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General Chemistry--Unit 2

04/25/09

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Concentration of Solutions


�Before You Begin:

To master this material you need to be able to perform chemical calculations, such as unit analysis. You should also be familiar with the properties of mixtures.

 


Solution Definitions

The most useful type of mixture for a beginning chemistry student is one in which a relatively small amount of one substance is blended with (dissolved in) a relatively larger amount of liquid, often water, to form a homogeneous mixture. These mixtures are the typical solutions that are used in freshman chemistry lab experiments. The solvent is the major component of a solution (the part that does the dissolving). The solute is the minor component of the solution (the part that gets dissolved).

Concentration Calculations

We often need a mixture with a specific composition, not only a particular solute and solvent but with specific relative amounts of each. There are several methods of stating the relative amounts of the components of a mixture. Each has its own advantages and disadvantages.

Percent Composition:

One of the easiest methods of making a mixture with a specific composition is to measure the masses of each of the parts and express the concentration as a percentage of solute to the mass of the entire mixture. This method is especially useful for mixtures of solids, like solder and alloys. For example, nichrome wire is a mixture of 75% nickel, 12% iron, 11% chromium, and 2% manganese by mass (data from Dr. Seeleyís web site), so 100 grams of the wire has 75 grams of nickel.

Mass/mass percentages are the best ways of expressing concentrations for applications in which the physical rather than the chemical properties of the mixture are most important. This is a poor way of expressing concentration for solutions used in chemical reactions, though, because mole ratio information is buried. We can convert the information to moles using the formula weight, if necessary.

Some solutions are so dilute, that it is more convenient to use parts per million, ppm, or even parts per billion, ppb, instead of percent (parts per one hundred, note). To find a concentration in parts per million, divide the mass of solute by the mass of solution and multiply by one million, 106. To find a concentration in parts per billion, divide the mass of the solute by the mass of the solution and multiply by one billion, 109.


4Concept Check: According to the organization SeaFriends, seawater has 904 ppm of sulfur. What is the mass percent of sulfur in seawater?

Answer: Assume one million grams of seawater. The number of grams of solute in one million grams is 904. This gives

                                           mass%=(904g/1EE6g)100=0.0904%

 

 


Mole Fraction:

The mole fraction, X, is the concentration expression most like percent composition yet includes mole ratio information. The mole fraction is the number of moles of one component of a mixture divided by the total number of moles in the mixture. This concentration expression is useful for situation in which the chemical behavior of the mixture is important and when the solute and solvent roles arenít clear. Raoultís Law is an example of an application that requires mole fraction concentrations: the vapor pressure of a liquid in a mixture depends on the mole fraction of that liquid. Henryís law, the solubility of a gas in a liquid, is sometimes stated in terms of mole fraction.


Concept Check: According to the alloys table from Dr. Seelyís website, Dentistís amalgam is 70% mercury and 30% copper, by mass. What is the mole fraction of copper in dentistís amalgam?

Answer: For the purpose of assigning significant figures, we will assume that the percentages are accurate to at least 1%. Assume 100 grams of amalgam, which contains 70 grams of mercury and 30 grams of copper, each value having at least two significant figures. Convert each to moles and find X. Note that X does not have units.

                                              Hg=70g/(200.59g/mol)=0.35mol
Cu=30g/(63.546g/mol)=0.47mol
XCu=0.47mol/(0.47+0.35)mol=0.57

 

 

 

 


Molarity:

The molarity, M, is defined as the number of moles of solute in each liter of solution. This is the most common concentration expression in the freshman chemistry lab. We use molarity when the focus is on the chemical behavior of the solute. We can measure the volume of the mixture and deduce the amount of solute present in a format that makes chemical computations as straightforward as possible. Mass/volume composition concentration would require an additional calculation to convert grams of solute into moles for any chemical calculations. In addition to laboratory labels, molarity is used extensively in physical chemistry calculations such as pH of acid solutions and equilibrium constant expressions.


Concept Check: A solution contains 5.7 grams of potassium nitrate dissolved in enough water to make 233 mL of solution. What is its molarity?

Answer: The formula weight of KNO3 is 101.103 g/mol.

                                                 5.7g/(101.103g/mol)=0.056mol
M=0.056mol/0.233L=0.24mol/L

 

 

 


Molality:

The molality, m, is defined as the number of moles of solute per kilogram of solvent. This is useful when the properties of the solvent are being studied rather than the properties of the solute. When we study colligative properties, we are observing the amount of solute and the nature of the solvent. Freezing point depression is an example of a colligative property. The freezing point of a liquid is lowered by the presence of impurities. The disadvantage to molality is that we need density information to determine the amount of the mixture.

