concentration acids & bases

8/4/2019 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 @



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)



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



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


Molarity(mol/ dm3)

X Molar Mass

÷ Molar Mass



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.


Molarity(mol/ dm3)

X Molar Mass

÷ Molar Mass



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)

concentration acids & bases

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