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ACID BASE
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ACIDS
What is acid?
An acid is a chemical substance, which ionizes in water to produce hydrogen ions, H+
When acids dissolved in water, the hydrogen atoms in acids are released as hydrogen ions, H+.
The hydrogen ion combined with water molecule, H2O to form a hydroxonium ion, H3O+. However, this ion can be written as H+.
We can classify an acid as a monoprotic acid or a diprotic acid based on basicity.
Basicity of an acid is the number of mole of H+ ion that can be produced by one mole of acid when it dissolves in water.
Monoprotic acidIonisation
Hydrochloric acid, HCl
Ethanoic acid, CH3COOH
Diprotic acidIonisation
Sulphuric acid, H2SO4
BASESWhat is base?
A base is a chemical substance that can neutralize an acid to produce a salt and water.
Examples of bases are;
i. metal hydroxidesii. metal oxide
Most bases are not soluble in water. Bases that are soluble in water are known as alkalis.
Bases that are insoluble in waterBases that are soluble in water
Zinc oxide, ZnOSodium oxide, Na2O
Zinc hydroxide, Zn(OH)2Sodium hydroxide, NaOH
Copper(II) oxide, CuO2Potassium oxide, K2O
Copper(II) hydroxide, Cu(OH)2 Potassium hydroxide, KOH
Calcium hydroxide, Ca(OH)2
Ammonia, NH3
Role of water and the properties of acids
Results
AcidConditionObservationInference
Ethanoic acid(molecule)Glacial (dry)- No colour change in the litmus paper- Light bulb is not lighted upDoes not show any acidic properties
Aqueous - Blue litmus paper is changed to red- Light bulb is lighted upShows acidic properties
Hydrogen chloride(molecule)Dissolved in methylbenzene- No colour change in the litmus paper
- Light bulb is not lighted upDoes not show any acidic properties
Aqueous- Blue litmus paper is
changed to red
- Light bulb is lighted upShows acidic properties
Discussion
The hydrogen ions, H+ are responsible for acids to show their properties1. Glacial/dry ethanoic acid does not show acidic properties because without water, ethanoic acid
remains or exists as molecules and there are no hydrogen ions present. 2. Aqueous ethanoic acid shows acidic properties because in water, ethanoic acid ionises to form hydrogen ions, H+.4. Hydrogen chloride gas in methylbenzene does not show acidic properties and does not conduct
electricity, because it exists as covalent molecules.
5. Hydrogen chloride gas in water shows acidic properties and conduct electricity. This is because
hydrogen chloride ionises in water to form hydrogen ions, H+ thus hydrochloric acid is formed.6. Aqueous ethanoic acid and hydrochloric acid can conduct electricity because in the solutions the ions can move freely. Conclusion
Chemicals Properties of Acid
a. Acids react with bases to form salts and water. Examples;
i. H2SO4 + ZnO
ii. 2HCl + CuO
b. Acids react with alkalis to form salts and water. Examples;
i. HCl + NaOH
ii. H2SO4 + Ca(OH)2
c. Acids react with reactive metal to produce salts and hydrogen gas.
Examples;
i. 2HCl + Mg
ii. H2SO4 + Zn
Reactive metals: Mg, Al, Zn (use this metals only)
d. Acids react with carbonate compound to produce salts, water and carbon
dioxide.
Examples;
i. 2HCl + PbCO3
ii. H2SO4 + CuCO3 Role of water and the properties of alkalis
Results
AcidConditionObservationInference
Ammonia
(molecule)In tetra chloromethane- No colour change in
the litmus paper
- Light bulb is not lighted upDoes not show any alkalis properties
Aqueous - Blue litmus paper is
changed to red
- Light bulb is lighted upShows alkalis properties
Discussion
The hydroxide ions, OH- are responsible for alkalis to show their properties1. Ammonia liquid in tetrachloromethane does not show alkalis properties because it exists as molecules.2. Aqueous ammonia acid shows acidic properties because in water, ethanoic acid ionises to form hydroxide ions, OH-.
