concentration acids & bases
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Chapter 7Acids and Bases
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Chapter 7:
Acids and Bases Learning Objectives6.1 Meaning of acid, base and
alkali- Properties of acids and
bases6.2 Strength of acids and alakalis-
pH Scale
6.3 Concentration –
Preparation ofStandard Solution
6.4 Neutralizationn Process
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Acid-Base Reactions
Acid + Base Salt + Water
(Aqueous Solution)
Solute + Solvent Solution
e.g: salt + water salt solution
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Concentrationof Acids and Alkalis
1. The concentration of a solutiondetermine the amount of solute dissolvedin a given volume of solution.
2. The greater the amount of solute
dissolved, the higher the concentration ofthe solution.
3. Three types of solution:
a) dilute solution : 0.001 mol dm
-3
.b) concentrated solution:1.0 mol dm-3 c) saturated solution : maximumsolute dissolved
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Concentrationof Acids and Alkalis
The concentration of a solution refers to thequantity of solute (gram or moles) in agiven volume of solution (dm3).
Concentration = Mass of solute (g)(g/ dm3) Volume of solution (dm3)
Concentration = Mass of solute (mol)(mol/ dm3) Volume of solution (dm3)
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Molarity
The concentration unit that is widely used bychemist is molarity (mol/ dm3) ormolar concentration (M)
Molarity is the number of moles of solute
that are present in1 dm-3 of solution.
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Find the molarity of the followingsolution.
1. 0.5 mole of sodium chloride
dissolved in 1 dm3
of distilled water.2. 0.2 mole of copper (II) sulphatedissolved in 250 cm3 of distilledwater.
3. 0.1 mole of sodium hydroxidedissolved in 500 cm3 of distilledwater.
Concentration = Mass of solute (mol)(mol/ dm3) Volume of solution (dm3)
C @
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Questions Answers
1. 0.5 mole ofsodium chloridedissolved in 1 dm3 of distilled water.
2. 0.2 mole ofcopper (II)sulphate dissolvedin 250 cm3 ofdistilled water.
3. 0.1 mole ofsodium hydroxidedissolved in 500cm3 of distilledwater.
Concentration@Molarity = Mass of solute (mol)(mol/ dm3) Volume of solution (dm3)
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1. 0.2 mole of sodium hydroxide isdissolved in 250 cm3 of water.(Relative formula mass of NaOH,
40). Calculate the concentration ofthe solution obtained in units of :a) g/ dm3 b) mol/ dm3
Concentration(g/ dm3)
Molarity(mol/ dm3)
X Molar Mass
÷ Molar Mass
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2. 0.98 g of sulphuric acid is dissolvedin 200 cm3 of water. (Relativeformula mass of H2SO4, 98). What
is the concentration of the sulphuricacid obtained in units of :a) g/ dm3 b) mol/ dm3
Concentration(g/ dm3)
Molarity(mol/ dm3)
X Molar Mass
÷ Molar Mass
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3. Calculate the mass of sodium nitrateto be added in 100 cm3 of water toproduce a solution with
concentration of 0.02 mol dm
-3
.(Relative formula mass : 85)4. 0.45 g of glucose is added in to 100
cm3 to prepare a glucose solution
with the concentration of 0.025 moldm-3 . Calculate the relativemolecular mass of glucose.
Concentration(g/ dm3)
Molarity(mol/ dm3)
X Molar Mass
÷ Molar Mass
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Molarity = Number of moles of solute (mol)(mol/ dm3) Volume of solution (dm3)
M = n/V
n = MV
M = Molarity (mol/ dm3)
n = number of moles of solute (mol)V = Volume of solution (dm3)