The Molar Concentration

Many chemical determinations are made using solutions, e.g., titrations. In the calculations which result from these measurements, the molar concentration of the solution is always an important quantity in the calculation. Often the object of a determination is to measure the molar concentration of a solution.

The term "molar concentration" used in this context has a very exact meaning. The molar concentration of a substance X in solution is defined as the amount of X dissolved per unit volume of solution. Algebraically, this can be expressed as the equation:

Here c_X is the concentration of X, while n_X is the amount of X. The volume of the solution in which n_X is dissolved is denoted by the symbol V_soln. The IUPAC recommended term for the quantity c_X, is 'amount concentration.' We have preferred to use the term ,'molar concentration' as being more acceptable to Anglophone ears. In those circumstances in which it is clear that other kinds of concentration (like the mass concentration) are not intended, the abridged term 'concentration' is perfectly acceptable. An example of this usage is the term 'hydrogen ion concentration.'

Like the other proportionality constants discussed above, the concentration can be used to interconvert the two quantities it connects:

Given the volume of a solution, we can find the chemical amount of X dissolved in it by multiplying by the concentration. Alternatively, we can find the volume of a solution containing a given chemical amount of X by dividing this chemical amount by the concentration.

The term Molarity is discussed in a separate section:

The 'Molarity'

Most introductory texts use the term "molarity" in place of the term "concentration". The chief objection to the use of this older term is that it is always defined in unit-specific terms, e.g. by the equation:

This definition is very algebra-unfriendly. When we use unit-specific terms like "number of moles" and "number of liters, It becomes very difficult to write down equations equivalent to the unit-neutral relationships:

There is always the temptation to multiply or divide by 1000 in order to convert moles into millimoles or liters into milliliters. This might well be why algebra is seldom used in textbooks when discussing how to do titration calculations.

Use of the term "number of moles" is particularly inappropriate for titration calculations, in a second way. The volumes involved in conventional titrations are usually between 25 and 45 milliliters and the chemical amounts involved are usually within the range 0.5 to 25 millimoles. Inside these parameters, working in terms of millimoles and milliliters is much easier than in terms of moles and liters. Using the older terminology not only raises the uncertainty of whether the number of liters is the same as the number of milliliters. It also leaves us in doubt about whether we can properly describe the number of millimoles per milliliter as the molarity or not. All these ambiguities disappear as soon as we use unit-neutral definitions and symbols.

A third difficulty with use of the older terminology, is the use of the symbol M in two senses. In the first sense, M is used to indicate the quantity molarity, as in the equation:

M = 0.516 mol HCl / L

In the second sense, M is used to indicate the units moles per liter as in:

molarity = 0.516 M HCl

Not entirely surprisingly, some books even write:

M = 0.516 M

implying, presumably, that we can equate one and 0.516 !

Finally, we should note that the use of the symbol M in this context overworks this letter of the alphabet. The letter M is also used to indicate the molar mass. There are even students who interpret M to mean the unit mole. The symbol c has none of these difficulties.

Although it is very tempting to write 0.1M HCl rather than 0.1 mol HCl / L, especially when labeling bottles of solution, it is worth trying to avoid doing this. Writing out the units explicitly, though more trouble, is also unambiguous. The units are also easier to cancel correctly. You will, however, frequently encounter reagent bottles and problems using the symbol M. You should teach students that when this occurs, they should convert the symbol M to an appropriate set of units. For example if a solution is labeled 0.5 M NaCl, one should translate this to 0.5 mmol NaCl/mL or perhaps into 0.5 mol NaCl / L. In special circumstances, more uncommon units like micromol / microliter may be appropriate.

Titrations in General

Whenever we perform a titration, we add a solution of reactant B from the buret to a solution of F in the titration flask until an end-point is signaled. The end-point corresponds to a situation in which the chemical amount of B added, n_B, and the chemical amount of F originally in the flask, n_F, are in the exact stoichiometric ratio. Algebraically this corresponds to the equation:

We can also easily generalize about the solution in the buret. Either the concentration of B is known or it is not. If we know the concentration of B we can easily find the chemical amount of B added by multiplying the volume added by this concentration. Once we know n_B, The Stoichiometric ratio, _F / _B will then allow us to calculate n_F. This argument gives us the partial road-map:

Titrations Involving Two Solutions

In a titration involving two solutions, not only is the volume of solution B in the buret known, but also the volume of solution F added to the flask from a pipet or perhaps a buret. In this type of titration there is a very simple relation between the two concentrations c_B and c_F. One of these will be known while the other must be calculated. The chemical amount of B in a volume V_B is given by the equation n_B = c_B V_B while the chemical amount of F is given by n_F = c_F V_F. We have just seen that:

