2.1 Simple equilibria and acid-base reactions

Question 1

heating of hydrated copper(II) sulfate, CuSO4⋅5H2O

Answer 1

Heating blue hydrated copper(II) sulfate crystals causes a colour change to white anhydrous copper(II) sulfate.

This is an endothermic process.

Water is released and instantly turns to steam due to the heat.

Some of the steam condenses on the cooler parts of the boiling tube.

Adding water to white anhydrous copper(II) sulfate turns it back to blue hydrated copper(II) sulfate.

A hissing sound is heard as some water boils from the heat of the reaction.

This is an exothermic reaction.

In a reversible reaction, if one direction is endothermic, the reverse is exothermic.

Question 2

Dynamic equilibrium

Answer 2

is the equilibrium that exists when the rate of the forward reaction equals the rate of the reverse reaction.

Question 3

A chemical system is in dynamic equilibrium when:

Answer 3

it is dynamic at the molecular or ionic level
both forward and reverse rates are equal
it is a closed system, i.e. a system where substances cannot leave or enter
it has macroscopic properties, e.g. concentration of the reactants and the products remain the same.
The reaction is reversible

Question 4

when does equilibrium lie to left and right

Answer 4

Increasing the concentration of a reactant:
the position of equilibrium moves to the right-hand side, forming more products, e.g. more ethyl ethanoate and water
the concentrations of the additional ethanol and ethanoic acid will decrease as a result of the change.

Increasing the concentration of a product:
the position of equilibrium will move to the left-hand side, forming more reactants, e.g. more ethanol and ethanoic acid
the concentrations of the additional ester and water will decrease as a result of the change.

Question 5

Equilibrium is a dynamic state

Answer 5

at a molecular level change is constantly taking place in both directions at equal rates. Macroscopic properties such as temperature, pressure, mass and volume are constant.

Question 6

The position of equilibrium can be altered by changing one or more of the following in the system:

Answer 6

concentration of the reactants or products
pressure in reactions involving gases
temperature.

Question 7

Le Chatelier’s principle.

Answer 7

When a system in dynamic equilibrium is subjected to a change, the position of equilibrium will shift to minimise the change.

Question 8

Answer 8

Question 9

The effect of a catalyst on equilibrium

Answer 9

In a system at equilibrium, a catalyst speeds up the rate of the forward and reverse reactions equally. Therefore, a catalyst has no effect on the position of the equilibrium. It does, however, allow the state of equilibrium to be established more quickly.

Question 10

The effect of changing the temperature on the position of equilibrium depends on the enthalpy sign (ΔH):

Answer 10

If the reaction is exothermic (ΔH is negative), increasing temperature shifts equilibrium to the left (towards the reactants).

If the reaction is endothermic (ΔH is positive), increasing temperature shifts equilibrium to the right (towards the products).

Decreasing temperature has the opposite effect:

Shifts equilibrium right for exothermic reactions.

Shifts equilibrium left for endothermic reactions.

Question 11

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Question 12

equilibrium equation

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Question 13

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Question 14

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Question 15

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Question 16

When Kc is large

Answer 16

If Kc is very large (Kc&raquo_space;1) the equilibrium lies to the RHS so the reaction mixture contains mostly products

Question 17

When Kc is small

Answer 17

If Kc is very small (Kc &laquo_space;1) the equilibrium lies to the LHS so the reaction mixture contains mostly reactants

Question 18

Kc close to 1

Answer 18

If Kc is close to 1 the mixture contains a similar concentration of both reactant and products

Question 19

When stating the value for Kc for a particular reaction, it is important to indicate the equation on which the constant is based.

Answer 19

Question 20

The units for Kc depend upon the number of reactants and products and their stoichiometric quantities. For this general example given in the first equation above:

Answer 20

Question 21

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Question 22

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Question 23

Calculate the Kc value for experiments 2 – 4 from the table on the previous screen. Is Kc constant?

Answer 23

Question 24

Acid

Answer 24

something that produces an excess of H+ ions in an aqueous solution.

Question 25

Alkali

Answer 25

something that produces an excess of OH- ions in an aqueous solution.

Question 26

Brønsted-Lowry (B-L) definition acid

Answer 26

Acid: donates a proton during chemical reactions: In the first example, water is the base as it accepts a proton from the hydrogen chloride.

Question 27

Brønsted-Lowry (B-L) definition base

Answer 27

Base: accepts a proton during chemical reactions; a base is called an alkali if it dissolves in water: In the second example, water acts as an acid donating a proton to the ammonia. Compounds or molecules that can act as both an acid and a base are called amphoteric.

Question 28

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Question 29

strong acid

Answer 29

will dissociate fully (100%).

