volume is 1L if not given as in this case.

Week 4 Homework Question #6: Distinguish between molar solubility product constant (or solubility product constant) and molar solubility.

Week 4 Homework Question #7: Explain how Arrhenius’s equation and the Rate Law helps explain the factor that affect the value of the equilibrium constant for a given chemical reaction.

True or False Questions

True/False: There is an equilibrium constant for every chemical equation that is different, for example from the dissolution of sodium chloride in water to the Haber Process. 

True/False: If you multiple all the co-efficients of species involved in an equilibrium reaction, you will alter the value of the equilibrium constant. For example, the Haber Process can be expressed as: N2(g) + 3H2 <-> 2NH3 (g). If the reaction was multipled by two to become: 2N2(g) + 6H2 <-> 4NH3(g), then the equilibrium constant value will change. 

Yes/No: Suppose there is an equilibrium and water is formed as a product but was no present on the reactants. Do you include water in your equilibrium constant? Yes or no? Explain your answer and give an example of such equilibrium reaction. Will you answer differ if water is present in both the reactant and product side of the equation? Yes or no? Explain your answer and give an example of such equilibrium reaction.

Week 4 Curveball Questions

Curveball Question #1: Why is the dissolution of ionic compounds considered a physical change and not a chemical change?

Curveball Question #2: Explain why there is a reaction between sulfuric acid and sodium hydrogen carbonate despite the products initially appear being soluble? (HINT: Observe to see whether or not you can break down the products further. hehe)

Curveball Question #3: Suppose you have 200 millilitres of calcium nitrate at a concentration of 0.30 M and 200 millilitres of sodium fluoride at 0.50 M. Can you determine whether or not a precipitate will form. Furthermore, determine the concentration of fluoride ions when the system has reached equilibrium. The Ksp for CaF2 is 3.9 x 10-11. HINT: You can either use the dilution formula or calculate moles directly then dividing by new volume.

Curveball Question #4: Suppose the molar solubility product constant for magnesium hydroxide is 1.2 x 10-11. Calculate the molar solubility of magnesium hydroxide, the pH and pOH of the magnesium hydroxide solution. Determine whether the molar solubility of magnesium hydroxide will increase or decrease when pH is greater than 10.45 and less than 10.45. 

Feel free to refer to earlier week’s notes on the formula for pH and pOH or use google. 

Curveball Question #5: If Qsp is greater than Ksp, then a precipitate will always be produced. True or false? Explain why or why not. (HINT: Re-read the cobalt (II) chloride equilibrium equation in earlier weeks’ notes, I slid in a hint that will help you answer this question over there).

Curveball Question #6: Suppose, for some reason, you have a beaker containing solution with 0.7M of barium ions and 0.7M of strontium ions (Sr2+) in solution. A fellow chemist accidentally poured potassium chromate into your beaker. You saw a yellow precipitate being formed in your solution. You knew in class that the solubility product constant for the formation of barium chromate is 1.2 x 10-10 and the Ksp for the formation of strontium chromate is 3.5 x 10-5. Which of the two, barium chromate or strontium chromate, will precipitate first? Also, determine the concentration of chromate ions required to precipitate barium chromate and determine the amount strontium chromate precipitated from solution. 

Week 4 Extension Questions

These EXTRA extension question are for people who wish to get Band 6.

These are released outside of the normal upload schedule and complement curveball Questions.

We will try get them out as often as possible 🙂

Extra Module 5 Question #1: Suppose you have two Bronsted-Lowry acids, hydrogen fluoride and acetic acid. Write out a chemical equation for dissociation of each of the acid in water. Suppose the acid dissociation constant for HF is greater than acetic acid. Explain what can you deduce about the strength of HF compared to acetic acid. In your response, include the explanation of water’s capacity to ‘attract’ and bond with a proton (hydrogen ion).

Extra Module 5 Question #2: Calculate the equilibrium constant for the oxidation of sulfur dioxide to form sulfur trioxide. Suppose you initially have 0.5 moles of SO2 and 0.5 moles of O2 in the 5L reaction vessel. At equilibrium, you realise that you have 0.25 moles of SO2 left. 

