Solution Concentrations

Mr. Bouyer
Day 1 | Day 2 - 3 | Day 4 | Lab | Skills Test


Describe the parts of a solution. Use molarity to describe solution concentration. Use factor-label to calculate molarity. Use calculations to determine solution dilutions. Express solution concentration is terms other than molarity.

Solutions:
homogeneous mixtures of solute and solvent.

There will be a formal test before Lab Day. You will demonstrate on this test that you do molarity calculations. Your paper periodic table and a calculator may be used during the test, but nothing else.

Solvent - the most abundant substance in a solution.

Solute - the other substance in a solution.

It is possible for a solution to have more than one solute. Air is an example.

Concentration - comparison of the amounts of solute and solvent.

Describing a solution as "stong" or "weak" does give you some comparison of the amounts of solute and solvent, but it is only a general idea. Even the terms "dilute" and "concentrated" do not give enough information to make quantitative calculations. To be able to compare solutions quantitatively, we must know "how much" solute and solvent are present. Moles can be used to compare solutions, giving us the most common units of concentration.

Molarity (M) = moles of solute per dm³ of solution.

Remember the following:

total solution

Molarity Calculation Examples:
The factor-label solution is shown for each of the following sample problems. Study how the problem is solved, understanding each step in the conversion process. When you understand, use your calculator to find each answer with the proper number of significant digits.

In the first two examples, the problem gives you a mass of solute in a volume of solution. This may be written as mass/volume, the form needed for molarity. The only thing you have to do is convert mass to moles and the volume to dm³.

What is the molarity of a liter of solution containing 100g of copper (II) chloride?

Calculate the molarity of 100ml of solution contains 25g of silver nitrate.

The next problem is different. It is asking "how do you prepare?" a certain volume of solution with a certain molar concentration. Even though it asks a different question, it is still a problem that converts units, therefore it is worked with factor-label. As in any factor-label problem, the first step is to write down what is given in the problem. Look closely at how that information is used.

Prepare 250ml of 0.5M salt water.

The key difference in the two types of molarity problems above is that one asks for a concentration and the other provides a concentration.
You absolutely must be able to work and understand these types of problems. Work these two practice problems. When you have both of them set up in factor-label form, show them to your science facilitator.

Molarity Practice Problems:

Calculate the M of 200cm³ of a solution containing 50g NaOH.
How much calcium sulfate is needed to make 100cm³ of 0.25M CaSO₄?
How many grams of HCl are in 100ml of a 0.5M hydrochloric acid solution?

Day 2 - 3
Making dilutions:

A solution can be made less concentrated by dilution with solvent. The number of moles of solute does not change when more solvent is added to the solution. If a solution is diluted from V₁ to V₂, the molarity of that solution changes according to the equation:

M₁ V₁ = M₂ V₂
original solution 1 = diluted solution 2
The volume units must be the same for both volumes in this equation.

Dilution calculation example:

How do you prepare 100ml of 0.40M MgSO₄ from a stock solution of 2.0M MgSO₄?

There are two solutions involved in this problem. Notice that you are given two concentrations, but only one volume. Solution #1 is the one for which you have only concentration - the solution that is already sitting on the shelf. Solution #2 is the one for which you have both concentration and volume - the solution that you are going to prepare.
At least until you are comfortable with this type of problem, it may be helpful to write out what numbers go with what letters in our equation.
M₁ = 2.0M MgSO₄ V₁ = unknown
M₂ = 0.40M MgSO₄ V₂ = 100ml

Step 1 - write the equation:

₁

₂

Step 2 - manipulate the equation:

₁

₂

₁

Step 3 - put numbers into the equation:

₁

Step 4 - do the calculation:

₁

Step 5 - describe preparing the solution:

₄

Test Your Concept Understanding:
Work these practice problems.

Use this webtest to practice molarity calculations.

Day 4

Although molatiry is the most common type of solution concentration used in general chemistry, there are several situations when a different comparison between solute and solvent is needed. Some of these have specialized uses, but you should be familiar with the following:

Other solution concentrations:

molality (m) = moles solute / Kg solvent

Colligative properties depend on the number of particles of a substance involved. Chemists do calculations dealing with changing vapor pressure, boiling point elevation, and freezing point depression in solutions. A solution concentration that compares moles of solute and kilograms of solvent is most useful in these calculations.

normality (N) = equivalents of solute / dm³ solution

Normality is a useful concentration unit to use during neutralization reactions (titrations). One mole of hydrogen ions reacts with one mole of hydroxide ions to produce water. But that doesn't mean that one mole of any acid will neutralize one mole of any base. Chemists need a unit for the amount of acid (or base) that will give one mole of hydrogen (or hydroxide) ions.
One equivalent is the amount of an acid (or base) that will give one mole of hydrogen (or hydroxide) ions.
The numerical values of normality and molarity are equal for acids and bases that give 1 equivalent of H⁺ or OH ^- per mole. For example, a solution containing 1 mole of NaOH per dm³ is 1M and also 1N. A solution containing 1 mole of H₂SO₄ per dm³ is 1M, but it is 2N because it contains 2 equivalents of H⁺ per mole.

mole fraction = the ratio of the number of moles of one substance to the total number of moles of all substances in the solution.

Mole fraction is a dimensionless quantity, like a ratio. The sum of the mole fractions of all the components in a solution must equal to ONE.
In a solution containing n_A moles of solute and n_B moles of solvent, the mole fraction of solute, X_A, and the mole fraction of the solvent, X_B, can be expressed as follows:

X_A = n_A / n_A + n_B X_B = n_B / n_A + n_B

mass percent = the percent of a solution's total mass that is solute.
Many commercial solutions are labeled with mass percent. This solution concentration compares the mass of the solute to the total mass of the solution. For example, a 10% salt solution contains 10 grams of salt in each 100 grams of solution.

Test Your Concept Understanding:
Work these practice problems:

What is the molality of a solution consisting of 45g of MgCl₂
dissolved in 620g of H₂O?
What is the mole fraction of ethanol in a solution made of 92g of C₂H₅OH
and 144g of H₂O?
What is the mole fraction of H₂O in question #2?
What is the mass percent of ethanol in question #2?
What is the normality of a 2M solution of H₃PO₄?

There will be a formal test tomorrow. Use the skills test as a review.

Research References:

Solution Preparations - Frontier High School
Molarities of Concentrated Acids and Bases - Frontier High School

Chemistry Class

answers:

6N H₃PO₄