Concentrations of Acids
and Alkalis
1.Concentration can be measured in
two units:
(a) mol dm-3
(b) g dm-3
2.Concentration in mol dm-3 is the amount of substance
dissolved in 1 dm-3 of solvent.
3.Concentration in g dm-3 is the mass of substance
dissolved in 1 dm-3 of solvent.
4.The concentration of the solution in
mol dm-3 is also called the molarity with the unit of M.
SAMPLE 1:
5 g of sodium hydroxide is dissolved in
250 cm3 water. What is the concentration
of sodium hydroxide in
(i) mol dm-3
(ii) g dm-3
Solution:
(i) Volume of solution = 250/1000
= 0.25 dm-3
Concentration in g dm-3 = 5/0.25
=
20 g dm-3
(ii) Relative
molecular mass of NaOH = 40
Number of moles of NaOH
= 5/40
= 0.125
Concentration = 0.125/0.25
= 0.5 mol dm-3
SAMPLE 2:
What is the mass of potassium hydroxide
required to prepare 100 cm3 of 0.2 mol dm-3 solution?
Solution:
Volume of solution = 100/1000
= 0.1 dm-3
0.2 = number of moles/0.1
Number of moles = 0.1 x 0.2
= 0.02 mol
Relative molecular mass of KOH =
56
Mass of KOH = 0.02 x 56
=1.12 g
Standard
solution
1. A standard solution is a solution of known concentration.
2. A standard solution can be prepared by
(a) Dissolving the substance into a fix volume of water
(b) Diluting a concentrated solution.
Dilution
1.Dilution is a process of obtaining a less concentrated solution
by adding water into a more
concentrated solution.
2.Dilution will lower the concentration but will not change the
number of moles of solute
dissolved in the solution.
3.Therefore, the mass of the solute dissolved in the solution
remains unchanged.
4. The concentration of the diluted solution can be calculated using
the following formula:
M1V1 = M2V2
Where
•M1 = concentration of the
concentrated
solution
•V1 = volume of the concentrated
solution
•M2 = concentration of the diluted
solution
•V2 = volume of the diluted solution
Sample 3:
50 cm3 of 2.0 mol dm-3 solution of sodium hydroxide is
added with 200 cm3 of
distilled
water. What is the concentration of the solution produced?
Solution:
M1V1 = M2V2
2(50) = M2 (200+50)
M2 = 0.4 mol dm-3
Sample 4:
Calculate the volume of 1.0 mol dm-3 hydrochloric acid needed to
prepare 200 cm3 of
0.2 mol dm-3 hydrochloric acid.
Solution:
M1V1 = M2V2
1 x V1 = 0.2 x 200
V1 = 40 cm3
pH value with molarity of acid and alkali
1.When the pH value of an acid decreases, the molarity of the acid increases.
2.When the pH value of an alkali increases, the molarity of the alkali increases.
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