reaction of acids Flashcards
write ionic equations for displacement reactions and identify which species are oxidised from a symbol or half equation.
Ionic Equations for Displacement Reactions
In a displacement reaction, a more reactive metal displaces a less reactive metal from its compound in solution. These reactions involve the transfer of electrons (redox reactions), so we can break them down into half-equations to identify which species are oxidized and reduced.
Here’s how to write ionic equations for some common displacement reactions and identify which species are oxidized.
- Displacement of Copper(II) Sulfate by Zinc
Reaction:
Zinc displaces copper from copper(II) sulfate solution.
Zn(s)
+
CuSO
4
(aq)
→
ZnSO
4
(aq)
+
Cu(s)
Zn(s)+CuSO
4
(aq)→ZnSO
4
(aq)+Cu(s)
Ionic Equation:
Zn(s)
+
Cu
2
+
(aq)
→
Zn
2
+
(aq)
+
Cu(s)
Zn(s)+Cu
2+
(aq)→Zn
2+
(aq)+Cu(s)
Half-equations:
Oxidation (at the zinc metal):
Zn(s)
→
Zn
2
+
(aq)
+
2
𝑒
−
Zn(s)→Zn
2+
(aq)+2e
−
Zinc loses electrons, so it is oxidized.
Reduction (at the copper(II) ions):
Cu
2
+
(aq)
+
2
𝑒
−
→
Cu(s)
Cu
2+
(aq)+2e
−
→Cu(s)
Copper ions gain electrons, so they are reduced.
Species Oxidized and Reduced:
Oxidized: Zinc (Zn) is oxidized to zinc ions (Zn²⁺).
Reduced: Copper ions (Cu²⁺) are reduced to copper metal (Cu).
2. Displacement of Magnesium from Magnesium Sulfate by Calcium
Reaction:
Calcium displaces magnesium from magnesium sulfate solution.
Ca(s)
+
MgSO
4
(aq)
→
CaSO
4
(aq)
+
Mg(s)
Ca(s)+MgSO
4
(aq)→CaSO
4
(aq)+Mg(s)
Ionic Equation:
Ca(s)
+
Mg
2
+
(aq)
→
Ca
2
+
(aq)
+
Mg(s)
Ca(s)+Mg
2+
(aq)→Ca
2+
(aq)+Mg(s)
Half-equations:
Oxidation (at the calcium metal):
Ca(s)
→
Ca
2
+
(aq)
+
2
𝑒
−
Ca(s)→Ca
2+
(aq)+2e
−
Calcium loses electrons, so it is oxidized.
Reduction (at the magnesium ions):
Mg
2
+
(aq)
+
2
𝑒
−
→
Mg(s)
Mg
2+
(aq)+2e
−
→Mg(s)
Magnesium ions gain electrons, so they are reduced.
Species Oxidized and Reduced:
Oxidized: Calcium (Ca) is oxidized to calcium ions (Ca²⁺).
Reduced: Magnesium ions (Mg²⁺) are reduced to magnesium metal (Mg).
3. Displacement of Hydrogen from Hydrochloric Acid by Zinc
Reaction:
Zinc displaces hydrogen from hydrochloric acid, producing hydrogen gas.
Zn(s)
+
2
HCl(aq)
→
ZnCl
2
(aq)
+
H
2
(g)
Zn(s)+2HCl(aq)→ZnCl
2
(aq)+H
2
(g)
Ionic Equation:
Zn(s)
+
2
H
+
(aq)
→
Zn
2
+
(aq)
+
H
2
(g)
Zn(s)+2H
+
(aq)→Zn
2+
(aq)+H
2
(g)
Half-equations:
Oxidation (at the zinc metal):
Zn(s)
→
Zn
2
+
(aq)
+
2
𝑒
−
Zn(s)→Zn
2+
(aq)+2e
−
Zinc loses electrons, so it is oxidized.
Reduction (at the hydrogen ions):
2
H
+
(aq)
+
2
𝑒
−
→
H
2
(g)
2H
+
(aq)+2e
−
→H
2
(g)
Hydrogen ions gain electrons to form hydrogen gas, so they are reduced.
Species Oxidized and Reduced:
Oxidized: Zinc (Zn) is oxidized to zinc ions (Zn²⁺).
Reduced: Hydrogen ions (H⁺) are reduced to hydrogen gas (H₂).
4. Displacement of Copper(II) Sulfate by Iron
Reaction:
Iron displaces copper from copper(II) sulfate solution.
Fe(s)
+
CuSO
4
(aq)
→
FeSO
4
(aq)
+
Cu(s)
Fe(s)+CuSO
4
(aq)→FeSO
4
(aq)+Cu(s)
Ionic Equation:
Fe(s)
+
Cu
2
+
(aq)
→
Fe
2
+
(aq)
+
Cu(s)
Fe(s)+Cu
2+
(aq)→Fe
2+
(aq)+Cu(s)
Half-equations:
Oxidation (at the iron metal):
Fe(s)
→
Fe
2
+
(aq)
+
2
𝑒
−
Fe(s)→Fe
2+
(aq)+2e
−
Iron loses electrons, so it is oxidized.