A solution of H₂O₂ was pipetted into an Erlenmeyer flask using a 25.00 mL pipet. An excess of H₂SO₄ was then added. Finally, the acidified solution in the flask was titrated with a solution containing 0.0413 mol KMnO₄ / L. An end-point was obtained when 26.32 mL had been added. Calculate the concentration of the H₂O₂ solution. The balanced reaction for the reaction is:

2KMnO₄ + 5H₂O₂ + 3H₂SO₄ = 2MnSO₄+ K₂SO₄ + 5O₂ + 8H₂O

Substituting the values given in the problem into the formula gives us:

and finally:

Some readers will object to the above method for solving the problem on the grounds that we have simply plugged numbers (and quantities) into a formula without any thought about the logic of the problem. There is some merit to this objection, particularly if beginning students are involved. Accordingly, we will now perform the same calculation using a road-map of the logic.

Since we are given the concentration of KMnO₄ and asked to find the concentration of H₂O₂ our road-map needs to involve the step:

Titrations Involving a Solid

When a titration involves adding a solution from the buret to react with a weighed solid sample, the calculation procedure depends on whether the sample is pure or not. If it is pure, then the object of the titration is usually to find c_, the concentration of the solution. On the other hand, if the solid is impure, then the object of the titration is usually to find its percentage purity.

Pure Solid

If the solid F is pure, we can easily find its chemical amount from its mass and molar mass using the formula: n_F = m_F M_F. Multiplying n_F by the stoichiometric ratio then gives us n_B, the amount of B added from the buret. Dividing n_B by V_B, the volume delivered from the buret gives the concentration of B. Diagraming this logic gives the following road-map:

When 0.2832 g of Na₂CO₃ was titrated with a solution of HCl, 32.62 mL of acid was required. Find the concentration of the HCl.

With the aid of the road-map we readily calculate:

Impure Solid

If the object of a titration is to find the percentage purity of a solid, then we need to find "how much" of the pure substance F there is in the impure solid. We first find 'how much' in terms of the chemical amount of F, but we need to convert this to the corresponding mass, m_F. We need this mass because we want to find out what fraction of the mass is F (i.e. we want to find the mass fraction, w_F = m_F / m_total). Next we must ask, "how do we find the chemical amount of F in the flask?" We find it from the amount of in the solution delivered from the buret and the simple whole number stoichiometric ratio. The chemical amount of B, in turn, we find from the product of the concentration and the volume of solution . The corresponding road-map reads:

An example of this type of calculation now follows:

A solid sample of impure Na₂S₂O₃ weighing 0.6907 g was titrated with a solution of I₂ in order to find its percentage purity. Exactly 26.83 mL of a solution containing 0.05186 millimole of I₂ per milliliter was required to achieve an end-point using a starch indicator. Find the mass fraction of Na₂S₂O₃ in the sample and hence its "percentage purity"

2 Na₂S₂O₃ (aq) + I₂ (aq) = Na₂S₄O₆ (aq) + 2 NaI (aq)

Using the above road-map as a guide, it is easy to write out the calculations performed for each step:

The Right Language

We conclude this presentation by considering three more examples. These have been chosen partly for their utility, but mainly because they illustrate the ability of an algebra-based unit -neutral convention to spell out the logic of a stoichiometric situation with complete clarity using both words and symbols. In a word, they show that we have the right language for discussing stoichiometry.

Rinsing a pipet

A common problem encountered in teaching students volumetric analysis is the difficulty of explaining to them why it is necessary to rinse a pipet. This is particularly difficult in unit-specific language. We seldom pipet as much as a mole of solute or a liter of solution.

Let us suppose that the solution to be pipetted has a known concentration c_F, and that the volume of the pipet is V_F. In such a case, the object of pipetting the solution is to introduce into the flask a known chemical amount of F given by the formula

n_F = c_F x V_F
Presumably, the pipet is accurate. Hopefully, trhe same can be said for the student's ability to use it to deliver the correct volume. In such a case, the only source of inaccuracy is the concentration of the solution in the pipet. If the pipet has been washed with water, but not rinsed, then the water, which may not even be visible, will dilute this solution of F when the pipet is filled. Accordingly, the actual concentration of F in the pipet will be less than c_F. Hence the actual chemical amount of F delivered into the titration flask will be less than the calculated value of n_F. Only if we rinse the pipet several times with the solution being pipetted, will we wash out this remaining water. After this rinsing, the solution will have the right concentration and the amount of F actually pipetted have the desired value of n_F.

It is also worth pointing out to students that once a known amount of F has been introduce into the flask, the addition of water will have no effect on the value of the end-point.

Preparing a Stock Solution

Our next example involves a non-linear operation, namely the addition of two masses. As a result, it is difficult to handle using "dimensional analysis"

A concentrated solution of silver nitrate was prepared by dissolving 40.0 g of pure crystalline AgNO₃ in 60.0 g of water. The density of the resulting solution was 1.447 g cm^-3 at room temperature. Find the concentration of the solution at this temperature. Give you results in moles per liter.

