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8/14/2019 Conc of Acids n Bases
http://slidepdf.com/reader/full/conc-of-acids-n-bases 1/17
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)
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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)
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Molarity (mol dm
-
3) Concentration (gdm-3)
x Molar mass
÷ Molar mass
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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…
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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
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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
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Preparation of standardsolution
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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!
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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…
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DILUTION
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100cm3 of originalsolution
1000cm3 of diluted
solution
Add
water
10times
dilution
Soluteparticle
Original concentration:10 particles per 100cm3
New concentration:1 particles per 100cm3
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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
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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