Students beware: the two words (molarity and molality) are very similar, and the two variables (M and m) are really close! These two concentration expressions are easily confused.

 


Concept Check: What is the molality of dentistís amalgam? Recall that dentistís amalgam is 70% mercury and 30 % copper, by mass.

Answer: Assume 100grams of amalgam and two significant figures, as in the previous example. By mass, mercury is the major component, so it must be the solvent. 100 grams of amalgam contains 70 grams of mercury, which is 0.070 kg of mercury. It also contains 30 grams of copper, which is 0.47 moles of copper. This gives

                                                   m=0.47mol/0.070kg=6.7mol/kg

 

 

Note that we have used the mass of the solvent rather than the mass of the mixture.


Normality:

Normality, N, is similar to molarity, moles of solute per liter of solution. However, instead of the entire solute, the normality is based on the number of moles of the active part of the solute, called a chemical equivalent. For an acid, the chemical equivalent is the number of moles of H+1 ion. For a base, the chemical equivalent is the number of moles of OH-1  ions. For an oxidation-reduction solution, the chemical equivalent is the number of moles of electrons transferred.

The normality of hydrochloric acid, HCl, is the same as the molarity of hydrochloric acid, because there is one mole of H+1 ions for every one mole of hydrochloric acid. The normality of sulfuric acid, H2SO4, is twice the molarity because there are two moles of H+1 ions per mole of sulfuric acid. The advantage to using normality is that it gives an effective concentration (3M sulfuric acid is twice as acidic as 3M hydrochloric acidóthis is clear if they are labeled 6N and 3N, respectively). The disadvantage to using normality is that it is redundant information, since we can use the molecular formula to find the relative amount of the effective part of the solute. A bigger drawback to normality is that it is situational, so it can cause confusion. This is especially true with oxidation-reduction solutions. Some compounds can gain or lose different numbers of electrons depending on the electrochemical aggressiveness of the other reactant. For example, copper (II) ion can gain two electrons to become copper metal or it can gain one electron to become copper (I) ion. A solution that has one mole of copper (II) ion for every liter of solution is 1.0 M, but the same solution could be 1.0 N or 2.0 N depending on the reactant to which it is added.

 

Formality:

Formality, F, is the number of formula weight units of solute per liter of solution. Remember that one mole of a compound has a mass equal to the formula weight in grams. The number of formula weight units is equal to the number of moles for molecular substances. The purpose of formality is to distinguish the number of moles of a compound from the number of moles of ions in solutions of ionic compounds or weak electrolytes. If we dissolve one mole of calcium nitrate in enough water to make a liter of solution, the formality is one. The molarity (as it is commonly used) of calcium nitrate is the same as the formality, but that is a little sloppy because, once dissolved, the calcium nitrate ionizes completely so there isnít really any Ca(NO3)2 in the solution. The molarity of the nitrate ions is two molar, because one formula unit has two nitrate ions. The formality of nitrate ion doesnít exist because it is defined as the number of moles of the entire compound. So the only difference between formality and molarity is that you can express the molarity of the different ions individually and the formality is the entire compound irrespective of ionization. Formality is somewhat old fashioned. Modern more casual usage is to use molarity for both ideas and label with the appropriate solute. For example 1.0 F solution of Ca(NO3)2 is 2.0 M NO31- or 1.0 M Ca(NO3)2.

 

Summary of the Concentration Types

Concentration Type

Mathematical Formula

Use

% composition

ppm

ppb

mass%A=(massA/mass total)100
ppmA=(massA/mass total)1EE6
ppbA=(massA/mass total)1EE9

For ease of making a solution.

When physical properties are more important than chemical properties.

Mixtures of solids.

When the concentrations are very low, use ppm or ppb.

 

 

 

Mole fraction

X=mol A/mol total

Mixtures of solids or gases

Mixtures in which the roles of solute and solvent are not clear

 

 

Molarity

M=n mol solute/L solution

Most common laboratory solutions

 

 

Molality

m=n mol solute/kg solvent

Situations in which the properties of the solvent are studied.

 

 

Normality

N=n mole equivalents/L solution

Acids and bases

Oxidizing agents

Reducing agents

Formality

F=n formula weight units/L solution

Ionic compounds

 

 

 

Conversions among Units

If we examine the definitions of the various concentration expressions, we can see that molality is the really different one. The denominator of the expression is the mass of only one component of the mixture. All the rest have a denominator that is a measure of the mixture as a whole. The general strategy for any other system to molality is to determine the mass of the solute as separate from the mass or volume of the mixture.

Molarity, normality, and formality all have liters of solution in the denominator. We need to remember the definitions to use in the numerator, but these fairly straightforward to convert, second only to mass %, ppm, and ppb. Those conversions require only a shift in the decimal place.