Ionisation equation for ammonia;
3. Hydroxide ion, OH- in ammonia solution / solution hydroxide cause solution shows alkalis properties.
4. Ammonia gas in methylbenzene does not show alkalis properties and does not conduct electricity because it exits as a molecules.
5. Ammonia gas in water shows alkalis properties and conduct electricity because in the solutions the ions can move freely.
Conclusion
Chemicals Properties of Alkalisa. Alkalis react with acids to form salts and water.
Examples;
i. NaOH + HCl
ii. Ca(OH)2 + H2SO4 b. When a mixture of an alkali and an ammonium salt is heated, ammonia gas is, NH3 is liberated.
Examples;
i. NaOH + NH4Cl
ii. Ca(OH)2 + 2NH4Cl The Strength of Acids and Alkalis
1. The pH scale us used to indicate the degree of acidity or alkalinity of a solutions.
Strong and Weak Acids/Alkalis1. The strength of an acid or alkali depends on the degree of ionisation or dissociation of the acid or alkali in water.
2. Strong Acid: An acid which ionises completely in water. Example; HCl
(monoprotic acid)
H2SO4
(diprotic acid)
3. Weak Acid: An acid which ionises partially in water.
Example; CH3COOH
4. Strong Alkali: An alkali which ionises completely in water. Example; NaOH 5. Weak Alkali: An alkali which ionises partially in water.
Example; NH3 + H2O NH4 + + OH-
Concentrations of Acids and Alkalis
1. Concentration of solution can be expressed in gdm-3 or mol dm-3.
Example. 1:
A student dissolves 50.0 g of anhydrous copper(II) sulphate in water to make a 250 cm3 of solution.
What is the concentration of the solution in mol dm(3?
Solution:Mass of anhydrous copper(II) sulphate, CuSO4 = 50.0 g
Volume of solution = = 0.25 dm3
( Concentration of CuSO4 solution = = 200.0 g dm(3 200 gdm-3 = 200 mol dm-3
64 + 32 + 4x16
molar mass
Example 2:
28.0 g of potassium hydroxide is dissolved in water to make 200 cm3 of solution.
Calculate the molarity of potassium hydroxide solution obtained. (Ar: H, 1; O, 16; K, 39) Ans: 2.5 mol dm(3Solution:
Question 1
Find (a) the concentration of a sodium in grams per dm3 when 36.5 g of hydrogen chloride, HCl is dissolved in water to make up 500 cm3 of solution. (Ans:73.0 g dm-3 ) (b) the molarity of a solution which is prepared by dissolving 0.30 mol of sodium hydroxide,
NaOH in distilled water to make up 250 cm3 of solution. (Ans: 1.20 mol dm-3)Solution:
(a) (b) Question 2The molarity of a bottle of nitric acid, HNO3 solution is 2.0 mol dm-3. What is the concentration of sodium in g dm-3
[Relative atomic masses: H, 1; N, 14; O, 16] Ans: 126 g dm-3Solution;
Question 3
Calculate the molarity of a sodium sulphate, Na2SO4 solution with a concentration of 28.4 g dm-3. [Relative atomic masses: O, 16; Na, 23; S, 32] Ans: 0.2 mol dm-3Question 4
A student pipettes 25.0 mol dm3 of sodium hydroxide, NaOH solution into a conical flask. The concentration of the alkali was 1.5 mol dm-3. Calculate the number of moles of sodium hydroxide, NaOH in the flask. Ans: 0.0375 molPreparation of Standard Solutions by Dilution MethodStandard solution is a solution in which its concentration is accurately known.
Use this formula ok..
