conc of acids n bases

8/14/2019 Conc of Acids n Bases

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Concentration The quantity of solute in a given volume of

solution, which is usually 1 dm3

Concentration (g dm-3)

Mass of solute (g)

Volume of solution (dm3)



Molarity The number of moles of solute that are

present in 1 dm3 of solution

Molarity (mol dm-3)

Number of moles of solute (mol)

Volume of solution (dm3)



Molarity (mol dm

-

3) Concentration (gdm-3)

x Molar mass

÷ Molar mass



The concentration of nitric acid, HNO3 is 126 g dm-3.

what is its molarity? [Relative atomic mass: H=1,N=14, O=16]

Molar mass of HNO3

= 1 + 14 + 3(16) g mol-1

= 63 g mol-1

Molarity of HNO3

= 126 g dm-3

63 g mol-1

= 2.0 mol dm-3

Examples…



How many moles of solute are in 20 cm3 of solutionwith a molarity of 0.4 mol dm-3?

0.4 mol dm-3 of a solution contains 0.4 moles of solute in 1 dm3 solution.

20 cm3 solution equals to 0.02 dm3 solution.

Therefore 0.02 dm3 contains

= 0.02 x 0.4 moles of solute

= 0.008 moles of solute



Calculate the volume of a 4 mol dm-3 solutioncontaining 2 moles of solute?

4 moles of solute are contained in 1 dm3 solution.Therefore 2 moles of solute are contained in

= (1 dm3 / 4 moles) x 2 moles

= 0.5 dm

3

= 500 cm3



Preparation of standardsolution



Activity of preparation of standard solution

Put away any chemistry books!

Form 2 groups

You have to arrange the step of preparation of standardsolution in the correct sequence.

Select a presenter for each group to present your groupworks in front of the class.

You only have 5 minutes to complete the task.

Start now!



Calculate the amount of sodium hydroxide to be

used for the preparation of 100 cm3 of sodiumhydroxide solution with a concentration of 2.0 moldm-3.

Molar mass of NaOH= 23 + 16 + 1 = 40 g

Mass of NaOH required

= number of mole NaOH x Molecular mass of NaOH= (MV / 1000) x 40 g

= (2.0 / 1000) x 100 x 40 g

= 8.0 g

Example…



DILUTION



100cm3 of originalsolution

1000cm3 of diluted

solution

Add

water

10times

dilution

Soluteparticle

Original concentration:10 particles per 100cm3

New concentration:1 particles per 100cm3



M1 x V1 = M2 xV2

M1 = molarity of the solution before water is

addedV1 = volume of the solution before water is

addedM2 = molarity of the solution after water is

addedV2 = volume of the solution after water is

added



Example…

Calculate the volume of 2.0 mol dm-3 sulphuric acid,

H2SO4 needed to prepare 100 cm3 of 1.0 mol dm-3 sulphuric acid, H2SO4.

The volume of H2SO4 needed= (1.0 x 100) / 2.0

= 50 cm3

conc of acids n bases

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