Arbitrary And Quantitative Methods Of Expressing Concentration Of Solutions
A concentration of solution expresses the amount of solute per unit amount of solution or solvent. Concentration may be expressed in several ways, arbitrarily or quantitatively. If the basic concept of concentration is to be understood, interconversion of the ways under quantitative method is a matter of simple mathematics.
I. Arbitrary (not precise)
A. Dilute or concentrated - these are not very specific ways of expressing concentration of a solution.
Examples:
• At room temperature, 100 grams of water can dissolve nearly 40 grams of sodium chloride.
• Thus,if 1 gram or 2 grams of sodium chloride is dissolved in 100 grams water, the solution is dilute.
• If 35 grams of NaCl is dissolved in 100 grams of water, then the solution is concentrated.
B. Saturated, unsaturated, supersaturated
1. A saturated solution is one that holds all the solute it possibly can at any given temperature.
Example:
The solubility of sodium chloride in water is 37 grams per 100 grams of water. To make a saturated solution of sodium chloride, dissolve 37 grams of NaCl in 100 grams of water
2. Any solution containing less than the maximum possible amount of solute is said to be unsaturated.
Example:
A solution with 5 grams of sodium chloride (or 10 grams or 20 grams or 30 grams) in 100 grams of water is unsaturated.
3. A supersaturated solution is one that holds more solute than is possible at some given temperature. To make a supersaturated solution, make a saturated solution at some higher temperature and then let the solution cool very carefully.
II. Quantitative - numerical way of stating precisely how much solute is dissolved in a solvent.
A. Weight of the solute in a given volume of solvent or solution (volume concentration units)
1. Molarity
Molarity is equal to the moles of solute divided by the liters of solution.
Briefly stated in the form of an equation:
Examples:
Thus, a molar(1M) solution contains 1 mole (1 gram-molecular weight) of the solute dissolved in sufficient water to make a liter of solution.
If two moles of NaCl is dissolved in 4 Lit of water, the concentration is 2/4=0.5 molar solution and is represented by M.
Molarity is the most commonly used method of expressing concentration of solution.
The molecular weight expressed in grams is the gram-mol. wt.
In a problem where moles is given instead of grams, use the molar mass of then compound to convert them to moles.
If the volume is given in “ml”, remember that 1 ml = 0. 001 liter
2. Normality
A normal solution contains 1 gram-equivalent weight of solute in a liter
of solution.
A gram equivalent weight(G.E.W.) of an acid or a base is the weight in
Grams of the solute required to react with, replace, or furnish one mole
or 1 gram formula weight of H3O+ or OH-.
Equivalent weight = molecular weight/n
Where n is the number of hydronium ion or hydroxyl ion
G. E. W. of acid = mole weight/no. of replaceable hydronium ion or hydroxyl ion
Examples:
(a) Calculate the G. E. W. of a normal solution of HCl.
H = 1
Cl = 35.46
36.46 g/1 = 36.46 g.
Thus, a solution of 1N HCl will furnish 1 g of hydrogen ions and
it contains 36.46 g HCl.
(b) Calculate the G. E. W. of a 1 normal solution of sulfuric acid.
2H = 2
S = 32
4O = 64
98 g/2 = 49 g of sulfuric acid in a liter of solution, since only
½ gram-formula weight is required to supply 1 g hydronium ions.
In REDOX REACTIONS, the equivalent weight is defined in terms of electrons lost or gained in the reaction:
Equivalent weight = molecular weight/no. of electrons lost or gained(as determined
from the redox reaction)
Example:
In an acid solution, the permanganate ion may be reduced to the Mn++ ion,
indicating a gain of five electrons since the oxidation number of the
manganese atom is changed from +7 to +2. Therefore,
Eq. wt. of potassium permanganate = 158.08/5 = 31.61 g
Hence, a normal solution of potassium permanganate, when used as an
oxidizing agent in an acid solution, contains 31.61 g of the solute in a
liter of solution.
The number of equivalents in any number of grams of a substance is:
Equivalents = grams/equivalent weight
The normality or normal concentration, N, is the number of equivalents of solute per
liter of solution.
N = equivalents of solute/liter of solution = milliequivalents of solute/ milliliters of solution
“The terms “equivalents” and “normality” tend to be ambiguous, because some substances may react with or furnish hydronium ion or hydroxyl ion in more than one way. In order that equivalents and normality make sense, the chemical equation by which the substance reacts must be explicitly stated, understood, or derivable from the context.” (Blaedel & Meloche)
3. Titer
The titer of a solution is the weight of a substance equivalent to a unit volume of the solution. The substance need not be the solute substance itself, but may be a sought-for substance titrated by the solution, or a substance derivable from the solution.
