14. Solution Concentration
Solutions are homogeneous mixtures of solute and solvent.
- Solvent - the most abundant substance in a solution. In a liquid solution, the solvent does the dissolving.
- Solute - the other substance in a solution. In a liquid solution, the solute is dissolved.
It is possible for a solution to have more than one solute, air is an example, but a solution can have only one solvent.
- Solutility - The amount of solute that can be dissolved in a given amount of solvent at a given temperature. Different substances have different solubilities.
Preparing a Saturated Salt Solution:
Sodium chloride has a solutility of 35.7 grams in 100 cm3 of COLD water. That would be 357 grams of NaCl in one liter of H2O.
Concentration is a comparison of the amounts of solute and solvent.
- Weigh out 357 grams of NaCl.
- Add the salt to a 1 liter volumetric flask.
- Add H2O to the graduation line and stir until dissolved.
- You now have "saturated" salt water.
- What is the volume of the solution?
- How many moles of NaCl are in the sample?
- How many particles of sodium chloride are in this sample?
- How many sodium ions, Na+, are in this sample?
Describing a solution as "stong" or "weak" gives 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, "how much" solute and solvent must be known.
The most common units of solution concentration involves moles.
Molarity (M) = moles of solute per cubic decimeter (dm3) of solution.
Remember the following:
- Volume refers to the volume of the total solution, not just the volume of the solvent.
- One cubic decimeter (dm3) = 1000 cm3 = 1 liter = 1000 ml
Molarity Calculation Examples:
The dimensional analysis 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.
Note: In examples 1 and 2, the problem gives a mass of solute in a volume of solution. This may be written as "mass over volume", the form needed for molarity. The only thing needed is to convert mass to moles and volume to dm3.
1. What is the molarity of a liter of solution containing 100 g of copper (II) chloride?
2. Calculate the molarity of 100 ml of solution containing 25 g of silver nitrate.
Note: Example 3 is different. It asks "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 dimensional analysis.
As in any dimensional analysis problem, the first step is to write down what is given in the problem. Look closely at how that information is written to begin dimensional analysis.
3. Prepare 250 ml of 0.5 M salt water.
The key difference in the two types of molarity problems above is that one type asks for a concentration and the other type provides a concentration.
Molarity Calculations #1
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 100 ml of 0.40 M MgSO4 from a stock solution of 2.0 M MgSO4?
There are two solutions involved in this problem. Notice that you are given two concentrations, but only one volume.
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 problem.
- 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.
M1 = 2.0 M MgSO4 V1 = unknown
M2 = 0.40 M MgSO4 V2 = 100 ml
To solve the problem, do the following:
- Write the equation: M1 V1 = M2 V2
- Manipulate the equation: V1 = M2 V2 / M1
- Put numbers into the equation: V1 = (0.40M) (100 ml) / 2.0 M
- Do the calculation: V1 = 20 ml
- Describe preparing the solution as follows: Add 80 ml of distilled water to 20 ml of the 0.40 M MgSO4 solution.
Molarity Calculations #2
Although molarity 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:
- 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 nA moles of solute and nB moles of solvent, the mole fraction of solute, XA, and the mole fraction of the solvent, XB, can be expressed as follows:
XA = nA / nA + nB XB = nB / nA + nB
- 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.
Concentration Calculations #3