Reduction (at the copper(II) ions):
Cu
2
+
(aq)
+
2
𝑒
−
→
Cu(s)
Cu
2+
(aq)+2e
−
→Cu(s)
Copper ions gain electrons, so they are reduced.
Species Oxidized and Reduced:
Oxidized: Iron (Fe) is oxidized to iron(II) ions (Fe²⁺).
Reduced: Copper ions (Cu²⁺) are reduced to copper metal (Cu).
Key Points:
Displacement reactions involve a more reactive metal displacing a less reactive metal or ion from its compound.
The metal that loses electrons is oxidized (the oxidation half-reaction).
The ion that gains electrons is reduced (the reduction half-reaction).
The ionic equations show only the ions and atoms that are involved in the displacement, excluding spectator ions (ions that do not change during the reaction).
By writing out the ionic equations and identifying the half-reactions, you can determine which species are oxidized and reduced in any displacement reaction.
explain in terms of gain or loss of electrons that the reactions between acids and some metals are redox reactions, and identify which species are oxidised and which are reduced.
Acid-Metal Reactions as Redox Reactions
When metals react with acids, they often undergo redox reactions. In these reactions, the metal loses electrons (oxidation) and the hydrogen ions from the acid gain electrons (reduction). Let’s break it down in terms of gain or loss of electrons and identify which species are oxidised and which are reduced.
General Reaction Between Metals and Acids
The general reaction between a metal (M) and a dilute acid (e.g., HCl) produces a metal salt and hydrogen gas (H₂).
For example, when zinc reacts with hydrochloric acid (HCl), the reaction is:
Zn(s)
+
2
HCl(aq)
→
ZnCl
2
(aq)
+
H
2
(g)
Zn(s)+2HCl(aq)→ZnCl
2
(aq)+H
2
(g)
In this reaction:
Zinc (Zn) reacts with hydrochloric acid (HCl) to form zinc chloride (ZnCl₂) and hydrogen gas (H₂).
Identifying Oxidation and Reduction in Acid-Metal Reactions
Let’s consider the oxidation states and electron transfer during the reaction:
- Oxidation of the Metal:
Zinc (Zn) starts in its elemental form (oxidation state 0) and ends up as zinc ions (Zn²⁺) in zinc chloride (ZnCl₂) (oxidation state +2).
To go from Zn (0) to Zn²⁺ (aq), zinc loses two electrons.
Oxidation half-reaction:
Zn(s)
→
Zn
2
+
(aq)
+
2
𝑒
−
Zn(s)→Zn
2+
(aq)+2e
−
This is oxidation because zinc loses electrons.
Species Oxidised:
Zinc (Zn) is oxidized to zinc ions (Zn²⁺).
2. Reduction of Hydrogen Ions:
Hydrogen ions (H⁺) in the acid (such as HCl) gain electrons to form hydrogen gas (H₂).
Each H⁺ ion gains one electron to form half a molecule of H₂.
Reduction half-reaction:
2
H
+
(aq)
+
2
𝑒
−
→
H
2
(g)
2H
+
(aq)+2e
−
→H
2
(g)
This is reduction because hydrogen ions gain electrons to form hydrogen gas.
Species Reduced:
Hydrogen ions (H⁺) are reduced to hydrogen gas (H₂).
Summary of Redox Process:
In a reaction between a metal and an acid, like the reaction of zinc with hydrochloric acid, we have the following:
Oxidation (loss of electrons): The metal (e.g., Zn) is oxidized to form metal ions (e.g., Zn²⁺), losing electrons.
Reduction (gain of electrons): The hydrogen ions (H⁺) in the acid are reduced to form hydrogen gas (H₂), gaining electrons.
Example with Magnesium and Hydrochloric Acid
When magnesium reacts with hydrochloric acid, the reaction is:
Mg(s)
+
2
HCl(aq)
→
MgCl
2
(aq)
+
H
2
(g)
Mg(s)+2HCl(aq)→MgCl
2
(aq)+H
2
(g)
1. Oxidation of Magnesium:
Magnesium (Mg) loses two electrons to form magnesium ions (Mg²⁺).
Oxidation half-reaction:
Mg(s)
→
Mg
2
+
(aq)
+
2
𝑒
−
Mg(s)→Mg
2+
(aq)+2e
−
Magnesium is oxidized because it loses electrons.
2. Reduction of Hydrogen Ions:
The hydrogen ions (H⁺) from the acid gain electrons to form hydrogen gas (H₂).
Reduction half-reaction:
2
H
+
(aq)
+
2
𝑒
−
→
H
2
(g)
2H
+
(aq)+2e
−
→H
2
(g)
Hydrogen ions (H⁺) are reduced because they gain electrons.
Summary of Oxidation and Reduction:
Oxidized: Magnesium (Mg) is oxidized to magnesium ions (Mg²⁺).
Reduced: Hydrogen ions (H⁺) are reduced to hydrogen gas (H₂).
