James Richard Fromm
In a heterogeneous mixture the components are present as entities much larger than molecules, and thus in principle a heterogeneous mixture can be separated by a manual selection of one type of particle. A homogeneous mixture which cannot be separated into its components by direct physical selection is called a solution; the selection would have to operate at a molecular level because individual molecules or ions are the form in which the solute is present. Solutions can, however, be separated into their component pure substances by other physical or chemical methods. For example, a salt water solution can be separated into its two components by boiling the water off and recondensing it elsewhere. Many of the procedures developed by alchemists and early chemists for the preparation of pure substances, such as distillation and recrystallization, were methods of separating the component substances of naturally occurring solutions.
Solutions may be solids, liquids, or gases. Bronze and brass are solid solutions of metals in each other, usually called alloys. The fillings of teeth are liquid or solid solutions of gold or silver in mercury which are called amalgams. The air we breathe is a more or less homogeneous solution of gases in each other. Sea water is a more or less homogeneous solution of salts in water.
Most of the solutions encountered in nature are solutions which are liquid, at least at the temperatures and pressures at which we will study them. The component or components present in largest quantity in such solutions is called the solvent, and in liquid solutions the pure solvent component is almost always a liquid. The other components are called solutes. Solutes are said to dissolve in the solvent to form the solution.
Chemical reactions often take place in solutions. A solution is a homogeneous mixture of one major component, the solvent, and at least one minor component, a solute. For nearly all reactions which take place in solution, the solvent does not react; the reaction takes place between solutes which react when two solutions are mixed. The stoichiometry of these reactions can be treated just as the stoichiometry of reactions which take place between pure substances. The moles which are significant in solution reactions, however, are those of the solutes. The moles of solvent, which are often present in great excess, are usually not directly involved in the reaction. The number of moles of a solute can be calculated from the quantity and composition of the solution.
The most fundamental way to specify the composition of any mixture of compounds, including uniform mixtures like solutions, is to give the composition in terms of amount of each different substance present. The composition of a solution in terms of one of its components, then, should be given as the ratio of amount of that component, in moles, to the sum of the amounts of all of the components, in moles. This ratio is called the mole fraction of the component in the solution. The mole fraction, multiplied by 100, is called the mole per cent. A solution of one mole of NaCl in nine moles of water would have a mole fraction of 0.100 and would be called 10.0 mole per cent NaCl. Mole fraction is a common method of specifying the composition of more concentrated solutions. It is less commonly used for solutions which are more dilute.
The composition of a solution can also be specified by mass, and this is often done because the solution can easily be prepared simply by weighing out the appropriate mass of each component and mixing them. Metal alloys are examples of solutions whose composition is normally specified by mass; the casting alloy of 50:50, or 1:1, lead:tin is made up directly by mass. Chemists generally specify compositions of solutions by mass much as they do for compositions of solutions by amount of substance. The mass ratio is the mass of solute per mass of solvent and the mass fraction is the mass of solute per total mass of the solution. Mass fraction, like mole fraction, can be multiplied by 100 to give mass per cent or per cent by mass. The composition of the alloy above could be specified as 1/1 mass ratio, 0.5 mass fraction, or 50 per cent by mass.
The composition of a solution can in principle be specified by any physical quantity which can be related to amount of substance. One of the few quantities other than mass which is used is volume, since the volume of any pure substance can be related to its mass if the density is known. Again, chemists generally specify compositions of solutions by volume much as they do for compositions of solutions by amount of substance or by mass. The volume ratio and volume fraction are uncommon but the volume per cent or per cent by volume, 100 times the volume of the solute per volume of the solute plus solvent, is used for solutions of high concentration or solutions which include organic components. A related unit is proof, which is defined as twice the volume per cent of ethanol in water. Proof is used almost exclusively for alcoholic beverages; a 50% ethanol by volume solution is 100 proof vodka.