M1 V1 = M2 V2 M1 = molarity of the solution before water is added
V1 = volume of the solution before water is added
M2= molarity of the solution after water is added
V2= volume of the solution after water is added
a) Preparation of standard solution
- prepared by using a volumetric flask b) Preparation of a solution by dilution method
- adding distilled water to a concentrated solution
- changes the concentration of the solution- does not change the amount of soluteExample 1:Find the volume of 2.0 mol dm3 sulphuric acid, H2SO4 needed to prepare 100 cm3 of 1.0 mol dm-3 sulphuric acid, H2SO4.
Solution;
M1 V1 = M2 V2 2.0 mol dm-3 V1 = 1.0 mol dm-3 100 cm3 V1 = 1.0 mol dm-3 100 cm3 = 50 cm3 2.0 mol dm-3 Example 2:
Volume of 2.0 mol dm(3 nitric acid needed to be diluted with distilled water to make 250 cm3 of
0.5 mol dm(3 nitric acid?
Solution: Example 3:
50 cm3 of water is added to 200 cm3 of a 2 mol dm(3 solution of sodium hydroxide.
Determine the molarity of the diluted solution.
NeutralizationDefinition: The reaction between an acid and a base to produce a salt and water only. Example : i. HCl + NaOH NaCl + H2O
ii. H2SO4 + CuO CuSO4 + H2O The ionic equation for neutralization;
H+ + OH- H2O
Acid-Base Titration
IndicatorColour in alkalisColour in neutralColour in acids
Methyl orangeYellow OrangeRed
PhenolphthaleinPinkColourlessColourless
LitmusBluePurpleRed
Calculation involving neutralization using balanced equations.
Question 1: A student pipettes 25.0 mol dm-3 of sodium hydroxide, NaOH solution into a conical flask and filled a burette with 0.10 mol dm-3 hydrochloric acid, HCl to carry out titration. He obtained an average volume of 22.0 cm3 hydrochloric acid, HCl. What was the molarity of the sodium hydroxide, NaOH solution? [ans: 0.088 mol dm-3]Question 2:
What is the volume of 0.5 mol dm-3 sulphuric acid, H2SO4 needed to neutralize 25.0 cm3 of 0.8 mol dm-3 ammmonia, NH3 solution?[Ans: 20.0 cm3]Question 3:
A sample of copper(II) oxide, CuO was found to completely neutralize 100 cm3 of 0.5 mol dm-3 hydrochloric acid. Calculate the mass of sample. [Ans : 2.0 g]
Alkalis: only bases that soluble in water
Bases: all metal oxides or metal hydroxides
Switch
Switch
Glacial Ethanoic acid
Carbon
Carbon
Light bulb
Aqueous
Ethanoic acid
Carbon
Carbon
x molar mass (g mol-1)
g dm -3
Increasingly alkaline
Increasingly acidic
Neutral
pH
Ammonia in water
14
13
12
Bases and alkalis only show its alkalis properties when dissolved in water.
Ammonia
in tetrachloromethane
NH3 (l) NH4+ (aq) + OH (aq)
Ammonia ammonium ion hydroxide ion
+ H2O (l)
Light bulb
mol dm-3
MOLARITY
An acid only shows its acidic properties when dissolved in water.
Switch
Switch
11
Carbon
Carbon
Light bulb
10
Carbon
Carbon
Light bulb
9
8
7
6
5
4
3
2
1
0
Concentration
(mol /dm3)
Number of mole of solute (mol)
=
Volume of solution (dm3)
Concentration
(g /dm3)
Mass of solute/substance (g)
=
Volume of solution (dm3)
Retort stand
Burette
Hydrochloric acid, HCl
Conical flask
25 cm3 sodium hydroxide, NaOH + phenolphthalein
MA VA
MB VB
=
a
b
MA = Molarity of alkaliMB = Molarity of base
VA = Volume of alkaliVB = Volume of base
a = number of mole of acid from balanced chemical equation
b = number of mole of base from balanced chemical equation
Calibration mark
molar mass (g mol-1)
1
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