The concentration of a chemical solution refers to the amount of solute that is dissolved in a solvent. We normally think of a solute as a solid that is added to a solvent (e.g., adding table salt to water), but the solute could just as easily exist in another phase. For example, if we add a small amount of ethanol to water, then the ethanol is the solute and the water is the solvent. If we add a smaller amount of water to a larger amount of ethanol, then the water could be the solute!

**Units of Concentration**

Once you have identified the solute and solvent in a solution, you are ready to determine its concentration. Concentration may be expressed several different ways, using

**percent composition by mass**,**volume percent**,**mole fraction**,**molarity**,**molality**, or**normality**.**Percent Composition by Mass (%)**This is the mass of the solute divided by the mass of the solution (mass of solute plus mass of solvent), multiplied by 100.

**Example:**Determine the percent composition by mass of a 100 g salt solution which contains 20 g salt.

Solution:

20 g NaCl / 100 g solution x 100 = 20% NaCl solution

**Volume Percent (% v/v)**Volume percent or volume/volume percent most often is used when preparing solutions of liquids. Volume percent is defined as:

v/v % = [(volume of solute)/(volume of solution)] x 100%

Note that volume percent is relative to volume of solution, not volume of*solvent*. For example, wine is about 12% v/v ethanol. This means there are 12 ml ethanol for every 100 ml of wine. It is important to realize liqud and gas volumes are not necessarily additive. If you mix 12 ml of ethanol and 100 ml of wine, you will get less than 112 ml of solution.

As another example. 70% v/v rubbing alcohol may be prepared by taking 700 ml of isopropyl alcohol and adding sufficient water to obtain 1000 ml of solution (which will not be 300 ml).

**Mole Fraction (X)**This is the number of moles of a compound divided by the total number of moles of all chemical species in the solution. Keep in mind, the sum of all mole fractions in a solution always equals 1.

**Example:**What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water? (molecular weight water = 18; molecular weight of glycerol = 92)

Solution:

90 g water = 90 g x 1 mol / 18 g = 5 mol water

92 g glycerol = 92 g x 1 mol / 92 g = 1 mol glycerol

total mol = 5 + 1 = 6 mol

x_{water}= 5 mol / 6 mol = 0.833

x_{glycerol}= 1 mol / 6 mol = 0.167

It's a good idea to check your math by making sure the mole fractions add up to 1:

x_{water}+ x_{glycerol}= .833 + 0.167 = 1.000

**Molarity (M)**Molarity is probably the most commonly used unit of concentration. It is the number of moles of solute per liter of solution (not necessarily the same as the volume of solvent!).

**Example:**What is the molarity of a solution made when water is added to 11 g CaCl_{2}to make 100 mL of solution?

Solution:

11 g CaCl_{2}/ (110 g CaCl_{2}/ mol CaCl_{2}) = 0.10 mol CaCl_{2}

100 mL x 1 L / 1000 mL = 0.10 L

molarity = 0.10 mol / 0.10 L

molarity = 1.0 M

**Molality (m)**Molality is the number of moles of solute per kilogram of solvent. Because the density of water at 25°C is about 1 kilogram per liter, molality is approximately equal to molarity for dilute aqueous solutions at this temperature. This is a useful approximation, but remember that it is only an approximation and doesn't apply when the solution is at a different temperature, isn't dilute, or uses a solvent other than water.

**Example:**What is the molality of a solution of 10 g NaOH in 500 g water?

Solution:

10 g NaOH / (4 g NaOH / 1 mol NaOH) = 0.25 mol NaOH

500 g water x 1 kg / 1000 g = 0.50 kg water

molality = 0.25 mol / 0.50 kg

molality = 0.05 M / kg

molality = 0.50 m

**Normality (N)**Normality is equal to the*gram equivalent weight*of a solute per liter of solution. A gram equivalent weight or equivalent is a measure of the reactive capcity of a given molecule. Normality is the only concentration unit that is reaction dependent.

**Example:**1 M sulfuric acid (H_{2}SO_{4}) is 2 N for acid-base reactions because each mole of sulfuric acid provides 2 moles of H^{+}ions. On the other hand, 1 M sulfuric acid is 1 N for sulfate precipitation, since 1 mole of sulfuric acid provides 1 mole of sulfate ions.

**Dilutions**

You

**dilute**a solution whenever you add solvent to a solution. Adding solvent results in a solution of lower concentration. You can calculate the concentration of a solution following a dilution by applying this equation:
M

_{i}V_{i}= M_{f}V_{f}
where M is molarity, V is volume, and the subscripts i and f refer to the initial and final values.

**Example:**

How many millilieters of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?

Solution:

5.5 M x V

V

V

V

5.5 M x V

_{1}= 1.2 M x 0.3 LV

_{1}= 1.2 M x 0.3 L / 5.5 MV

_{1}= 0.065 LV

_{1}= 65 mL
So, to prepare the 1.2 M NaOH solution, you pour 65 mL of 5.5 M NaOH into your container and add water to get 300 mL final volume.

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