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

We 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

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 solute_concentrated = mols solute_dilute

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

this equation simplifies to

or

M_concentrated * volume_concentrated = M_dilute * volume_dilute

M_cV_c = M_dV_d

Sample Molarity Problem #3:

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

Here are two additional problems which I did not have time to do in class. Give them a try and I'll post the answers next week.

Additional solution stoichiometry problems;

• How many grams of Ba(OH)₂ are contained in a 25.00 mL solution if 16.52 mL of 0.850 M HCl are required to completely neutralize the sample?
• If 25.0 mL of 0.625 M HBr are mixed with 42.0 mL of 0.352 M NaOH, will the resulting solution be acidic or basic?