Key Points:
The reaction between a metal and an acid is a redox reaction because there is both oxidation (loss of electrons) and reduction (gain of electrons).
The metal is oxidized because it loses electrons and forms positive ions.
The hydrogen ions (H⁺) in the acid are reduced because they gain electrons to form hydrogen gas (H₂).
explain that acids can be neutralised by alkalis, bases and metal carbonates and list the products of each of these reactions.
Neutralisation Reactions:
Neutralisation occurs when an acid reacts with a base, alkali, or metal carbonate, resulting in the formation of water and a salt. The general formula for a neutralisation reaction is:
Acid
+
Base/Alkali/MetalCarbonate
→
Salt
+
Water(ifacidandalkali)
+
CarbonDioxide(ifmetalcarbonate)
Acid+Base/Alkali/MetalCarbonate→Salt+Water(ifacidandalkali)+CarbonDioxide(ifmetalcarbonate)
The H⁺ ions from the acid react with the OH⁻ ions from the base or alkali, or with the CO₃²⁻ ions from the metal carbonate.
- Neutralisation Between Acids and Alkalis:
An alkali is a soluble base that dissolves in water to release OH⁻ (hydroxide) ions.
Example Reaction:
When hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH), the products are water and sodium chloride (NaCl).
HCl(aq)
+
NaOH(aq)
→
NaCl(aq)
+
H
2
O(l)
HCl(aq)+NaOH(aq)→NaCl(aq)+H
2
O(l)
Acid: Hydrochloric acid (HCl)
Alkali: Sodium hydroxide (NaOH)
Salt: Sodium chloride (NaCl)
Water: H₂O
Products:
Water (from the neutralisation of H⁺ and OH⁻)
Salt (e.g., sodium chloride, NaCl)
General Equation for Neutralisation with Alkalis:
Acid
+
Alkali
→
Salt
+
Water
Acid+Alkali→Salt+Water
2. Neutralisation Between Acids and Bases:
A base is a substance that can accept protons (H⁺) or donate electron pairs. Bases are typically insoluble in water, unlike alkalis which are soluble.
Example Reaction:
When sulfuric acid (H₂SO₄) reacts with copper(II) oxide (CuO) (a base), the products are water and copper(II) sulfate (CuSO₄).
H
2
SO
4
(aq)
+
CuO(s)
→
CuSO
4
(aq)
+
H
2
O(l)
H
2
SO
4
(aq)+CuO(s)→CuSO
4
(aq)+H
2
O(l)
Acid: Sulfuric acid (H₂SO₄)
Base: Copper(II) oxide (CuO)
Salt: Copper(II) sulfate (CuSO₄)
Water: H₂O
Products:
Water (from the neutralisation of H⁺ and OH⁻)
Salt (e.g., copper(II) sulfate, CuSO₄)
General Equation for Neutralisation with Bases:
Acid
+
Base
→
Salt
+
Water
Acid+Base→Salt+Water
3. Neutralisation Between Acids and Metal Carbonates:
A metal carbonate reacts with an acid to produce a salt, water, and carbon dioxide (CO₂). The carbonate ions (CO₃²⁻) react with the hydrogen ions (H⁺) from the acid to produce carbonic acid (H₂CO₃), which then decomposes into water and carbon dioxide.
Example Reaction:
When hydrochloric acid (HCl) reacts with sodium carbonate (Na₂CO₃), the products are sodium chloride (NaCl), water (H₂O), and carbon dioxide (CO₂).
HCl(aq)
+
Na
2
CO
3
(aq)
→
2
NaCl(aq)
+
H
2
O(l)
+
CO
2
(g)
HCl(aq)+Na
2
CO
3
(aq)→2NaCl(aq)+H
2
O(l)+CO
2
(g)
Acid: Hydrochloric acid (HCl)
Metal Carbonate: Sodium carbonate (Na₂CO₃)
Salt: Sodium chloride (NaCl)
Water: H₂O
Carbon Dioxide: CO₂ (gas)
Products:
Salt (e.g., sodium chloride, NaCl)
Water (from the neutralisation of H⁺ and CO₃²⁻)
Carbon Dioxide (CO₂) (gas)
General Equation for Neutralisation with Metal Carbonates:
Acid
+
MetalCarbonate
→
Salt
+
Water
+
CarbonDioxide
Acid+MetalCarbonate→Salt+Water+CarbonDioxide
Summary of the Three Types of Neutralisation Reactions:
Acid + Alkali → Salt + Water
Example: HCl + NaOH → NaCl + H₂O
Acid + Base → Salt + Water
Example: H₂SO₄ + CuO → CuSO₄ + H₂O
Acid + Metal Carbonate → Salt + Water + Carbon Dioxide
Example: HCl + Na₂CO₃ → NaCl + H₂O + CO₂
Conclusion:
Acids neutralize alkalis to form salt and water.
Acids neutralize bases to form salt and water.
Acids neutralize metal carbonates to form salt, water, and carbon dioxide.
