Download Expressing the concentration of solutions

Expressing the concentration of solutions
Definitions:
 A solution is a homogeneous mixture
 Solvent is the liquid in which the solute is dissolved
 an aqueous solution has water as solvent
 A saturated solution is one where the concentration is at a maximum - no
more solute is able to dissolve.
 A solution is composed of a solvent which is the dissolving medium and a
solute which is the substance dissolved.
 In a solution there is an even distribution of the molecules or ions of the
solute throughout the solvent.
 The concentration of a solution can be expressed in a variety of ways
(qualitatively and quantitatively).
 Formerly, the concentration of a solution can be expressed in four ways:
– Molarity(M): moles solute / Liter solution
– Mass percent: (mass solute / mass of solution) * 100
– Molality (m) - moles solute / Kg solvent
– Mole Fraction(cA) - moles solute / total moles solution
Qualitative Expressions of Concentration
A solution can be qualitatively described as:


dilute: a solution that contains a small proportion of solute relative to
solvent, or
concentrated: a solution that contains a large proportion of solute relative to
solvent.
Semi-Quantitative Expressions of Concentration:
A solution can be semi-quantitatively described as
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Unsaturated: a solution in which more solute will dissolve, or
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-
Saturated: a solution in which no more solute will dissolve.
The solubility of a solute is the amount of solute that will dissolve in a
given amount of solvent to produce a saturated solution.
Percent Composition (by mass)


o
o
We can consider percent by mass (or weight percent, as it is sometimes
called)
We need two pieces of information to calculate the percent by mass of a
solute in a solution:
The mass of the solute in the solution.
The mass of the solution.
Use the following equation to calculate percent by mass:
Molality
Molality, m, tells us the number of moles of solute dissolved in exactly one
kilogram of solvent. (Note that molality is spelled with two "l"'s and represented
by a lower case m.)
We need two pieces of information to calculate the molality of a solute in a
solution:


The moles of solute present in the solution.
The mass of solvent (in kilograms) in the solution.
To calculate molality we use the equation:
Molarity
Molarity tells us the number of moles of solute in exactly one liter of a solution.
(Note that molarity is spelled with an "r" and is represented by a capital M.)
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We need two pieces of information to calculate the molarity of a solute in a
solution:


The moles of solute present in the solution.
The volume of solution (in liters) containing the solute.
To calculate molarity we use the equation:
Mole per liter solutions
A mole per litre (mol/l) solution contains one mole of a solute dissolved in, and
made up to 1 litre with solvent.
Mole:
• A counting unit
• Similar to a dozen, except instead of 12, it’s 602 billion trillion
602,000,000,000,000,000,000,000
• 6.02 X 1023 (in scientific notation)
• A Mole of Particles Contains 6.02 x 1023 particles
• 1 mole C= 6.02 x 1023 C atoms
• 1 mole H2O = 6.02 x 1023 H2O molecules
• 1 mole NaCl = 6.02 x 1023 NaCl “molecules”
• The mole is the SI base unit that measures an amount of substance.
• One mole contains Avogadro's number (approximately 6.023×1023) (number
of atoms or molecules).
Preparation of mol/l solution
To preparation a mol/l solution, use the following formula:
Number of grams to be dissolved in 1 litre of solution= Required mol/l
solution X Molecular mass of substance
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Examples:
Make liter of sodium chloride (NaCl), 1 mol/l
 To make 1 litre of sodium chloride (NaCl), 1 mol/l:
Required mol/l concentration= 1
Molecular mass of NaCl= 58.44
Therefore 1 lit NaCl, 1 mol/l contains:
1 X 58.44= 58.44 g of the chemical dissolved in 1 litre of solvent.
 To make 1 litre of sodium chloride, 0.15 mol/l (physiological saline):
…………………………………………………………………………
…………………………………………………………………………
…………………………………………………………………………
…………………………………………………………………….
 To make 50 ml of sodium chloride, 0.15 mol/l:
Required mol/l concentration= 0.15
Molecular mass of NaCl= 58.44
Therefore 50 ml NaCl, 0.15 mol/l contains:
0.15X 58.44 X 50 = 0.438 g of the chemical substance
1000
dissolved in 50 ml of solvent.
Conversion a percentage solution into a mol/l solution:
By the following formula:
Mol/l solution =
g% (w/v) solution X 10
Molecular mass of the substance
Examples:
 To convert a 4% w/v NaOH solution into a mol/l solution:
Gram % solution = 4
Molecular mass of NaOH= 40
Conversion to mol/l= 4X10 = 1
40
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Therefore 4% w/v NaOH is equivalent to NaOH, 1 mol/l
solution.

