# Solution Concentrations

Submitted by ChemPRIME Staff on Wed, 12/08/2010 - 23:50

In the laboratory, in your body, and in the outside environment, the majority of chemical reactions take place in solutions. Macroscopically a solution is defined as a homogeneous mixtureA mixture in which all regions have the same composition and state of matter. of two or more substances, that is, a mixture which appears to be uniform throughout. On the microscopic scale a solution involves the random arrangement of one kind of atom or moleculeA set of atoms joined by covalent bonds and having no net charge. with respect to another.

There are a number of reasons why solutions are so often encountered both in nature and in the laboratory. The most common type of solution involves a liquidA state of matter in which the atomic-scale particles remain close together but are able to change their positions so that the matter takes the shape of its container solventThe substance to which a solute is added to make a solution. which dissolves a solidA state of matter having a specific shape and volume and in which the particles do not readily change their relative positions. soluteThe substance added to a solvent to make a solution.. (The term solvent usually refers to the substance present in greatest amount. There may be more than one solute dissolved in it.) Because a liquid adopts the shape of its container but does not expand to fill all space available to it, liquid solutions are convenient to handle. You can easily pour them from one container to another, and their volumes are readily measured using graduated cylinders, pipets, burets, volumetric flasks, or other laboratory glassA solid material that does not have the long-range order of a crystal lattice; an amorphous solid. A glass melts over a range of temperatures instead of having the definite melting temperature characteristic of crystalline solids.-ware. Moreover, atomsThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume. or molecules of solids dissolved in a liquid are close together but still able to move past one another. They contact each other more frequently than if two solids were placed next to each other. This “intimacy” in liquid solutions often facilitates chemical reactions.

Since solutions offer a convenient medium for carrying out chemical reactions, it is often necessary to know how much of one solution will react with a given quantity of another. Examples in other sections have shown that the amount of substance is the quantity which determines how much of one material will react with another. The ease with which solution volumes may be measured suggests that it would be very convenient to know the amount of substance dissolved per unitA particular measure of a physical quantity that is used to express the magnitude of the physical quantity; for example, the meter is the unit of the physical quantity, length. volume of solution. Then by measuring a certain volume of solution, we would also be measuring a certain amount of substance.

One measure of the concentration c of a solute in a solution is often called molarityA concentration unit expressed as chemical amount of a substance per liter of solution., but it is probably better to call it "the concentration in molar units" or "molar concentration" (keeping the parameter concentration, and its unit, M for molar distinct). The molar concentration is the amount of the substance per unit volume (L or dm3) of solution(not solvent):

$\text{Concentration of solute, M}=\frac{\text{amount of solute, mol}}{\text{volume of solution, L}}$

$c_{\text{solute, M}}~=~\frac{n_{\text{solute, mol}}}{V_{\text{solution,L}}}\text{ (1)}$

The units moles per literA unit of volume equal to a cubic decimeter. (mol liter–1) or moles per cubic decimeter (mol dm–3) are used to express molar concentration. They are equivalent (since 1 dm–3 = 1 liter).

If a pure substance is solubleAble to dissolve in a solvent to a significant extent. in water, it is easy to prepare a solution of known concentration. A container with a sample of the substance is weighed accurately, and an appropriate massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object. of sample is poured through a funnel into a volumetric flask, as shown in the figure. The container is then reweighed. Any solid adhering to the funnel is rinsed into the flask, and water is added until the flask is about three-quarters full. After swirling the flask to dissolve the solid, water is added carefully until the bottom of the meniscusThe curvature of a liquid in a cylindrical vessel at the liquid-air interface. It is caused by attractive forces between the liquid and the walls of the vessel. coincides with the calibration mark on the neck of the flash. This process is shown in detail in Figure 1:

Figure 1. Making a 1 liter of 1 molar NaCl solution.

The following movie shows a difficult step in this process: adding water to dilute the solution to the proper concentration. In this movie, water is added to the volumetric flask until it is 2cm from the mark. A wash bottle is then used to bring the solution level to within a few millimeters of the mark. Finally, a dropper is used to fill to the mark and ensure the calibration mark is not overshot.

EXAMPLE 1 A solution of KI was prepared as described above. The initial mass of the container plus KI was 43.2874 g, and the final mass after pouring was 30.1544 g. The volume of the flask was 250.00 ml. What is the concentration of the solution?

Solution The concentration can be calculated by dividing the amount of solute by the volume of solution [Eq. (1)]:

$c_{\text{KI}}=\frac{n_{\text{KI, mol}}}{V_{\text{solution, L}}}$

We obtain nKI from the mass of KI added to the flask:

mKI = 43.2874 - 30.1544 g = 13.1330 g

nKI = 13.1330 g × $\frac{\text{1 mol}}{\text{166}\text{.00 g}}$ = 7.9115 × 10-2 mol

The volume of solution is 250.00 ml, or

Vsolution = 250.00 cm3 × $\frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}}$ = 2.5000 × 10-1 dm3

Thus

$c_{\text{KI}}=\frac{n_{\text{KI}}}{V_{\text{solution}}}=\frac{\text{7}\text{.9115}\times \text{10}^{\text{-2}}\text{ mol }}{\text{2}\text{.50 }\times \text{10}^{\text{-1}}\text{ dm}^{\text{3}}}=\text{3}\text{.1645 }\times 10^{^{\text{-1}}}\text{mol dm}^{\text{-3}}$

Note that the definition of concentration is entirely analogous to the definitions of densityThe ratio of the mass of a sample of a material to its volume., molar massThe mass of a mole of substance; the same as molecular weight for molecular substances., and stoichiometric ratio that we have previously encountered. Concentration will serve as a conversion factorA relationship between two units of measure that is derived from the proportionality of one quantity to another; for example, the mass of a substances is proportional to its volume and the conversion factor from volume to mass is density. relating the volume of solution to the amount of dissolved solute.