These reactions are important in many chemical processes, such as the treatment of acid indigestion with antacids (bases or alkalis) and the production of salts in laboratories and industry.
predict the salt produced in a neautralision reaction based of the acid used and the positive ions in the base, alkali or carbonate and the use the formulae of common ions to deduce the formulae of the salt.
Predicting the Salt Produced in a Neutralisation Reaction
In neutralisation reactions, the acid and the base/alkali/metal carbonate react to form a salt and, depending on the reactants, water and/or carbon dioxide. To predict the salt produced, we need to identify the acid and the base (or alkali or metal carbonate) and use the positive (cation) and negative (anion) ions to deduce the formula of the salt.
The formula of the salt will be determined by combining the cation from the base or alkali and the anion from the acid.
Steps to Predict the Salt Produced:
Identify the Acid:
An acid provides H⁺ ions (hydrogen ions).
Common acids:
Hydrochloric acid (HCl) → Cl⁻ ion (chloride).
Sulfuric acid (H₂SO₄) → SO₄²⁻ ion (sulfate).
Nitric acid (HNO₃) → NO₃⁻ ion (nitrate).
Phosphoric acid (H₃PO₄) → PO₄³⁻ ion (phosphate).
Identify the Base, Alkali, or Metal Carbonate:
A base or alkali provides OH⁻ ions (hydroxide ions) or a metal ion for the salt formation.
Example bases: Sodium hydroxide (NaOH) → Na⁺.
Example alkalis: Potassium hydroxide (KOH) → K⁺.
Example metal carbonates: Calcium carbonate (CaCO₃) → Ca²⁺ and CO₃²⁻.
Combine the Positive and Negative Ions to Form the Salt:
The positive ion (cation) from the base/alkali/metal carbonate combines with the negative ion (anion) from the acid to form the salt.
Balance the Charges to Find the Formula:
The charges of the ions must balance to form a neutral salt.
Examples:
1. Acid + Alkali Reaction:
Acid: Hydrochloric acid (HCl) → Cl⁻ (chloride).
Alkali: Sodium hydroxide (NaOH) → Na⁺ (sodium).
The salt produced will be Sodium chloride (NaCl).
Formula:
Combine Na⁺ and Cl⁻ to form NaCl.
HCl(aq)
+
NaOH(aq)
→
NaCl(aq)
+
H
2
O(l)
HCl(aq)+NaOH(aq)→NaCl(aq)+H
2
O(l)
2. Acid + Base Reaction:
Acid: Sulfuric acid (H₂SO₄) → SO₄²⁻ (sulfate).
Base: Copper(II) oxide (CuO) → Cu²⁺ (copper).
The salt produced will be Copper(II) sulfate (CuSO₄).
Formula:
Combine Cu²⁺ and SO₄²⁻ to form CuSO₄.
H
2
SO
4
(aq)
+
CuO(s)
→
CuSO
4
(aq)
+
H
2
O(l)
H
2
SO
4
(aq)+CuO(s)→CuSO
4
(aq)+H
2
O(l)
3. Acid + Metal Carbonate Reaction:
Acid: Nitric acid (HNO₃) → NO₃⁻ (nitrate).
Metal Carbonate: Calcium carbonate (CaCO₃) → Ca²⁺ (calcium) and CO₃²⁻ (carbonate).
The salt produced will be Calcium nitrate (Ca(NO₃)₂).
Formula:
Combine Ca²⁺ and NO₃⁻ to form Ca(NO₃)₂.
HNO
3
(aq)
+
CaCO
3
(s)
→
Ca(NO
3
)
2
(aq)
+
H
2
O(l)
+
CO
2
(g)
HNO
3
(aq)+CaCO
3
(s)→Ca(NO
3
)
2
(aq)+H
2
O(l)+CO
2
(g)
4. Acid + Alkali Reaction:
Acid: Phosphoric acid (H₃PO₄) → PO₄³⁻ (phosphate).
Alkali: Potassium hydroxide (KOH) → K⁺ (potassium).
The salt produced will be Potassium phosphate (K₃PO₄).
Formula:
Combine K⁺ and PO₄³⁻. To balance the charges, you need three potassium ions to balance the negative three charge on the phosphate ion.
H
3
PO
4
(aq)
+
3
KOH(aq)
→
K
3
PO
4
(aq)
+
3
H
2
O(l)
H
3
PO
4
(aq)+3KOH(aq)→K
3
PO
4
(aq)+3H
2
O(l)
General Guidelines for Predicting the Salt Formula:
From an Acid and an Alkali (Base) Reaction:
The salt’s cation comes from the alkali or base (e.g., Na⁺, K⁺, Ca²⁺).
The salt’s anion comes from the acid (e.g., Cl⁻ from HCl, SO₄²⁻ from H₂SO₄).
From an Acid and a Metal Carbonate Reaction:
The salt’s cation comes from the metal carbonate (e.g., Na⁺, Ca²⁺).
The salt’s anion comes from the acid (e.g., CO₃²⁻ from CaCO₃ or Na₂CO₃).
Balance the Charges:
The total positive charge from the cations must balance the total negative charge from the anions.
