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Chapter 2 – Solutions

Introduction

Solutions are all around us. Our atmosphere is a solution of gases in gases, carbonated beverages are solutions of gases in liquids, sweetened drinks are solutions of solids in liquids, and solder is a solution of solids in solids. Reactions are normally carried out in solutions of gases or liquids because the particles in these solutions are free to move and to collide. In this chapter, we define several terms relevant to the study of solutions and discuss some solution properties.

2.1. Concentration Units

Introduction

The properties of a solution depend upon the relative amounts of solvent and solute(s), which is given by the solute concentration in the solution. A concentrated solution is one in which the concentration of a solute is relatively large, while a dilute solution is one in which the solute concentrations are relatively low. However, the terms 'concentrated' and 'dilute' used in this manner are not quantitative. In this section, we present several different ways of reporting the concentrations of solutions in a quantitative manner.

Objectives

2.1-1. Definitions

A solution consists of a solvent and one or more solutes.
Before we start into the quantitative aspects of solutions, we review some of the terms introduced in Section 10.1 of Chemistry – A Molecular Science (Chapter 10). When ionic substances dissolve in water, they break up into their component ions, i.e., they ionize or dissociate in water. A substance that ionizes or dissociates completely in water is said to be a strong electrolyte. Ionic compounds are strong electrolytes. NaCl dissociates completely into Na1+ and Cl1– ions in water, so it is a strong electrolyte. Molecular substances do not dissociate when in water and are classified as nonelectrolytes. Sugar dissolves in water as sugar molecules (C12H22O11), so sugar is a nonelectrolyte. Acids and bases react with water to produce ions, so they too are electrolytes. Strong acids and bases react completely with water and are strong electrolytes, but weak acids and bases react only partially with water and are classified as weak electrolytes.
components of a solution
Figure 2.1: Components of a Solution
A solution consists of a solvent and at least one solute. Almost all of the solutions considered in this course are aqueous solutions, which are solutions in which water is the solvent.

2.1-2. Molarity (M)

The molarity (M) of a solute is the number of moles of the solute in one liter of solution or the number of millimoles of solute in one milliliter of solution.
( 2.1 )
M =
moles of solute
liters of solution
=
n
V
=
mmoles of solute
mL of solution
Molarity (M)
molarity
Figure 2.2: Molarity
Molarity relates moles of solute to the volume of solution.
The molarity of a solute allows us to convert between the number of moles (or millimoles) of that solute and the number of liters (or milliliters) of solution. To determine the number of moles of solute in a given volume of solution, we solve Equation 2.1
M =
moles of solute
liters of solution
=
n
V
=
mmoles of solute
mL of solution
for n, the number of moles:
n = MV
Alternatively, we can use the units of molarity and the factor-label method to determine the number of moles.
M mol solute
1 L solution
× V L of solution = MV mol solute
If V is given in milliliters, then MV is millimoles of solute. We will frequently use millimoles because volumes in a chemistry laboratory are most often given in milliliters. We can also solve Equation 2.1
M =
moles of solute
liters of solution
=
n
V
=
mmoles of solute
mL of solution
for the volume of solution to determine what volume of solution contains a given number of moles of solute:
V =
n
M
Or, using the factor-label method,
n mol solute
1 L solution
M mol solute
=
n
M
L solution

2.1-3. Preparing a Solution of Known Molarity

The denominator of molarity is the volume of the solution, not the volume of solvent. Thus, a solution of known molarity must be made by adjusting the volume after the solute and solvent have been thoroughly mixed. This is done with a volumetric flask of the appropriate volume (Figure 2.3) as follows.
a volumetric flask
Figure 2.3: A 500-mL Volumetric Flask

2.1-4. Preparing a Solution Exercise

Exercise 2.1:

How many mmoles of K2HPO4 are required to prepare 350. mL of 0.0865 M K2HPO4?
30.3_0__
350. mL of solution ×
0.0865 mmol K2HPO4
1 mL solution
= 30.3 mmol

or
0.350 L of solution ×
0.0865 mol K2HPO4
1 L solution
= 0.0303 mol

(0.0303 mol)(1000 mmol/mol) = 30.3 mmol
mmol
How many grams of K2HPO4 (Mm = 174.17 g/mol) must be dissolved to make the solution?
5.27_0__
30.3 mmol ×
174.17 mg K2HPO4
1 mmol K2HPO4
×
1 g
1000 mg
= 5.27 g K2HPO4
g
Molarity is the number of moles of solute per liter of solution not solvent, so the K2HPO4 would be dissolved in less than 350 mL of water (~250 mL). Water would then be added to adjust the volume of the resulting solution to 350 mL.
How many milliliters of the 0.0865 M solution of K2HPO4 are required to deliver 3.50 mmoles of K1+ ions?
V = 20.2_0__
3.50 mmol K1+ ×
1 mmol K2HPO4
2 mmol K1+
×
1 mL solution
0.0865 mmol K2HPO4
= 20.2 mL solution
mL