Combinations of the physical quantities of mass, volume, and amount of substance can also express the compositions of solutions, but in practice only two such combinations are used: molal concentration (molality) and molar concentration (molarity). Molality is by definition the quotient of the amount of solute substance present, in moles, and the mass of solvent present, in kg.
Although we shall use no symbol for molality in this section, its usual symbol is m, which unfortunately is also used as the symbol for mass. Molality is very useful because it is closely related to the mole fraction yet the amount of solvent is specified in the directly measurable unit of mass. The molar mass of the solvent need not be known in order to make up a solution of a desired molality, but the molar mass of the solute must be known if the amount of solute is measured by mass. Molality is used for dilute solutions, both aqueous and nonaqueous; solutions above a few molal usually have their composition given in mole fractions. Molality is used in the most precise solution measurements, however, because the measurement of mass can be carried out with greater precision than can the measurement of volume
Solutions of known molality can be prepared by two methods. If and only if the pure solute and the pure solvent are both available in easily weighable forms, direct measurements of their masses can be used to prepare the solution. The mass of the solute is measured and the number of moles of solute is calculated. The mass of the solvent is also measured, and the ratio of moles of solute to mass of solvent is expressed in the units appropriate to molality.
Example. Appropriate instructions for the preparation of approximately 500 cm3 of 0.375 molal aqueous sodium acetate solution can be developed as follows. The molar mass of sodium acetate is 82.0245 g/mol. The mass of sodium acetate should then be (0.375 mol/kg water)(0.500 kg water)(82.0245 g/mol) which is 30.759 g of sodium acetate, which is to be added to 500.000 g water. The exact mass of the resulting solution will be 530.759 g. The approximate volume of the solution will be 500 cm3 since water has a density of 1.00 g/mL and makes up the vast majority of the solution. The exact volume can be determined only by measurement of density or molar concentration.
If the solute or solvent cannot be conveniently weighed in pure form, a solution of known molality can be obtained by dilution of a previously prepared molal solution of greater concentration. The more concentrated solution must have a known molality, which might have been determined by a stoichiometric solution reaction or titration as discussed in a following section.
Example. A solution known to be 1.000 molal in NaCl was available when a chemist required a solution that was to be 0.3100 molal. The desired solution could be prepared by taking some amount (here 500.000 g) of the 1.000 molal solution. This would contain (1.000 mol/kg)(0.5000 kg) = 0.5000 mol NaCl. The molar mass of NaCl is 35.453 + 22.990 = 58.443 g/mol NaCl and so half of that, 29.2215 g NaCl, must be present in the 500 g of solution; the remaining 470.7785 g of the solution must be the solvent water. The desired solution is to contain 0.3100 mol NaCl/kg H2O, so from the ratio 0.3100/1.000 = 0.5000/x, the mass of water needed, x, is 1.621903 kg. There are already 0.470779 kg of water present, so 1.142133 kg of water would have to be added to obtain the desired concentration. By this procedure the chemist would prepare about 1.6 litres of the desired solution.
The molar concentration of a solute is by definition the quotient of the amount of solute substance present, in moles, and the volume of solution present, in litres (L or dm3). The molar concentration is usually known as the molarity and its usual symbol is M (a symbol unfortunately also used for molar mass). The general concentration symbol, c, which normally carries the general meaning of amount of solute substance divided by volume, will often also carry the specific meaning that the amount of solute substance will be measured in moles and the volume will be measured in litres (L or dm3). Another alternative commonly used, and which we shall follow, is to include the species whose molar concentration is being indicated in square brackets, as in the symbol [CO32-] which represents the molar concentration of the carbonate ion. This alternative is helpful when equations which contain the molar concentrations of several different substances must be written.
The volume used in calculations of concentration, molar or otherwise, is the total volume in which the solute is distributed, which is the volume of the solution. Except for extremely dilute solutions, which contain almost nothing but solvent, the volume of the solution is equal neither to the volume of the solvent nor to the sum of the volume of the solvent and the volume of the solute. This is particularly important in the case of aqueous solutions, since at 4oC the mass of 1.0 litre of water is almost exactly 1.0 kg. Even at considerably higher temperatures the volume occupied by 1.0 kg of water is approximately 1.0 litre as shown in the Table below.