If necessary, adjust the number of ions in the formula to ensure charge balance.
Conclusion:
To predict the salt produced in a neutralisation reaction, follow these steps:
Identify the acid and its anion (Cl⁻, SO₄²⁻, NO₃⁻, etc.).
Identify the cation from the base, alkali, or metal carbonate (Na⁺, K⁺, Cu²⁺, Ca²⁺, etc.).
Combine the cation and anion to form the salt, ensuring that the charges balance out.
By following these steps, you can predict the salt formed and deduce its formula in neutralisation reactions.
describe how soluble salts can be made from acids and how pure, dry samples of salts can be obtained.
Making Soluble Salts from Acids:
Soluble salts are salts that dissolve in water to form a solution. They can be made by reacting an acid with a base, alkali, or metal carbonate. The process involves neutralising the acid with the appropriate substance, and then isolating the salt from the resulting solution.
Here’s a step-by-step guide on how to make soluble salts from acids:
- Reaction of Acid with an Alkali or Base
General Reaction:
Acid + Base/Alkali → Salt + Water.
For example, when hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH), sodium chloride (NaCl) salt and water are produced.
HCl(aq)
+
NaOH(aq)
→
NaCl(aq)
+
H
2
O(l)
HCl(aq)+NaOH(aq)→NaCl(aq)+H
2
O(l)
Procedure for Making Soluble Salt:
Add Base/Alkali to Acid: Slowly add the base (e.g., sodium hydroxide) to the acid (e.g., hydrochloric acid) while stirring.
Continue Adding Until Neutralisation: Add the base until all of the acid has reacted, which can be checked using pH paper or an indicator (e.g., phenolphthalein).
Forming the Salt: The resulting solution will contain the salt (e.g., sodium chloride, NaCl) dissolved in water.
2. Reaction of Acid with a Metal Carbonate
General Reaction:
Acid + Metal Carbonate → Salt + Water + Carbon Dioxide.
For example, when sulfuric acid (H₂SO₄) reacts with copper(II) carbonate (CuCO₃), copper(II) sulfate (CuSO₄) salt, water, and carbon dioxide (CO₂) are produced.
H
2
SO
4
(aq)
+
CuCO
3
(s)
→
CuSO
4
(aq)
+
H
2
O(l)
+
CO
2
(g)
H
2
SO
4
(aq)+CuCO
3
(s)→CuSO
4
(aq)+H
2
O(l)+CO
2
(g)
Procedure for Making Soluble Salt:
Add Metal Carbonate to Acid: Slowly add copper(II) carbonate (or another metal carbonate) to sulfuric acid (or another acid) in excess.
React Until Carbonate is in Excess: Ensure that enough metal carbonate is added to neutralise all the acid.
Filter the Solution: After the reaction, filter the mixture to remove any unreacted excess carbonate.
Evaporate to Get the Salt: The solution contains the dissolved salt (e.g., copper(II) sulfate). To get the salt, heat the solution to evaporate the water, leaving the salt behind.
How to Obtain Pure, Dry Samples of Soluble Salts:
Once the soluble salt has been formed, the next step is to purify and obtain a pure, dry sample. Here’s the method:
Steps to Obtain Pure, Dry Salt from a Soluble Solution:
1. Filtration (If Necessary)
If the reaction involves a metal carbonate, there may be insoluble residues (e.g., unreacted carbonate). You can filter out these impurities using filter paper and a funnel.
2. Evaporation to Crystallise the Salt
Heat the Solution: To obtain the salt, heat the filtered solution to evaporate the water. Use gentle heating (e.g., on a water bath) to avoid decomposition of the salt.
Crystals Form: As the water evaporates, the salt will start to crystallise. Once all the water has evaporated, solid crystals of the salt will remain.
3. Crystallisation (If Evaporation is Incomplete)
For more pure and large crystals of the salt, you can use crystallisation:
Heat the solution gently to remove most of the water, leaving the solution more concentrated.
Allow the solution to cool slowly. As it cools, the salt will crystallise.
Filter the Crystals: After cooling, filter out the salt crystals using filter paper.
Dry the Crystals: To obtain dry salt, place the crystals on a watch glass or filter paper and leave them to air dry. Alternatively, you can dry them in an oven at a low temperature.
required practical: preparation of pure, dry sample of a soluble salt from an insoluble oxide or carbonate using a bunsen burner to heat dilute acid and a water bath or electric heater to evaporate the solution.
Required Practical: Preparation of a Pure, Dry Sample of a Soluble Salt from an Insoluble Oxide or Carbonate
In this practical, we will prepare a pure, dry sample of a soluble salt by reacting an insoluble oxide or carbonate with an acid. We will use a Bunsen burner to heat the acid, a water bath or electric heater to evaporate the solution, and filtration to remove any excess insoluble material.
Objective:
To prepare a pure, dry sample of a soluble salt (such as copper(II) sulfate (CuSO₄)) from an insoluble carbonate (e.g., copper(II) carbonate).