2.1-5. Molarity to Solute Mass Exercise

Exercise 2.2:

Ocean water is typically 0.53 M in chloride ion. How many grams of chloride ion are contained in 500. mL of ocean water?
mass = 9.4_0__
500. mL solution ×
1 L solution
1000 mL solution
×
0.53 mol Cl1–
1 L solution
×
35.5 g Cl1–
1 mol Cl1–
= 9.4 g Cl1–
g Cl1–

2.1-6. Molarity of Ions Exercise

Exercise 2.3:

Ba(OH)2 is a strong electrolyte. What are the ion concentrations in a solution prepared by dissolving 0.210 g of Ba(OH)2 in enough water to make 450. mL of solution?
The molarity of the Ba(OH)2 solution is:
0.00272_0__
n = 210 mg Ba(OH)2 ×
1 mmol Ba(OH)2
171.3 mg Ba(OH)2
= 1.23 mmolBa(OH)2

M =
n
V
=
1.23 mmol Ba(OH)2
450 mL solution
= 0.00272 M
M
The ion concentrations are:
[OH1−]
=
0.00544_0__ There are two OH1– for each Ba(OH)2. M

[Ba2+]
=
0.00272_0__ There is one Ba2+ for each Ba(OH)2. M

The number of millimoles of OH1– in 275 mL of solution is:
1.50_0_3_
275 mL solution ×
0.00545 mmol OH1–
1 mL solution
= 1.50 mmol OH1–
mmol

The number of milliliters of solution required to deliver 0.300 mmol of Ba2+ is:
110._0_3_
0.300 mmol Ba2+ ×
1 mL solution
0.00272 mmol Ba2+
= 110 mL solution
mL

2.1-7. Common Prefixes

1 M solutions are considered fairly concentrated, and most of the solutions encountered in the chemistry laboratory are much less concentrated. Indeed, some processes require only very dilute concentrations. For example, the maximum allowed level of mercury in drinking water is 0.000002 M, and a testosterone level of only 0.00000001 M initiates puberty in human males. Prefixes are used to avoid the preceding zeros or exponents when expressing these very dilute solutions.
Unit Symbol Value
millimolar mM 10–3 M 1 mmol·L–1
micromolar µM 10–6 M 1 µmol·L–1
nanomolar nM 10–9 M 1 nmol·L–1
Table 2.1: Common Prefixes
Thus, the maximum allowed level of mercury in drinking water is 2 × 10–6 M = 2 µM, and the testosterone level that initiates puberty is 10 × 10–9 M = 10 nM.

2.1-8. Using Prefixes Exercise

Exercise 2.4:

How many micrograms of testosterone (C19H28O2, Mm = 288 g/mol) are in the blood (5.7 liters) of a person if the testosterone concentration is 12.5 nM?
mass = 21_0__
5.7 L blood ×
12.5 e−09 mol C19H28O2
1 L blood
×
288 g C19H28O2
1 mol C19H28O2
×
1e+06 µg C19H28O2
1 g C19H28O2
= 21 µg C19H28O2
µg testosterone

2.1-9. Mole Fraction (X)

The mole fraction (X) of a substance in a solution is the number of the moles of that substance divided by the total number of moles of all substances (solutes and solvent) in solution. Because all of the substances comprise the whole mixture, the sum of all mole fractions (including the solvent) must equal one; i.e.,
XA
A
= 1.
( 2.2 )
XA =
n
ntot
=
moles of A in mixture
total moles in mixture
Mole Fraction
mole fraction
Figure 2.4: Mole Fraction
The mole fraction of A equals the number of moles of A divided by the total number of moles of substances in the solution.

2.1-10. Mole Fraction Exercise

Exercise 2.5:

A sample of the alloy known as "yellow brass" is found to be 67.00% Cu and 33.00% Zn by mass. What are the mole fractions of Cu and Zn in the alloy?
Exactly 100 g of alloy contains:
1.054_0__
67.00 g Cu ×
1 mol Cu
63.546 g Cu
= 1.054 mol Cu
mol Cu

0.5047_0__
33.00 g Zn ×
1 mol Zn
65.39 g Zn
= 0.5047 mol Zn
mol Zn

1.559_0__ 1.054 mol Cu + 0.5047 mol Zn = 1.559 mol total mol total in mixture
The mol fractions are
XCu = 0.6763_0__
1.0544 mol Cu
1.5590 mol total
= 0.6763

XZn = 0.3237_0__
0.50466 mol Zn
1.5590 mol total
= 0.3237


Alternatively, XZn = 1.0000 – XCu = 1.0000 – 0.6763 = 0.3237

2.1-11. Mass Fraction (%, ppt, ppm)

The mass fraction relates the mass of a solute to the mass of the solution. It is commonly used for solutions where the solvent is not clearly defined.
( 2.3 )
mass fraction =
mass of solute
mass of solution
Mass Fraction
mass fraction
Figure 2.5: Mass Fraction
The mass fraction relates the mass of a solute to the mass of the solution.