 

Percent Composition to Molarity:

In order to convert mass percent to molarity or vice versa, you need to know the density of the solution. Assume 100 grams of solution; this gives you a mass of solution and a mass of solute. Use the density to find the volume of the solution. Use dimensional analysis to convert the mass of the solute to moles. Parts per million and billion can be worked the same way except for the decimal point.

 


Concept Check: Concentrated hydrochloric acid is 31% hydrochloric acid and 69% water, by mass. If the density of concentrated hydrochloric acid is 1.16g/mL, what is its molarity?

Answer: Assume 100 grams of concentrated hydrochloric acid; this has 31 grams of HCl and 69 grams of water. The formula weight of hydrochloric acid is 36.46 g/mol, so 31 grams is 0.85moles. Find the volume of the solution.

 

                                       D=M/V
V=M/D
V=100g/(1.16g/mL)=86.2mL=0.08262L

 

 

 

Find the molarity.

                                                    M=0,85mol/0.0826L=12mol/L

 

 

 


Mole Fraction to Molarity:

You need density information to find mole fraction if given molarity, or vice versa.

 


Concept Check: An aqueous solution of sulfuric acid has a molarity of 18 mol/L. if its density of 1.84 g/mL, what is the mole fraction of water in the solution?

Answer: Assume one liter of solution, which has 18 moles of sulfuric acid. The formula weight of sulfuric acid is 98.08 g/mol, so 18 moles has a mass of 1800 grams (1756 with two significant figures). Use the density formula to find the mass of the solution

                                        D=M/V
M=DV=1.84g/mL(1000mL)=1840g

 

 

 

If 1800 grams of the solution is sulfuric acid, 40 grams must be water. The formula weight of water is 18.015 g/mol, so 40 grams of water is 2 moles. The mole fraction of water is

                                                  For water: X=2mol/(18mol+2mol)=0.1

 

 


Molarity and Molality:

You need density information to find molarity if given molality, or vice versa.


Concept Check: An aqueous solution of hydrogen peroxide, H2O2, is 16.9 mol/L. if it has a density of 1.196 g/mL, what is its molality?

Answer: Assume one liter of solution; this has 16.9 moles of hydrogen peroxide. The formula weight of hydrogen peroxide is 34.015 g/mol, so the liter of solution contains 575 grams of hydrogen peroxide. Use the density formula to find the mass of one liter of solution:

                                       D=M/V
M=DV=1.196g/mL(1000mL)=1196g

 

 

 

If the mixture has 575 grams of hydrogen peroxide in 1196 grams of solution, it must have 621 grams of water (or 0.621 kg of solvent). The molality is

                                                  m=16.9mol/0.621kg=27.2mol/kg

 

 


Normality and Formality:

The most sensible way of doing this is to use molarity all the time then use definitions to convert to formality or normality, if your instructor uses them (which probably wonít happen until you reach sophomore analytical chemistry).


Concept Check: What is the normality and formality of concentrated sulfuric acid, 18 M?

Answer: Sulfuric acid is an acid. There are two hydrogen ions in its formula, so one mole of solute has two moles of equivalents. Concentrated sulfuric acid has a normality of 36N.

The number of gram formula weights for sulfuric acid is the same as the number of moles of solute, so concentrated sulfuric acid has a formality of 18F.


Dilutions

A solution is diluted when extra solvent is added. We can calculate the concentration that results when a solution is diluted using the equation C1V1=C2V2, where C is the concentration, usually in molarity, and V is the volume. Any volume units will work in this formula as long as they are consistent. When solutions have a fairly low concentration and/or the solute is solid, the volume of the solution is roughly equal to the volume of the solvent. This is not true for solutions with a high concentration.

 


Concept Check: 25mL of 6M sodium hydroxide are diluted to 1.3 L. What is the resulting concentration?

Answer:

                                           C1V1=C2V2
C2=C1V1/V2=6.0M(0.025L)/1.3L=0.12L



Concept Check: A student wishes to make 0.5 L of a 3.0 M sulfuric acid solution using concentrated sulfuric acid, 18 M. How should she do this?

Answer: Very Carefully! Diluting concentrated acids is dangerous. The process releases so much thermal energy that the water used in the dilution will boil, which may cause boiling hot acid burns. Ouch!

The amount of acid to use can be calculated by the equation above.

                                       C1V1=C2V2
V1=C2V2/C1=3.0M(0.5L)/18M=0.08L=80mL

 

 

 

The student needs to carefully add 80 mL of concentrated acid to roughly 400 mL of water then add enough water to make a total of 500 mL of solution. Always add the acid to the water. It takes more energy to boil a lot of water than to boil a little acid. Besides, if something has to splash out, you want it to be the water not the acid.


 

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