Apparatus and Chemicals:
Apparatus:
Bunsen burner
Beaker (for the acid)
Conical flask
Stirring rod
Filter paper
Funnel
Evaporating dish
Water bath or electric heater
Glass rod (for crystallisation)
Watch glass or filter paper (for drying)
Chemicals:
Dilute sulfuric acid (H₂SO₄) (or another suitable acid, depending on the salt to be made)
Copper(II) carbonate (CuCO₃) (or another insoluble oxide or carbonate)
Method:
Step 1: Reacting the Acid with the Insoluble Oxide or Carbonate
Measure the acid: Pour a measured volume of dilute sulfuric acid into a beaker.
Heat the acid: Place the beaker on a Bunsen burner and gently heat the acid to speed up the reaction. Do not let the acid boil.
Add the insoluble oxide or carbonate: Slowly add the copper(II) carbonate (or other insoluble oxide or carbonate) to the hot acid. Stir the mixture with a stirring rod.
Observe the reaction: The insoluble copper(II) carbonate will react with the acid to form soluble copper(II) sulfate (CuSO₄), water (H₂O), and carbon dioxide (CO₂) gas, which will bubble out.
Reaction equation:
H
2
SO
4
(aq)
+
CuCO
3
(s)
→
CuSO
4
(aq)
+
H
2
O(l)
+
CO
2
(g)
H
2
SO
4
(aq)+CuCO
3
(s)→CuSO
4
(aq)+H
2
O(l)+CO
2
(g)
Ensure excess carbonate: Continue adding copper(II) carbonate until the acid is fully neutralised. The reaction stops when no more bubbles of CO₂ are produced.
Step 2: Filtration
Set up filtration: Place a funnel with filter paper into a clean conical flask or beaker.
Filter the solution: Pour the reaction mixture into the filter paper. The insoluble copper(II) carbonate will remain on the filter paper, while the soluble copper(II) sulfate solution will pass through into the receiving flask.
Step 3: Evaporation
Transfer the solution: Transfer the filtered solution of copper(II) sulfate to an evaporating dish.
Heat the solution: Place the evaporating dish on a water bath or use an electric heater to gently heat the solution. Do not use a Bunsen burner directly under the evaporating dish, as the solution may boil and cause splashing.
Water bath: Set up a beaker of hot water in which the evaporating dish can sit without the solution boiling too violently.
Evaporate the water: Heat the solution until the volume is reduced and crystals of copper(II) sulfate begin to form. Allow the solution to continue evaporating until saturation occurs, and no more water can evaporate.
Step 4: Crystallisation and Drying
Allow the solution to cool: Let the remaining solution cool and crystallise at room temperature. This process may take several hours.
Filter the crystals: Once the crystals of copper(II) sulfate have formed, filter them using filter paper to remove any remaining solution.
Dry the crystals: To obtain dry copper(II) sulfate crystals, place them on a watch glass or filter paper and leave them to air dry. Alternatively, you can gently dry the crystals in an oven at a low temperature to speed up the drying process.
Safety Precautions:
Wear safety goggles and lab coat to protect from splashes.
Handle the Bunsen burner carefully to avoid burns. Keep flammable materials away from the flame.
Ensure you use dilute acid to minimize risks of burns.
When heating, ensure that the water bath is used for gentle heating to avoid boiling or splashing.
If using an oven to dry crystals, make sure the temperature is low to avoid decomposition of the salt.
Conclusion:
By following this procedure, you will have successfully prepared a pure, dry sample of a soluble salt (in this example, copper(II) sulfate) from an insoluble oxide or carbonate. The key steps include:
Reacting the acid with the insoluble oxide/carbonate.
Filtering the resulting mixture to remove excess insoluble material.
Evaporating the solution using a water bath or electric heater to obtain the salt crystals.
Drying the salt crystals to get a pure, dry sample.
This experiment demonstrates the process of preparing salts in the laboratory, using neutralisation, filtration, evaporation, and crystallisation techniques.
recall what the ph scale measures and describe the scales used to identify acidic, neutral or alkaline solutions.
The pH Scale:
The pH scale measures the acidity or alkalinity (basicity) of a solution. It is a scale from 0 to 14 that indicates the concentration of hydrogen ions (H⁺) in a solution. The pH scale helps us identify whether a solution is acidic, neutral, or alkaline.
Acidic solutions have a pH less than 7.
Neutral solutions have a pH of exactly 7.
Alkaline (basic) solutions have a pH greater than 7.
How the pH Scale Works:
A pH of 0 is the most acidic (e.g., strong acids like hydrochloric acid, HCl).
A pH of 7 is neutral (e.g., pure water).
A pH of 14 is the most alkaline (e.g., strong bases like sodium hydroxide, NaOH).
The pH scale is logarithmic, which means each whole number on the scale represents a tenfold difference in the concentration of hydrogen ions (H⁺). For example:
A solution with a pH of 3 is 10 times more acidic than a solution with a pH of 4.
A solution with a pH of 12 is 10 times more alkaline than a solution with a pH of 11.
Acidic, Neutral, and Alkaline Solutions:
1. Acidic Solutions (pH < 7):
pH range: 0 to 6.