$\text{Volume of solution}\overset{concentration}{\longleftrightarrow}\text{amount of solute}$      $V\overset{c}{\longleftrightarrow}n$

Because the volume of a liquid can be measured quickly and easily, concentration is a much-used quantity. The next two examples show how this conversion factor may be applied to commonly encountered solutions in which water is the solvent (aqueousDescribing a solution in which the solvent is water. solutions).

EXAMPLE 2 An aqueous solution of HCl [represented or written HCl(aq)] has a concentration of 0.1396 mol dm–3. If 24.71 cm3 (24.71 ml) of this solution is delivered from a buret, what amount of HCl has been delivered?

Solution Using concentration as a conversion factor, we have

$V\text{ }\xrightarrow{c}\text{ }n$

$n_{\text{HCl}}=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{0}\text{.1396 mol}}{\text{1 dm}^{\text{3}}}$

The volume units will cancel if we supply a unity factor to convert cubic centimeters to cubic decimeters:

$n_{\text{HCl}}=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{0}\text{.1396 mol}}{\text{1 dm}^{\text{3}}}\times \left( \frac{\text{1 dm}}{\text{10 cm}} \right)^{\text{3}}$

$=\text{24}\text{.71 cm}^{\text{3}}\times \frac{\text{0}\text{.1396 mol}}{\text{1 dm}^{\text{3}}}\times \frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}}$

= 0.003 450 mol

The concentration units of moles per cubic decimeter are often abbreviated M, pronounced molar. That is, a 0.1-M (one-tenth molar) solution contains 0.1 mol solute per cubic decimeter of solution. This abbreviation is very convenient for labeling laboratory bottles and for writing textbook problems; however, when doing calculations, it is difficult to see that

1 dm3 × 1 M = 1 mol

Therefore we recommend that you always write the units in full when doing any calculations involving solution concentrations. Also, it is sometimes easier to use the unit liter, which is equivalent to cubic deciliters:

$\text{1 dm}^{\text{3}}\times \text{1 }\frac{\text{mol}}{\text{dm}^{\text{3}}}=\text{1mol}$

$\text{1 L}\times \text{1 }\frac{\text{mol}}{\text{L}}=\text{1mol}$

Problems such as Example 2 are easier for some persons to solve if the solution concentration is expressed in millimoles per cubic centimeter (mmol cm–3) or millimoles per ml (1 ml = 1 cm–3) instead of moles per cubic decimeter. Since the SI prefix m means 10–3, 1 mmol = 10–3 mol, and

$\text{1 M}~=~\frac{\text{1 mol}}{\text{1 dm}^{\text{3}}} ~\times~ \frac{\text{1 dm}^{\text{3}}}{\text{1 L}} ~=~\frac{\text{1 mol}}{\text{L}}$

$\text{1 M} ~=~\frac{\text{1 mol}}{\text{L}} ~\times~\frac{\text{10}^{\text{-3}}\text{ L}}{\text{1 ml}} ~\times~\frac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~=~\frac{\text{1 mmol}}{\text{1 ml}}$

$\text{1 M} ~=~\frac{\text{1 mol}}{\text{1 dm}^{\text{3}}} \times \frac{\text{1 dm}^{\text{3}}}{\text{10}^{\text{3}}\text{ cm}^{\text{3}}} \times \frac{\text{1 mmol}}{\text{10}^{\text{-3}}\text{ mol}} ~=~ \frac{\text{1 mmol}}{\text{1 cm}^{\text{3}}}$

Thus a concentration of 0.1396 mol dm–3 (0.1396 M) can also be expressed as 0.1396 mmol cm–3, 0.1396 mol/L or 0.1396 mmol/mL. Expressing the concentration this way is very convenient when dealing with laboratory glassware calibrated in milliliters or cubic centimeters.

EXAMPLE 3 Exactly 25.0 ml NaOH solution whose concentration is 0.0974 M was delivered from a pipet. (a) What amount of NaOH was present? (b) What mass of NaOH would remain if all the water evaporated?

Solution

a) Since 0.0974 M means 0.0974 mol dm–3, or 0.0974 mmol cm–3, we choose the latter, more convenient quantity as a conversion factor:

$n_{\text{NaOH}}=\text{25}\text{.0 cm}^{\text{3}}\times \frac{\text{0}\text{.0974 mol}}{\text{1 cm}^{\text{3}}}=\text{2}\text{.44 mmol}=\text{2}\text{.44}\times 10^{\text{-3}}\text{ mol}$

b) Using molar mass, we obtain

$\text{m}_{\text{NaOH}}=\text{2}\text{.44}\times 10^{\text{-3}}\text{ mol}\times \frac{\text{40}\text{.01 g}}{\text{1 mol}}=9.\text{76}\times 10^{\text{-2}}\text{g}$

Note: The symbols nNaOH and mNaOH refer to the amount and mass of the solute NaOH, respectively. They do not refer to the solution. If we wanted to specify the mass of aqueous NaOH solution, the symbol mNaOH(aq) could be used.