Question 30

eg strong acids

Answer 30

Question 31

weak acid

Answer 31

only partially dissociates

Question 32

eg of weak acid

Answer 32

citric acid and ethanoic acid

Question 33

Acids undergo neutralisation reactions with

Answer 33

bases, alkalis and carbonates

Question 34

acids + alkalis eg nitric acid and sodium hydroxide full equation, ions present, ionic equation

Answer 34

Question 35

acids + carbonates eg sulfuric acid + sodium carbonate full equation, ions present, ionic equation

Answer 35

Question 36

acids + metals eg hydrocholric acid and magnesium full equation, ions present, ionic equation

Answer 36

Question 37

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Question 38

pH equation, log

Answer 38

Question 39

Calculating pH

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Question 40

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Question 41

primary standard

Answer 41

a solid eg Anhydrous sodium carbonate (Na₂CO₃) primary standard is a very pure chemical used to make a standard solution (a solution of known concentration). ✅ Must be: Very pure Stable (doesn’t react or break down easily) Not hygroscopic (doesn’t absorb water from the air) High Mr (to reduce weighing errors) 🧪 You weigh it accurately, then dissolve in a known volume of water to make a solution with a precise concentration.

Question 42

how to make a standard solution

Answer 42

Question 43

titration - to determine conc of an acid

Answer 43

The first step is to fill the burette, usually with the acid. Use a small funnel, making sure that the tip of the burette is filled. Remove the funnel after filling and read the level of the acid. Then proceed as follows: A. Use a pipette to add a measured volume (usually 25.0 cm3) of the other solution, in this case the alkali, to a conical flask. B. Add a few drops of indicator to the solution in the conical flask. C. Run acid from the burette to the solution in the flask, swirling as you do so. Stop when the indicator turns colour. This will be your ‘rough’ titration and you know that this is the maximum volume you need to add from the burette. D. Repeat the titration, this time adding the acid dropwise when you near the volume you got in part C, until the indicator just turns colour. This is the endpoint of your titration. Record the volume of acid used (the titre). Repeat until you have at least two readings that are within 0.20 cm3 (called concordant results) of each other and calculate a mean titre.

Question 44

acid base titration worked eg of calculating conc of standard solution

Answer 44

Question 45

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Question 46

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Question 47

🧪 Back Titration – Key Concept

Answer 47

In a back titration, an excess known amount of a reagent is added to a substance to ensure complete reaction. The unreacted excess is then titrated to determine how much was left over, and from this, you can calculate how much was used in the original reaction.

Question 48

eg of back titration and why it was used

Answer 48

Question 49

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Question 50

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Question 51

🧪 Double Titration – Concept

Answer 51

A double titration is used when a solution contains two different bases (often of differing strengths, like a strong base and a weak base), and we want to determine the concentration of each base individually. This method takes advantage of the fact that: Different acid-base indicators change colour at different pH values. This allows each base to be quantified separately by selecting the appropriate indicator for each one.

Question 52

Double Titration - How it works

Answer 52

Stage 1: Titration of Strong Base A strong acid is added to the solution containing both bases. The first indicator is chosen to change colour at a low pH (e.g. methyl orange), suitable for identifying the end-point of the strong base. At this point, only the strong base has reacted completely with the acid. Stage 2: Titration of Weak Base More acid is added to react with the remaining weak base. A second indicator is used (e.g. phenolphthalein) that changes colour at a higher pH, appropriate for detecting the end-point of the weak base. Now, both the strong and weak bases have reacted, and the titration is complete.

Question 53

🧠 Purpose and Application Double titration

Answer 53

From the volume of acid added in each stage: You can calculate the moles of acid reacting with each base. Using stoichiometry, determine the concentration of both bases in the original mixture.

Question 54

✅ Key Exam Tips Double titration

Answer 54

Know the pH range of indicators: Methyl orange: ~3.1–4.4 (red to yellow) → good for strong base–strong acid Phenolphthalein: ~8.3–10 (colourless to pink) → good for weak base–strong acid State clearly that the first base neutralised is the stronger base, due to the indicator used. Show separate balanced equations if needed for each reaction.

Question 55

🧪 Determining Concentrations of NaOH and Na₂CO₃ via Double Titration

Answer 55

A mixture of NaOH (sodium hydroxide) and Na₂CO₃ (sodium carbonate) can be analysed using a double titration with HCl (hydrochloric acid). Two indicators are used at different pH ranges: Phenolphthalein (changes from pink to colourless at pH ~9) – used in the first titration. Methyl orange (changes from yellow to red/orange at pH ~4) – used in the second titration. 🔁 Stage 1 – Phenolphthalein Titration (Strong base + half carbonate) Reactions during this stage: NaOH + HCl → NaCl + H₂O Na₂CO₃ + HCl → NaHCO₃ + NaCl End point (pH ~9): All NaOH and half of the Na₂CO₃ have reacted. 🔁 Stage 2 – Methyl Orange Titration (Remaining carbonate) Reaction: NaHCO₃ + HCl → NaCl + CO₂ + H₂O The second titre gives the moles of NaHCO₃, which is equal to the moles of Na₂CO₃ (since all Na₂CO₃ becomes NaHCO₃ in stage 1). ➗ Calculations Summary Moles of Na₂CO₃ = moles of HCl from second titre. Moles of NaOH = moles of HCl from first titre − moles of Na₂CO₃. Use the moles and volumes to calculate concentrations of NaOH and Na₂CO₃ in the original mixture.

Question 56

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