Titer = grams of substance/liters of solution
= milligrams of substance/milliliters of solution
The titer is a volumetric chemical factor, with which the amount of sought-for substance may be calculated from the volume of standard solution required for its titration, without any consideration or knowledge that occur in the analysis. The titer method of computation is useful as a time-saver in routine calculations, and in teaching the associated chemistry. But in studying chemical principles, the titer method should not be overemphasized.
4. Formality
The formal concentration (F) of a solute is the number of formula weights per liter of solution, without regard for the form in which the solute exists in solution. There is an advantage to the use of this term over the term “molarity.” Thus it may not always be clear whether the term “5 M sulfuric acid” represents the total sulfuric acid concentration or just the un-ionized sulfuric acid concentration. This ambiguity does not exist when formal concentration are used.
A formal solution contains 1 gram-formula weight of the solute dissolved in
sufficient water to produce a liter of solution.
The formula weight is the sum of the atomic weights of those elements in the
assumed formula of the solute. Usually, but not always, the assumed formula
is the simplest formula of an inorganic substance. This method of expressing
concentration applies equally well to either ionic substances or substances that
exist as molecules.
Example: A formal solution of NaCl contains 58.46 g (23 = 35.46) of NaCl in a liter
of solution.
Formality and molarity become equivalent terms when the formula weight is known
to be the same as the molecular weight.
F = gram-formula weight/liter of solution = wt. of solute/gfw x 100/ml. of solution
Sample problems:
(1) Express in formality the concentration of a solution containing
36.9 % KOH with a density of 1.37 g/ml.
100 ml x 1.37 g/ml x 0.369 = 1000 x 1.37 g x 0.369
= 505.530 g KOH in a liter of solution
K = 39; O = 16; H = 1
F = gFwt/liter of solution = 56/505.530 = 9.027 F
(2) If 100 g of NaOH is dissolved in water to give a solution with a volume of
250 ml., what is the formality, F of the solution?
V = 250 ml = 250/1000 = 0.25 liter
F = wt/Vl x Fwt
F = 100/(0.25 x 40)
F = 10 F
II. Weight of Solute in a given Weight of Solvent or Solution (weight concentration units)
1. Molality
A Molal solution contains 1 gram-molecular weight (1 mole) of the solute dissolved in 1,000 g (1 kg) of water. It differs from a molar solution in being a ratio of solute to solvent by weight.
1 m = 1 mole of solute/kg of solvent or wt of solute/g-mol. wt f solute
x 1000 g/g of solvent
2. Percentage Composition (parts/100 parts) - easy to calculate since
information about the chemical nature of the substance is not required.
The percentage of a solution is the weight in grams (or any other weight) of
the solute contained in 100 g (or corresponding unit) of solution.
(a) Mass percentage (% by wt.)
Instead of moles, if solute and solution are expressed as mass in grams, it is concentration in mass percentage. It is shown as w/w %.
wt –wt % = wt of solute/wt of solution x 100
Example: 10 % NaCl = 10 g NaCl/100 g solution
10 grams of NaCl in 100 g solution is a 10 % solution of NaCl
(b) wt – vol % = wt of solute/vol of solution x 100
Example: 21.2 g potassium phosphate is added to 480 g water to prepare 500 ml of potassium phosphate solution.
Find % wt. – vol. = 21.2 g/500 ml x 100
(c) Volume ? volume percentage (% by volume)
If the solute and solvent are both liquids, then the concentration can be expressed as volume of solute/volume of solution.
Vol. ? vol. % = vol. of solute/vol. of solution x 100
Examples: (1) Concentration of alcoholic beverage
A wine that is 10 % alcohol has 10 ml of alcohol per 100 ml of wine.
(2) If 20 ml ethanol is dissolved in 100 ml of water solution, it is 0.20 v/v % solution of ethanol.
References:
^Elementary Quantitative Analysis by J. Blaedel & V. W. Meloche; ©1978
^General and Quantitative Chemistry Classroom Notes and Laboratory Manuals, 1974
^http://www.scienceclarified.com/Ro-Sp/Solution.html ^http://misterguch.brinkster.net/molaritytutorial.html
^http://environmentalchemistry.com/yogi/chemistry/MolarityMolalityNormality.html
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