2.1-12. Mass Fraction Multipliers

Mass fractions represent a part of the whole, so they are always less than one. Indeed, mass fractions in dilute solutions can be so small that they are multiplied by powers of ten to eliminate the preceding zeros (see Table 2.2). Thus,
Name Multiplier Useful Units
mass fraction 1
g A
g mixture
mass percent (%) 100
g A
100 g mixture
parts per thousand (ppT) 1000
g A
1000 g mixture
parts per million (ppm) 106
g A
106 g mixture
parts per billion (ppb) 109
g A
109 g mixture
Table 2.2: Common Multipliers Used with Mass Fraction

2.1-13. Mass Fraction Exercise

Exercise 2.6:

A mixture contains 6.0 g N2, 16.0 g O2 and 2.0 g He.

What are the mass fraction, mass percent, and mole fraction of each gas?
Mass Fractions
Total mass = 24.0_0__1 6.0 g N2 + 16.0 g O2 + 2.0 g He = 24.0 g total. g

Mass fraction of N2 = .25_0__
6.0 g N2
24.0 g total
= 0.25

Mass fraction of O2 = .667_0__
16.0 g O2
24.0 g total
= 0.667

Mass fraction of He = .083_0__
2.0 g He
24.0 g total
= 0.083
Mass Percents
Mass percent of N2 = 25_0__ Multiply the mass fraction of nitrogen by 100 %

Mass percent of O2 = 66.7_0__ Multiply the mass fraction of oxygen by 100 %

Mass percent of He = 8.3_0__ Multiply the mass fraction of helium by 100 %
Moles
Moles of N2 = .21_0__
6.0 g N2 =
1 mol N2
28.0 g N2
= 0.21 mol N2
mol

Moles of O2 = .500_0_3_
16.0 g O2 =
1 mol O2
32.0 g O2
= 0.500 mol O2
mol

Moles of He = .50_0_2_
2.0 g He =
1 mol He
4.0 g He
= 0.50 mol He
mol

Total moles = 1.21_0__ 0.21 mol N2 + 0.50 mol O2 + 0.50 mol He = 1.21 mol total mol
Mole Fractions
XN2 = .17_0__
0.21 mol N2
1.21 mol total
= 0.17

XO2 = .412_0_3_
0.500 mol O2
1.21 mol total
= 0.412

XHe = .41_0__
0.50 mol Ne
1.21 mol total
= 0.41

2.1-14. Molality (m)

The molality (m) of a solute is the number of moles of that solute per kilogram of solvent.
( 2.4 )
m =
moles of solute
kg of solvent
Molality (m)
Figure 2.6: Molality
Molality relates the moles of solute to the mass of solvent.
The volume of a solution varies with temperature, so the molarity (M) of the solution changes with temperature as well. Thus, molarity (M) is not a good unit of concentration to use in experiments involving temperature change. While volume is temperature dependent, mass is not, so molality (m), which involves no volumes, is commonly used for the concentration of any solution used in experiments that involve measurements resulting from temperature changes.

2.1-15. Molality Exercise

Exercise 2.7:

A solution is prepared by mixing 3.75 g of glucose (C6H12O6, Mm = 180.2 g/mol) and 25.0 g of water. What is the molality of the resulting glucose solution?
moles of glucose in solution
0.0208_0__
3.75 g ×
1 mol
180.2 g
= 0.0208
mol C6H12O6
kilograms of solvent present in solution
0.0250_0_3_ 25.0 grams is 0.0250 kg. kg H2O
the molality of the solution
0.832_0__
0.0208 mol glucose
0.0250 kg solvent
= 0.832 m
m

2.2 Changing Concentrations Units

Introduction

It is sometimes necessary to change the units of concentration. This routine procedure is most often done when the concentration of a stock solution is given in units that are not convenient for a particular experiment. The procedure presented here entails converting the units of the numerator and denominator of the given concentration to those of the desired concentration and then dividing.