Characteristics:
High concentration of hydrogen ions (H⁺).
Have a sour taste (e.g., lemon juice, vinegar).
Turn blue litmus paper red.
Examples of acids: Hydrochloric acid (HCl), Sulfuric acid (H₂SO₄), Citric acid.
2. Neutral Solutions (pH = 7):
pH range: Exactly 7.
Characteristics:
Equal concentrations of hydrogen ions (H⁺) and hydroxide ions (OH⁻).
Neither acidic nor alkaline.
Pure water is neutral.
Neutral litmus paper stays purple.
3. Alkaline (Basic) Solutions (pH > 7):
pH range: 8 to 14.
Characteristics:
High concentration of hydroxide ions (OH⁻).
Have a bitter taste and feel slippery (e.g., soap, ammonia).
Turn red litmus paper blue.
Examples of alkalis: Sodium hydroxide (NaOH), Ammonia (NH₃), Potassium hydroxide (KOH).
Identifying Acidity, Neutrality, or Alkalinity:
You can identify whether a solution is acidic, neutral, or alkaline using the following methods:
- pH Indicator Paper:
pH indicator paper, also known as universal indicator paper, changes color depending on the pH of the solution. The paper can be compared to a color chart to determine the pH. - pH Probe or pH Meter:
A pH meter or pH probe gives an accurate reading of the pH value of a solution. - Litmus Paper:
Blue litmus paper turns red in acidic solutions and stays blue in alkaline solutions.
Red litmus paper turns blue in alkaline solutions and stays red in acidic solutions.
Summary:
The pH scale ranges from 0 to 14, measuring how acidic or alkaline a solution is.
Acidic solutions have a pH less than 7.
Neutral solutions have a pH of 7.
Alkaline (basic) solutions have a pH greater than 7.
The scale is logarithmic, so each unit represents a tenfold difference in hydrogen ion concentration.
define the terms acid and alkali in terms of production of hydrogen ions or hyrdoxide ions. define the term base.
Acid: Produces H⁺ ions when dissolved in water.
Alkali: Produces OH⁻ ions when dissolved in water.
Base: A substance that can neutralise an acid (may or may not dissolve in water).
describe the use of universal indicator to measure the approximate pH of a solution and use the pH scale to identify acidic or alkaline solutions
Summary:
Universal indicator is used to measure the approximate pH of a solution by changing colors.
The color of the solution is compared to a pH color chart to estimate the pH value.
Solutions with a pH less than 7 are acidic, solutions with a pH of 7 are neutral, and solutions with a pH greater than 7 are alkaline.
describe how to carry out titrations using strong acids and strong alkalis only
ConcentrationofNaOH
Step-by-Step Procedure:
Step 1: Prepare the Titration Setup
Fill the burette with the strong acid (e.g., hydrochloric acid). Use a funnel to fill the burette, making sure to rinse it with the acid solution first to avoid contamination.
Record the initial volume of the acid in the burette.
Pipette a known volume of the strong alkali (e.g., sodium hydroxide solution) into a clean conical flask. Usually, 25.0 cm³ of alkali is used.
Add a few drops of indicator to the alkali solution in the conical flask.
If using phenolphthalein, the solution will be pink in an alkaline solution.
If using methyl orange, the solution will be yellow in an alkaline solution.
Step 2: Carry Out the Titration
Place the conical flask on a white tile to make it easier to see the color change during the titration.
Slowly add the acid from the burette into the alkali solution while constantly swirling the conical flask to mix the solutions.
Add the acid dropwise near the endpoint, where you expect the color change to occur, to avoid overshooting the endpoint.
Observe the color change:
With phenolphthalein: The color will change from pink to colorless when the neutral point (equivalence point) is reached.
With methyl orange: The color will change from yellow to red as the pH moves from alkaline to slightly acidic.
Step 3: End the Titration
Once you observe the color change, note the final volume of acid in the burette.
The volume of acid used is the difference between the initial and final burette readings.
Step 4: Calculate the Concentration of the Alkali
Use the titration equation to calculate the concentration of the alkali:
For HCl (acid) and NaOH (alkali), the reaction is:
HCl(aq)
+
NaOH(aq)
→
NaCl(aq)
+
H
2
O(l)
HCl(aq)+NaOH(aq)→NaCl(aq)+H
2
O(l)
The equation shows a 1:1 molar ratio between the acid and alkali.
The formula to use is:
(
ConcentrationofHCl
)
×
(
VolumeofHCl
)
VolumeofNaOH
ConcentrationofNaOH=
VolumeofNaOH
(ConcentrationofHCl)×(VolumeofHCl)
Where:
Concentration of HCl is known.
Volume of HCl is the volume measured from the burette.
Volume of NaOH is the volume of alkali you used in the titration (usually measured by pipetting).
Repeat for Accuracy:
It’s important to repeat the titration several times (usually 3 times) to get consistent results. You can then average the results to improve accuracy.
Safety Precautions:
Wear safety goggles and a lab coat to protect from splashes.