Extra Module 5 Question #3: You are given that the acid dissociation constant for acetic acid (Ka) is 1.8 x 10-5. Calculate the [H+] and pH. Suppose the initial concentration of acetate ion from the addition of sodium acetate into the acetic acid is 0.50M and the initial concentration of acetic acid is 0.25M.

  • This question is a taste of some calculations involved in Module 6

Solution to Homework Set

Solution to Question 1

Solution to Question 2

By definition, the reaction quotient of a reaction is equal to the equilibrium constant of the same reaction when the system is at equilibrium. This means that, in order establish (or re-establish) equilibrium, the reaction quotient of the will move towards the Keq value. 

Suppose an equilibrium reaction has the generic chemical equation: A + B <-> C + D.

The reaction quotient can be expressed as, Qc = [C] [D] / [A] [B]. 

If Qc is > Kc, the system will shift its equilibrium position to the left, in favour of the reverse reaction, to re-establish equilibrium so that Qc = Kc. This means that the [C] and [D] will decrease and the [A] and [B] will increase.

If the Qc is < Kc, the system will shift its equilibrium position to the right, in favour of the forward reaction, to re-establish equilibrium so that Qc = Kc. This means that the [C] and [D] will increase and the [A] and [B] will decrease.

If the Qc = Kc, it would mean that the system is already at a state of equilibrium and there will be no shift in equilibrium position as long as there is no external disturbance being applied to the system. 

Solution to Question 3

Generally speaking, Le Chatelier’s Principle (LCP) is used to to predict shift in equilibrium position when only qualitatively information is given. 

For example, what happens when someone adds A into the beaker that has the equilibrium reaction: A + B <-> C + D. Well, the equilibrium position will shift to the right according to Le Chatelier’s Principle to minimise the increase in [A]. 

Comparatively, the reaction quotient is generally used to predict shift in equilibrium quantitatively when the reaction quotient of the reaction is compared to the reaction’s equilibrium constant (e.g. Qc with Kc, Qsp with Ksp, Qp with Kp, etc). 

For example, you may be given initial concentration of A, B, C and D and ask which way the equilibrium position will shift. 

Well, you need to use substitute the initial concentration of each species into the Qc formula and compute Qc value. Then, you compare the Qc value with the Kc value of the reaction to determine which way the equilibrium position will shift. Remember that you always put only the initial or current concentrations of reacting species in the equilibrium reaction into the Qc formula. 

Comparatively, you always put the equilibrium concentrations of the reacting species into the Keq formula! So even though the Q and Keq formulae are the same, the type of values that you should put into each formula are different! Don’t get them mixed up. 

In some questions, you may not be given sufficient information and so you may end up generalising and thus you can use either one (LCP or reaction quotient) to explain the shift direction in equilibrium position. However, be assured that in HSC, the question will provide you with sufficient information.

Solution to Question 4

The dissolution of ionic compounds can be explained using three processes. The first process is 

the breaking the interaction between solvent molecules (i.e. water molecules). The second process is breaking the interaction between solute molecules (i.e. ionic bonds in ionic compound). The third process is creating the interaction between the molecules of the solute and solvent (i.e. ion-dipole force). 

During the dissolution of ionic compounds in water, the some of the hydrogen bond between the water molecules are broken and the ionic bonds between the cation and anions of the ionic compounds are also broken. These hydrogen and ionic bonds are broken due to the ion-dipole forces that are formed between the cation and anions with water molecules. The ion-dipole weakens the electrostatic attraction and breaks ionic bonds that are holding the cation and anions together in the lattice, where this process is known as hydration. 

Successful dissolution of the ionic compound requires the hydration energy to be greater than the energy of the lattice energy of the ionic compound (i.e. the energy that is released when the ionic bonds are initially formed).