It might be thought that the molality and molarity of aqueous solutions would be the same. They are only approximately the same even for dilute solutions and considerably less so for concentrated solutions.
Making up a molal solution requires adding preweighed quantities; making up a solution of known molar concentration requires use of a mixing container in which the volume contained can be measured after the solute has dissolved in the solvent. Such a container is called a volumetric flask. Molarity is used more often than molality in the laboratory because the accurate measurement of volume is easier than is the accurate measurement of mass. Solution concentrations on the basis of molarity are used very extensively in chemistry. It is useful to remember that a solution which contains 1.0 mol of solute/litre of solution is exactly the same as one which contains 1.0 mmol of solute/mL of solution or 1.0 kmol of solute/m3 of solution.
The most common method of preparing a solution of known molar concentration is to weigh out an appropriate mass of solute, place it in a volumetric flask, and dilute to the mark with dissolution of the solute. The solute must dissolve before the final amount of the solvent is added, since the volume change on dissolution is not known. This procedure for preparing a molar solution can only be used when the pure solute is in an easily weighable form.
Example. Appropriate instructions for the preparation of 500 mL of 0.375 molar aqueous sodium acetate solution can be derived as follows. The molar mass of CH3COONa, sodium acetate, is 82.0245 g/mol. The mass of sodium acetate required would be (0.375 mol/L)(82.0245 g/mol)(0.500 L) g, which is 30.759 g. The procedure is then to place this mass in a 500 mL volumetric flask and add water to prepare the solution, with dissolution of the sodium acetate.
The simplest way to prepare a solution of known molarity is by dilution of a previously prepared solution of greater molar concentration. The preparation requires calculation of the amount of solvent to be added to a solution to prepare the desired solution.
Example. An aqueous solution which is 0.50 molar is available and 1.00 litres of 0.10 molar aqueous solution of the solute is required. Such a solution can be prepared as follows. The amount of solute in 1.00 L of 0.10 molar aqueous solution is 0.100 mol. This amount of solute would be contained in 0.100 mol/(0.50 mol/L) = 0.2 L or 200 mL. The appropriate procedure would then be to take 200 mL of the original solution and dilute it to 1000 mL with the solvent, water.
In order to compare concentrations on the molal and molar scales the ratio between mass and volume of the solution must be known. This ratio, the solution density, is in itself a useful tool in the measurement of concentration, especially of solutions whose concentration is high. The batteries used in modern automobiles contain concentrated aqueous solutions of sulfuric acid, and the concentration of the acid is directly related to the amount of stored charge in the battery. A hydrometer is used to measure the density of this solution in automobile service stations. A table of the densities and other properties of the strong acids and bases which are commercially available is given below.
|Acid or Base||Molar Mass||Density, g/ML||Molarity||Mass %|
Example. The concentrated nitric acid of commerce has a density of 1.41 g/mL. If the mass per cent HNO3 in such a solution is 69%, we can compute the molar concentration of HNO3. Then we can compute the molal concentration of HNO3. The total mass of one dm3 of concentrated nitric acid is 1410 g, of which (0.69)(1410 g) = 972.9 g HNO3/mL and the balance is water. The molar mass of HNO3 is 63.01 g/mol so 972.9/63.01 = 15.44 mol/mL is the molar concentration, c(HNO3) or [HNO3].
To calculate the molal concentration, it is necessary to obtain the mass of water per mL of solution, which must be (1.00 - 0.69)(1410 g) = 437.1 g H2O/mL. Since the molar concentration of HNO3 is 15.44 mol/L, the molal concentration of HNO3 is (15.44 g/mL)/(0.4371 kg H2O/mL) = 35.32 mol HNO3/kg water.