Be careful when handling the strong acid and strong alkali as they can cause burns.
Ensure good ventilation when performing titrations with strong acids or alkalis.
Summary:
In a titration using strong acids and strong alkalis, you add the acid from a burette to a measured volume of alkali in a conical flask, using an indicator to determine the endpoint (color change). By knowing the volume and concentration of the acid, you can calculate the concentration of the alkali using the titration formula. Repeating the titration ensures accuracy in the results.
calculate the chemical quantities in titrations involving concentrations in mol/dm3 and g/dm3
Number of moles = Concentration (mol/dm³) × Volume (dm³)
Concentration (mol/dm³) = Number of moles ÷ Volume (dm³)
Mass (g) = Concentration (mol/dm³) × Volume (dm³) × Molar mass (g/mol)
Concentration (g/dm³) can be calculated by converting from mol/dm³ using the molar mass.
These calculations allow you to work out the chemical quantities involved in titrations and solutions involving strong acids and strong alkalis.
required practical 2- determination of the reacting volumes of solutions of a strong acid and a strong alkali by titrations.
Step 1: Preparing the Titration Setup
Fill the burette with the strong acid (e.g., HCl). Use a funnel to fill the burette, but first rinse the burette with the acid solution to ensure it is clean.
Record the initial volume of acid in the burette.
Pipette a known volume of the strong alkali (e.g., NaOH) into a clean conical flask. Usually, 25.0 cm³ of alkali is used.
Add a few drops of indicator to the alkali solution in the conical flask. The indicator helps you identify the endpoint (when neutralization occurs).
If using phenolphthalein: The solution will turn pink in alkaline conditions and will become colorless when neutralized.
If using methyl orange: The solution will be yellow in alkaline conditions and will change to red when neutralized.
Step 2: Performing the Titration
Place the conical flask on a white tile to easily observe the color change.
Slowly add the acid from the burette to the alkali in the conical flask, while gently swirling the flask to mix the two solutions.
As you approach the endpoint, add the acid drop by drop to prevent overshooting.
Observe the color change:
With phenolphthalein: The solution will change from pink to colorless when all the alkali has been neutralized.
With methyl orange: The solution will change from yellow to red when the neutralization point is reached.
Step 3: Completing the Titration
Once you observe the color change, note the final volume of acid in the burette.
Calculate the volume of acid used by subtracting the initial volume from the final volume.
Step 4: Repeat for Accuracy
Repeat the titration at least three times and take the average of the results to ensure accuracy and reliability.
use and explain the terms dilute and concentrated in terms of amount and substance and weak and strong relation to adults. in key points only
Dilute and Concentrated refer to the amount of solute in a solution.
Weak and Strong refer to the degree of ionization or strength of the acid in water.
Dilute vs. Concentrated:
Dilute: A solution that has a small amount of solute (substance) compared to the solvent (usually water).
Example: A small amount of acid in a large volume of water.
Concentrated: A solution that has a large amount of solute compared to the solvent.
Example: A large amount of acid in a small volume of water.
Weak vs. Strong (Acid Strength):
Weak Acid: An acid that partially ionizes in water, meaning only some of its molecules release hydrogen ions (H⁺).
Example: Citric acid or acetic acid.
Strong Acid: An acid that completely ionizes in water, meaning most or all of its molecules release hydrogen ions (H⁺).
Example: Hydrochloric acid (HCl) or sulfuric acid (H₂SO₄).
explain how the concentration of an aqueous solution and the strength of an acid affects the pH of the solution and how pH is related to the hydrogen ion concentration of a solution
pH
. Concentration of an Aqueous Solution and pH:
Concentration refers to the amount of acid dissolved in a solution, typically measured in mol/dm³ (molarity). The higher the concentration of the acid, the lower the pH (more acidic).
In general:
Higher concentration of an acid → More H⁺ ions released → Lower pH (more acidic).
Lower concentration of an acid → Fewer H⁺ ions released → Higher pH (less acidic).
2. Strength of an Acid and pH:
Strength of an acid refers to the degree of ionization in water (how many H⁺ ions the acid produces).
A strong acid completely dissociates in water, releasing many H⁺ ions, resulting in a low pH (very acidic).
A weak acid only partially dissociates in water, releasing fewer H⁺ ions, resulting in a higher pH compared to a strong acid of the same concentration.
3. pH and Hydrogen Ion Concentration (H⁺):
pH is related to the concentration of hydrogen ions (H⁺) in a solution. The formula is:
−
log
[
H
+
]
pH=−log[H
+
]
where [H⁺] is the concentration of hydrogen ions in mol/dm³.
Lower pH means a higher concentration of H⁺ ions, indicating a more acidic solution.
Higher pH means a lower concentration of H⁺ ions, indicating a less acidic or more alkaline solution.
Summary:
Concentration: Higher concentration of an acid → lower pH (more acidic).
Strength: Strong acids → more H⁺ ions → lower pH; weak acids → fewer H⁺ ions → higher pH.
pH is a measure of the concentration of H⁺ ions: the higher the H⁺ concentration, the lower the pH.