We are asked to find the concentration of AgNO₃. This concentration is defined by the equation:

The chemical amount,, in the denominator is easily obtained by dividing the mass of AgNO₃ by its molar mass:

The difficulty in this problem is to find the final volume of the solution. The density is provided for this purpose. We know the total mass of the silver nitrate solution. It is the sum of the 40.0 g of salt and 60.0 g of water, namely 100 g of solution. Accordingly the volume of the solution is given by:

Finally, the concentration of the silver nitrate solution is given by:

Here we have ignored the very small difference between the liter and the cubic decimeter.

It is also possible to draw a road-map for this type of problem. However, the road-map is no longer a simple line of proportionality steps. We need to calculate and separately, and then combine these to give the final concentration:

It is worth noting that no stoichiometric ratio is involved because no chemical reaction is involved,

Finding a Solubility Product

As a third example of the flexibility of our new language, we will calculate the solubility product of magnesium oxalate. This determination is currently a second semester freshman lab experiment at Emporia State University. As in most equilibrium problems, the calculation of K_sp here, is not a simple linear set of proportionalities. In consequence, it does not lend itself to a "dimensional analysis" treatment.

In an experiment to determine the solubility product of magnesium oxalate, 35.0 mL of a solution containing 0.500 mol MgSO4 per liter and 25.0 mL of a solution containing 0.500 mol K₂C₂O₄ (potassium oxalate) per liter were mixed and allowed to stand overnight. The crystals of magnesium oxalate which formed were filtered off. A 20.00 mL aliquot of the resulting clear solution was then acidified with H₂SO₄ and titrated with a solution containing 0.004 mol KMnO₄ per liter, according to the equation:

5 C₂O₄^2- + 2 MnO₄^- + 16 H⁺ = 2 Mn²⁺ + 10 CO₂ + 8 H₂O

Exactly 49.90 mL of the oxidizing agent was needed to obtain an end-point. Calculate the solubility product of Mg₂C₂O₄:

MgC₂O₄ = Mg²⁺ + C₂O₄²

In discussing this problem, we will distinguish between various quantities before and after precipitation with appropriate labels.

The object of this experiment is to determine the solubility product of magnesium oxalate given by the expression:

The purpose of the permanganate titration was to determine the concentration of oxalate in the solution after precipitation. We will use the two-solution titration formula:
The concentration of Mg²⁺ after precipitation is more difficult to calculate. We have first to realize that when the magnesium oxalate precipitates, this decreases not only the oxalate concentration but also the magnesium concentration. Since the stoichiometric ratio of Mg²⁺ to Ox^2- in the precipitate is 1/1, equal chemical amounts of each ion will be removed from the solution when precipitaion occurs. Thus both concentrations will decrease to the same extent:
We can easily calculate from the difference in the magnesium ion concentrations before and after precipitation:
Knowing , we can calculate the concentration of Mg²⁺ after precipitation from its value before precipitation:
The solubility product of magnesium oxalate follows immediately:
In the above calculation, we have made no attempt to allow for the severe effect of ionic interactions but have treated activities as equal to concentrations.

A noteworthy feature of the above calculation is that two stoichiometric ratios are involved. This is because two chemical reactions are involved: the reaction of KMnO₄ with oxalate and the precipitation reaction. Inevitably, if we proceed substance-wise from permanganate to oxalate to magnesium, we have no choice but to use these two stoichiometric ratios.

We can also diagram a road-map for this calculation. Since the logic involves two subtractions it cannot be arranged in a single line of logic. The diagram given below is partly determined by the equation editor in Word Perfect but it also represents an attempt to accentuate that the decrease in concentration on precipitation of the anion and the cation is the same. Readers may have different ideas on how this diagram should be drawn.

Conclusion

It has not been the intention in this presentation to produce a complete account of how to teach stoichiometry. Many topics, for example gases and empirical formulae, have been omitted. Rather, the object has been to argue that the way stoichiometry is currently taught is full of inadequacies and inconsistencies, and that it is time that we changed it. At the same time the essentials of an alternative approach stressing quantities and their algebraic symbols has been proposed. The chief value of this approach is that it enables us to discuss the logic of each step in a stoichiometric calculation both in words and in symbols.

A point that has been deliberately excluded from this discussion is the Avogadro constant. We simply do not need to know the number of entities per mole in order to do stoichiometric calculations. In the author's opinion, too much emphasis is placed on this constant (usually in the form of the Avogadro number) in the early chapters of introductory texts. Certainly the Avogadro constant is important, but not in stoichiometry. This constant harping on 6.023 x 10²³ distracts students from the central logic of all stoichiometric calculations, namely the law which states that in any chemical reaction, the amounts of substances consumed or produced are always in simple whole number ratios.

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