THE CONCEPT OF MOLE
A mole is an amount of a substance, also it is a specific number of molecules, formula units, ions, etc. of a substance. A practical definition for a mole is that one mole of a substance is the molecular mass (weight) or formula weight of the substance expressed in grams. The formal definition for mole is that it is the amount of a substance that contains exactly the number of molecules or formula units as the number of atoms in 12 g of carbon-12 (12C). This number is 6.02 x 1023 and is called Avogadro’s number. One mole of a molecular compound therefore contains 6.02 x 1023 molecules or Avogadro’s number of molecules. The number of moles in a particular mass of a substance is:
____________________________mass of a substance_____________________________________ = moles molecular mass (weight) or formula weight of a substance expressed in grams (g/mole)
In practice, the term mole is also commonly used with other species such as ions, elements, electrons, and so forth. For example, one gram-atom of sodium is often called a mole of sodium and contains 6.02 x 1023 atoms. The species under discussion must be specifically stated. For example, one gram-atom of nitrogen contains one mole of nitrogen atoms (N) but 1/2 mole of N2molecules. Therefore it must be stated whether the nitrogen atom or the N2 molecule is the species under discussion.
CONCENTRATIONS BASED ON THE VOLUME OF SOLVENT
1. Molarity (M) is the number of moles of solute dissolved in one liter of solution. Molarity is also called molar concentration. The molarity of a solution can be calculated when the number of grams of the substance in the solution, the volume of the solution, and the molecular weight or formula weight of the substance are known. Conversely, the number of grams of the substance needed to prepare a solution of a particular molarity can be calculated if the final volume of the solution is defined and the molecular weight or formula weight of the substance is known. Dilute solutions are named by using the appropriate prefix. For example:
10-3M = 1mM (millimolarity) = 1 mmole L-1 = 1µmole mL-1
10-6M = 1µM = 1 µmole L-1 = 1 nmole mL-1
The equation defining molarity is:
molarity (M) = moles of solute / liters of solution
or rearranged
molarity (M) x liters of solution = moles of solute
The following are true for a solution that is diluted. Before dilution
molarity (M1) x liters (V1) = moles (m1)
M1,V1 = molarity, volume before dilution
After dilution (number of moles does not change)
M2 x V2 = m1
M2,V2 = molarity, volume after dilution
therefore
M1 x V1 = M2 x V2
Any units can be used for volume and molarity but V1 and V2 must be in the same units as must M1 and M2.
2. Normality (N) is the number of equivalents of a solute dissolved in one liter of solution. For an acid or base, an equivalent is the molecular weight or formula weight of the acid or base expressed in grams divided by the number of moles of hydronium (H3O+) or hydroxyl ions produced by this amount of acid or base. For a compound that undergoes oxidation or reduction, an equivalent is the molecular weight or formula weight of the reductant or oxidant expressed in grams divided by the number of faradays of electrons released or accepted by this amount of reductant or oxidant. A faraday is a mole of electrons, i.e. 6.02 x 1023 electrons. Equivalents can be calculated from moles.
equivalents = moles / n
n = number of hydronium or hydroxyl ions produced from one molecule (ion) of an acid or base
or
n = number of electrons released or accepted from one molecule (ion) of an reductant or oxidant.
The term equivalent weight is used and related to molecular weight and formula weight by
equivalent weight = molecular weight / n = formula weight / n
An equivalent is also defined as the equivalent weight expressed in grams. The number of equivalents in a certain mass of a substance is calculated from:
equivalents = mass (grams) of a substance / equivalent weight (g/eqiv.)
Normality and molarity are related by the formula, N = nM, where n has the meanings stated above.
BUFFERS
A buffer is a solution that resists changes in pH when acid or base are added to it. A buffer consists of a solution of a weak acid and the corresponding conjugate base. One can also state it consists of a solution of a weak base and its conjugate acid. These two statements say the same thing. A buffer may be prepared from either a weak acid or a weak base. For example, a 0.05 M acetate solution at pH 5.0 is a buffer. It is prepared from acetic acid by converting some of the acetic acid to acetate ion with strong base. The conjugate weak acid/weak base pair is acetic acid/acetate ion. The word “acetate” in acetate buffer refers to both the conjugate acid (in this example, acetic acid) and the conjugate base (acetate ion). These two species together make up the acetate buffer. Each species is present in the buffer at a specific concentration and the concentration given for the acetate buffer is the sum of the concentrations of these two species, in this example 0.05 M. The ratio of the concentrations of the two species in the buffer is obtained from the Henderson-Hasselbalch equation.
How do we know what weak acid or base to select for the preparation of a buffer? First, the pH of the buffer to be prepared must be decided. This is dictated by what pH is needed in the experiment being performed. Secondly the pKa of the weak acid of the conjugate acid/base pair should be at or near the pH to be used in the experiment. Recall from the discussion of titration curves that the pH of a solution of a weak acid changes least at the pKa of the acid. When the pH is equal to the pKa of the weak acid, the ratio of A–/HA is 1/1. Finally the weak acid and conjugate base should not interfere in any way in the experiment, such as by altering the activity of an enzyme, or by reacting chemically with any of the substances present in the experimental solution. As an example, assume the activity of an enzyme is to be tested at pH 5.0. What buffer should be used? A weak acid whose pKa is about 5.0 should be used. Acetic acid has a pKa of 4.77 and hence an acetate buffer can be used.
How is a buffer prepared? All buffers should be prepared by dissolving the appropriate weak acid (or weak base) in about 3/4 of the total volume of water needed for the buffer and adjusting the solution to the desired pH by using a pH meter and either a strong base (or acid). The solution is then quantitatively transferred to a volumetric flask and brought to volume.
How does a buffer work? Consider the acetate buffer above. When an acid (H3O+) is added to the buffer, a change in pH is resisted by Ac– reacting with the acid:
In this way, H3O+ is neutralized. But all of the added H3O+ is not neutralized because as neutralization occurs, the [HAc] increases and the [Ac–] decreases and neutralization only continues until equilibrium is again restored. The [H3O+] is somewhat higher in concentration at the new equilibrium because the new equilibrium concentration of Ac– is lower. For this reason a buffer only resists changes in pH it does not completely prevent a change in pH. Buffers therefore do change in pH when an acid or base is added. They simply change much less than a solution with no buffer. When base (OH–) is added to the acetate buffer the following reaction occurs:
Again all the added OH– is not neutralized because the new equilibrium concentration of HAc is lower and therefore a higher concentration of OH– is required to maintain the new equilibrium.
Buffers have maximum buffering capacity against added acid and base when equal molar quantities of the weak acid/weak base conjugate pair are present in the buffer solution. If more of the conjugate acid is present in the buffer than conjugate base, the buffer has greater buffering capacity against added bases than acids. Conversely if more conjugate base is present than conjugate/acid, the buffer has greater buffering capacity against added acid than base.