To convert a 0.9% w/v NaCl solution into a mol/l solution:
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
……………………………………………………………………………….
Conversion of a normal solution into a mol/l solution:
By the following formula:
mol/l solution= Normality of solution
Valence of substance
Examples:
 To convert 0.1 N (N/10) HCl into a mol/l solution:
Normality of solution= 0.1
Valence of HCl= 1
Conversion to mol/l= 0.1/1= 0.1
Therefore 0.1 N HCl is equivalent to HCl, 0.1mol/l solution.
 To convert 1 N Na2CO3 into a mol/l solution (valence =2):
………………………………………………………………………………
………………………………………………………………………………
………………………………………………………………………………
……………………………………………………………………………….
How to dilute solutions and body fluids
In the laboratory it is frequently necessary to dilute solutions and body fluids to
reduce its concentrations.
Diluting solutions:
A weaker solution can be made from a stronger solution by using the following
formula:
Volume (ml) of stronger solution required= R X V
O
Where: R= concentration of solution required
V= volume of solution required.
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O= strength of original solution.
Examples:
 To make 500 ml of NaOH, 0.25 mol/l from a 0.4 mol/l solution:
C= 0.25 mol/l, V= 500 ml, S=0.4 mol/l
ml of stronger solution required:
0.25 x 500 = 312.5
0.4
Therefore, measure 312.5 ml NaOH, 0.4 mol/l and make up to 500 ml with
distilled water.
 To make 1 litre of HCl, 0.01 mol/l from a 1.0 mol/l solution:
………………………………………………………………………………
………………………………………………………………………………
……………………………….
 To make 100 ml glucose, 3 mmol/l in 1 g/l benzoic acid from glucose 100
mmol/l solution :
C= 3 mmol/l V= 100 ml S= 100 mmol/l
ml of stronger solution required= 3 x 100 = 3
100
Therefore, measure 3 ml of glucose, 100 mmol/l and make up to 100 ml
with 1 g/l benzoic acid.
 To make 500 ml H2SO4, 0.33 mol/l from concentrated H2SO4 which has an
approximate concentration of 18 mol/l:
C= 0.33 mol/l, V= 500 ml S= 18 mol/l
ml of stronger solution required= 0.33 x 500 = 9.2
18
Therefore, measure 9.2 ml conc. H2SO4, and slowly add it to
about 250 ml of distilled water in a volumetric flask. Make
up to 500 ml with DW.
Diluting body fluids and calculating dilutions
- To prepare a dilution or series of dilutions of a body fluid:
Examples:
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 To make 8 ml of 1 in 20 dilution of blood:
Volume of blood required= 8/20 = 0.4 ml Therefore, to prepare 8 ml of a
1in 20 dilution, add 0.4 ml of blood to 7.6 ml of diluting fluid.
 To make 4 ml of a 1 in 2 dilution of serum in physiological saline:
Volume of serum required= 4/2 = 2 ml
Therefore, to prepare 4 ml of a 1in 2 dilution, add 2 ml
of serum to 2 ml of physiological saline.
- To calculate the dilution of a body fluid:
Examples:
 Calculate the dilution of blood when using 50 µl of blood and 950 µl
of diluting fluid:
Total volume of body fluid and diluting fluid= 50+950=1000 µl
Therefore, dilution of blood: 1000/50= 20
i. e. 1 in 20 dilution.
 Calculate the dilution of urine using 0.5 ml of urine and 8.5 ml of
diluting fluid (physiological saline):
Total volume of urine and diluting fluid= 8.5+0.5=9 ml
Therefore, dilution of urine: 9.0/0.5= 18
i. e. 1 in 18 dilution.
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