Solution, Concentration and Solution Stoichiometry

In the laboratory chemical reactions are generally performed in solution. For our purposes all reactions will be performed in water, or aqueous solutions. To maintain the quantitative aspect of chemical reactions in solution we must be able to express the amount of a substance in water. To do this we use the term concentration. While there are many ways to express the concentration of a solution, we will focus on only one at this point. But before we introduce this concentration expression, we need to consider some definitions.

An aqueous solution will consist of water and a dissolved substances. For example, an aqueous solution of sodium chloride contains NaCl and water. One of the components of the solution is the solute the other is the solvent. By definition the solute is the component present in the smallest amount. The solvent is the component present in the largest amount, or the component whose phase is the phase of the solution, or water.

The concentration expression of interest is called molarity. Molarity has the following definition-

1 Molar (1M) HCl contains 1 mol per liter of solution. 1M HCl, 1M NaOH, 1M NaCl, 1M C6H12O6 each contain 1 mole of solute in 1 liter. 2M HCl contains 2 mole of HCl in 1 liter. A 1M solution could also contain 0.5 mol in 0.5 liters.

Sample Molarity Problem #1:

Calculate the molarity of a solution containing 0.875 mol NaCl in;

a) 1 liter Answer

b) 500. mL Answer

c) 100 mL Answer

d) 6 liters Answer

We need to spend a few minutes discussing how solutions of particular concentrations (molarity) are prepared. The normal situation encountered in the laboratory requires the preparation of a particular volume of a solution of a particular concentration. Generally, the solution can be prepared two different ways. Which method depends on the what chemicals and/or solutions are available.

Solutions of a particular concentration can

either be prepared by mixing a specified mass of solute with water, or by diluting a more concentrated solution of the same reagent. We will discuss both methods.

The calculations and the procedure are both important when preparing a solution of a particular concentration.

Sample Molarity Problem #2

How would you prepare 500.00 mL of a 0.500 M solution of Cu(NO3)2?


To begin we must calculate the amount of Cu(NO3)2 contained in 500. mL of a 0.500 M solution. To do this we calculate,

To prepare the solution 47.0 g of Cu(NO3)2 are measured using an analytical balance. This mass of solute is dissolved in enough water so the final volume is 500. mls.

It is important to note that in this problem the mass of solute is not added to 500. mLs, but enpough water is added to dissolve the Cu(NO3)2 and reach a final volume of 500. mLs. A 500 mL volumetric flask is used to prepare the solution. The solid is transferred to the flask and approximately 300 mLs of distilled water are added. This mixer is stirred/swirled until all of the Cu(NO3)2 dissolves, then the remaining distilled water is added to reach the final volume.

Another important procedure in preparing solution is dilution. Suppose after preparing the above solution you found that you needed 200 mLs of a 0.150M Cu(NO3)2 solution. We can obtain 200 mLs of 0.150 M solution by diluting the more concentrated solution.

In a dilution the moles of solute remain constant, one is only adding water to an intial volume of concentrated solution to dilute it to a new concentration. That is,

mols soluteconcentrated = mols solutedilute

If we multiply this equation by 1 we can change it to a more useful relationship for solving dilution problems. But the factor we multiply by will be . The equation becomes

this equation simplifies to


Mconcentrated * volumeconcentrated = Mdilute * volumedilute

McVc = MdVd

Sample Molarity Problem #3:

Calculate the volume of 0.500 M Cu(NO3)2 needed to prepare 200. mLs of a 0.150 M Cu(NO3)2 solution.


To solve this problem we will use the relationship,

McVc = MdVd

Rearranged the equation becomes;

This solution is prepared by measuring 60 mLs of 0.500 M Cu(NO3)2 into a 500. graduated cylinder. Enough water is added to reach a final volume of 200 mLs.

We've been doing stoichiometry calculations with the amounts of reactants given in grams or moles. Now we can solve problems when the reaction occur in solutions. Here is a sample problem of this new type;

Sample Molarity (stoichiometry) Problem #4

One of the reaction types described earlier was the neutralization reaction. This is a reaction between an acid and a base. A common technique used in introductory chemistry courses to determine concentrations of solution is titration. Titration involves the addition of a measured amount of solution (base) to a known volume of another solution (acid). This technique will work as long as the two compounds react with each other, and there is a reagent which can be added to indicate when the reaction is over. An acid/base titration can be used to determine the amount of acid in a solution. To do this a base of known concentration is added using a volumetric buret to a known volume of acid. An indicator is used to signal when the equivalence point is reached in the titration. At the equivalence point the moles of acid equal the moles of base. Both stoichiometry and molarity calculations are used to arrive at an answer. Let's consider a problem as an example,


Sample Molarity (stoichiometry) Problem #5:

Calculate the molarity of sulfuric acid in a 20.00 mL sample which is neutralized by 18.50 mLs of 0.750 M NaOH.


To do this problem, like any stoichiometry problem we need a balanced chemical equation. The equation is;

H2SO4(aq) + 2NaOH(aq) ---> Na2SO4(aq) + H2O(l)

This is basically a stoichiometry problem. To calculate the molarity of sulfuric acid we need to know the mols of sulfuric acid. We already know the volume. To determine the mols of H2SO4 we will use the balanced equation and the information about NaOH. The equation says 1 mol of H2SO4 reacts with 2 mols of NaOH, so if we can determine the mols of NaOH that react we can convert to mols of H2SO4.

To calculate the mols of NaOH;

Now we calcualte the mols of H2SO4 required to react with 0.0139 mols NaOH;

So now we know the mols of H2SO4 we can calculate the concentration of H2SO4;