Solution to Question 5


Solution to Question 6

Molar solubility of a substance is the referred to as the amount of moles a substance can dissolve in one litre of water before the solution is saturated (i.e. no more of the solute can dissolve in the solution when more solute is added). 

Molar solubility product constant (or solubility product constant) can be denoted as Ksp. It is a equilibrium constant value for a solution containing a sparingly soluble solid and its constituent ions, measuring the extent of dissociation of the solid.

Solution to Question 7

The equilibrium constant expression is derived from the rate law for the forward and reverse reaction in an equilibrium reaction. The rate constant for both the forward and reverse reaction can be determined using the Arrhenius equation. 

The ideal gas constant, activation energy of the forward and reverse reaction as well as the frequency factor (A) are constant in a given reaction. This means that only the temperature at which the equilibrium reaction is occurring is a variable factor which results in the equilibrium constant being dependent on temperature of reaction (or system). 

  • Side Note: that frequency factor does in fact change when you change the temperature as the frequency factor is the rate at which the reacting particles of the reaction collide with each other. However, the change in the frequency factor is insignificant when compared to changing the unit of temperature by one. Thus, since the frequency factor have very minimal dependency on temperature, we treat it as a constant. 

Answer to True/False Questions



Yes, water should be included if it only appears on the product side of the reaction. This is because water is not a solvent for the reaction and, in the question, water only appears on the product side of the chemical equation and not on the reactants side. This means disregarding water as a reacting species will produce an inaccurate equilibrium constant value for the chemical reaction. 

Vice versa, if water appears on both side of the chemical reaction (e.g. acting as a solvent), its concentration can be disregarded. This is because it is acting as a pure liquid (i.e. constant concentration) as water will appear both as a numerator and denominator in the equilibrium expression. Thus, disregarding the concentration of water will not affect the equilibrium constant value for the chemical reaction. 

Solution to Week 4 Curveball Questions

Solution to Curveball Question #1

The dissolution of ionic compounds is a physical change because the charged ions that make up the ionic compound remains to be cations and anions (charged species/ions) before and after dissolution. 

Solution to Curveball Question #2

The reaction between sulphuric acid and sodium hydrogen carbonate can be expressed as: H2SO4(aq) + NaHCO3(aq) -> H2CO3(aq) + Na2SO4(aq). While from this chemical reaction, it appear that all reactants and products are ionic and therefore no driving force and no reaction will occur.. 

However, we know that H2CO3 decomposes into CO2(g) and H2O(l) which is a gas and liquid respectively, so not all products are aqueous ions. Therefore, a reaction will occur.

Solution to Curveball Question #3

Solution to Curveball Question #4


Solution to Curveball Question #5

True. If the Qsp is greater than the Ksp, it means that the concentration of ions in solubility is exceeds the extent at which the sparingly soluble solid dissolves in solution (i.e. greater than the concentration of ions at equilibrium). Therefore, to re-establish equilibrium, the concentration of ions in solution must decrease and so precipitation must occur. Precipitation will stop when Qsp = Ksp, i.e. equilibrium has been reached so that the rate of precipitation is equal to the rate of dissolution. 

Note that at Qsp = Ksp, ‘trace precipitate’ are indeed formed but no precipitate is formed. For HSC Chemistry, there is a difference between trace precipitate (very small quantities) and precipitate (appreciable quantities). So watch out for the term ‘trace’ in the exam. 

Solution to Curveball Question #6


Solution to Extra Questions for Module 5

Solution to Module 5 Extra Question #1

HF(aq) + H2O(l) <-> H3O+ (aq) + F(aq)

CH3COOH(aq) + H2O(l) <-> H3O+(aq) + CH3COO(aq)

Since the Ka of HF is greater than CH3COOH it means that more of H+ ions are donated HF to water compared to the H+ ions from CH3COOH. This means that water is able to more strongly attract and act as a stronger base relative to HF than compared to CH3COOH.*

  • *More detailed explanation is that CH3COOH have a resonance structure which you will learn in the organic chemistry module.

Solution to Module 5 Extra Question #2

Solution to Module 5 Extra Question #3

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