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Chapter 13- Volumetric Analysis; Acid-Base (Expressing the concentration…
Chapter 13- Volumetric Analysis; Acid-Base
Expressing the concentration of solutions
Solute + Solvent = Solution
A solution
is a completely perfect mixture of a solute and a solvent.
Homogeneous
means there is no boundaries between the particles. i.e. the particles are in the same phase.
(1) If there is a large amount of solute relative to the quantity of solvent, the solution is said to be
concentrated
.
(2) If there is only a small amount of solute relative to the quantity of solvent, the solution is said to be
dilute
.
The concentration
of a solution is the amount of solute that is dissolved in a given volume of solution.
(A) Percentage of Solute
There are 3 ways to express the percentage of solute;
(1)
Weight per weight (w/w)
(example)(a solution labelled 10%w/w NaCl means there are 10g of sodium chloride per 100g of solution.)
(2)
Weight per volume (w/v)
(example)(A solution labelled 10%w/v NaCl means there are 10g of sodium chloride per 100cm3 of solution.)
(3)
Volume per volume (v/v)
(example)(a solution of ethanol marked 40%v/v means there is 40cm3 of ethanol per 100cm3 of solution.)
(B)Parts per Million (ppm)
This method of expressing the concentration of a solution is only used for very dilute solutions. It's particularly useful in water analysis where you are dealing with very low concentrations of certain substances.
(example)
(1mg per litre= 1mg per 1000 grams of
water (the density of water is 1g/cm3)
=1mg per 1000x1000mg
=1mg per million mg
=1ppm
...
1ppm=1mg/l
(C)Moles of Solute per Litre of Solution (molarity)
the molarity
of a solution is the number of moles of solute per litre of solution. i.e.
1M NaOH= 1mol/l NaOH
a 1m (one molar)
solution is one that contains one mole of the solute dissolved in one litre of solution.
(example)(a 1mol/l solution of sodium hydroxide contains 40g of NaOH (because the Mr of NaOH is 40)per litreof solution)
These solutions are made up in volumetric flasks.
This is a Volumetric flask.
Calculations Involving Solutions
(A)Converting Moles per Litre to Grams per Litre.
(example)
(How many grams of NaCl per litre are present in a solution marked 0.25M NaCl)
(Na=23, Cl=35.5)
(Rel. Molecular Mass of NaCl=23+35.5=58.5
A 1M NaCl solution contains 58.5 grams per litre.
...a 0.25M NaCl solution contains 58.5 x 0.25
=14.625g/l
Answer: 14.625 grams NaCl per litre.)
(B)Converting Grams per Litre to Moles per Litre (Molarity)
(example)
(What is the molarity of a solution that contains 3.68g of NaOH per litre of solution?)
(H=1, O=16, Na=23)
(Rel. Molecular Mass of NaOH= 23+1+16=40
40g in 1L solution --> 1M NaOH solution
1g in 1L solution --> 1/40M NaOH solution
3.68g in 1L solution --> 3.68/40M NaOH solution
=0.092M NaOH solution
Answer: 0.092M NaOH)
1 mole/1000cm3 = 1 mole/L
1 mole/500cm3 = 2 moles/L
1 mole/250cm3 = 4 moles/L
1 mole/100cm3 = 10 moles/L
(C)Calculating the Number of Moles Given the Volume and the Molarity of a Solution
(example)
(How many moles of NaOH are present in 25cm3 of 0.55M NaOh)
(1 litre of 0.55M NaOH contains 0.55 moles NaOH
i.e. 1000cm3 of 0.55M NaOH contains 0.55 moles NaOH
1cm3 of 0.55M NaOH contains 0.55/1000 moles NaOH
25cm3 of 0.55M NaOH contains 25 x 0.55/1000moles NaOH
=0.014moles NaOH
Answer: 0.014moles NaOH)
Number of moles(in cm3)= volume x molarity
...............................................------------------------
......................................................1000............
Reaction Between a Solution and a Solid
(example)
(What mass of magnesium will react with 50cm3 of 0.5M H2SO4 (Mg=24)? The balanced equation for the reaction is:
Mg + H2SO4 --> MgSO4 + H2
1mole 1 mole
0.025mole 0.025mole
No. of moles of acid= vol x mol / 1000
=50 x 0.5 / 1000
=0.025 moles
From the balanced equation, one mole of sulfuric acid reacts with one mole of magnesium. Therefore, moles of magnesium which reacts= 0.025
0.025 x 24g
=0.6g
Answer: 0.6g Magnesium
Standard Solutions
a standard solution
is a solution whose concentration is accurately known
a primary standard solution
is a substance that can be obtained in a stable, pure and soluble solid form so that it can be weighed out and dissolved in water to give a solution of accurately known concentration.
Volumetric Analysis: Titrations
A titration
is a laboratory procedure where a measured volume of one solution is added to a known volume of another solution until the reaction is complete.
Apparatus Used in Volumetric Analysis:
(A) Graduated cylinder
(B)Volumetric Flask
(C)Pipette
(D)Burette
(E)Conical Flask
Solving Volumetric Problems
(1)Calculating the concentration of a Solution of Unknown Concentration from Titration Data
Va x Ma = Vb x Mb
.................. ..............
Na ---------- Nb ----
(2)Calculating the Amount of Water of Crystallisation in a Compound and the Relative Molecular Mass of the Compound from Titration Data
(example)
a 20cm3 sample of vinegar was diluted to 100cm3 in a volumetric flask. This diluted solution was then titrated against 20cm3 of 0.12M NaOH solution. The balanced equation for the reaction is:
CH3COOH + NaOH --> CH3COOH + H2O
The average titration figure was 14.2cm3. Calculate the concentration of ethanoic acid in the original vinegar in (a)mol/L (b) g/L (c)%w/v.
(H=1, C=12, O=16)
Va= 14.2 Ma=? Na=1 Vb=20 Mb=0.12 Nb=1
14.2 x Ma/1= 20 x 0.12/1
Ma= 20 x 0.12/ 14.2
=0.169 moles per litre
Since the solution was diluted by a factor of five, concentration of ethanoic acid in original vinegar= 0.169 x 5 moles per litre
=0.845mol/L
0.845 x 60g/L (Mr CH3COOH=60)
=50.7g/L
Since there are 50.7g of ethanoic acid in a litre of vinegar,there are 5.07g of ethanoic acid in 100cm3 of vinegar.
%w/v= 5.07%
Answer: (a)0.845M (b)50.7g/L (c)5.07%w/v