Objectives

2.2-1. Converting Concentration Units

To change concentration units, convert the numerator and denominator separately.
All concentrations are ratios of two quantities. The numerator indicates the amount of solute and the denominator the amount of solution or solvent. Use the following steps to convert one concentration unit into another: The numerators and denominators of the concentration units that we have dealt with are:
molarity mass fraction mole fraction molality
moles of solute
liters of solution
mass of solute
mass of solution
moles of solute
total moles (all solutes + solvent)
moles of solute
kilograms of solvent

To convert between any of the above you need only some combination of the molar masses of solute and/or solvent, and/or the density of the solution.
Conversion Requires
mol solute ↔ mass solute molar mass of the solute
liters solution ↔ mass solution density of solution
liters solution ↔ mol solute + mol solvent density of solution, molar masses of solute and solvent
liters solution ↔ kg solvent solution density, molar mass of solute
mass solution ↔ mol solute + mol solvent molar masses of solute and solvent
mass solution ↔ kg solvent molar mass of solute
mol solute + mol solvent kg solvent molar masses of solute and solvent
Table 2.3: Converting Concentration Units

2.2-2. Molarity to Mass Percent Exercise

Exercise 2.8:

A solution of concentrated sulfuric acid is 18.00 M and has a density of 1.839 g/cm3. What is the mass percent of H2SO4 in the concentrated acid solution? Molar mass of sulfuric acid = 98.08 g/mol.
The given units are molarity, while the desired units are mass fraction, which must be multiplied by 100 to obtain mass percent. Molarity has units of (mol solute)/(L solution), while mass fraction has units of (mass solute)/(mass solution). Thus, we must make the following conversions.
  • numerator: 18.00 mol H2SO4 yields ↔ x grams H2SO4
  • denominator: 1 L solution yields ↔ y grams solution
Convert the given numerator into the desired numerator.
18.00 mol H2SO4 = 1765_0__
18.00 mol H2SO4 ×
98.08 g H2SO4
1 mol H2SO4
= 1765 g H2SO4
g H2SO4
Convert the denominator of the given concentration units into those of the desired concentration units.
1 L solution = 1839_0__
1 L soln ×
1000 mL soln
1 L soln
×
1.839 g soln
1 mL soln
= 1839 g soln
g solution
The mass fraction of H2SO4 in the solution = 0.9600_0_4_
X =
1.76544e+03 g H2SO4
1.839e+03 g soln
= 0.9600
The mass percent of H2SO4 in the solution = 96.00_0_4_ mass percent = (mass fraction)(100%) = (0.9600)(100%) = 96.00% %

2.2-3. Molarity to Molality Exercise

Exercise 2.9:

Concentrated sulfuric acid is 18.00 M and has a density of 1.839 g/cm3. What is the molality of H2SO4 in the concentrated acid solution?
The given units are molarity, while the desired units are molality. Molarity has units of (mol solute)/(L solution), while molality has units of (mol solute)/(kg solvent). Thus, we must make the following conversions:
  • numerator: 18.00 mol H2SO4 18.00 mol H2SO4
  • denominator: 1 L solution y kilograms solvent
The numerators are the same (18.00 mol), so we need only change the denominator and then divide.
1 L solution contains
.074_0__ From the previous part, we know that 1839 g soln contains 1765 g H2SO4. Thus, the solution must contain 1839 – 1765 = 74 g water. The denominator must be kilograms, so the answer is 0.074 kg. kg solvent
Divide the numerator by the denominator to obtain molality:
240_0_2_
18.00 mol H2SO4
0.074 kg solvent
= 2.4e+02 m
m

2.2-4. ppm to Molarity Exercise

Exercise 2.10:

What is the molarity of fluoride ion in drinking water that is 7.0 ppm F1–?
The given units are ppm, (grams solute)/(106 g solution), while the desired units are molarity, (mol solute)/(L solution). Thus, the following conversions are required.
  • numerator: 7.0 g F1– x mol F1–
  • denominator: 106 g solution y liters of solution
Convert the given numerator into the desired numerator.
7.0 g F1– = 0.37_0_2_ The molar mass of fluoride ion is 19.0 g/mol, so
7.0 g F1− =
1 mol F1−
19.0 g F1−
= 0.37 mol.
mol F1–
Next, convert the denominator of the given concentration into that of the desired concentration. This solution is very dilute, so you can assume that the density of the solution is the same as the solvent (1.0 g/mL).
106 g solution = 1e3_0_2_
106 g solution ×
1 mL soln
1.0 g soln
×
1 L soln
1000 mL soln
= 1.0e3 L soln
L solution
Finally, divide the two to obtain the desired concentration units.
[F1−]
=
3.7e-4_0__
0.37 mol F1−
1e3 L soln
= 3.7e−04 M
M

2.2-5. Mass Percent to Molarity Exercise

Exercise 2.11:

What is the molarity of a 30.0% sulfuric acid solution if its density is 1.218 g/mL?
  • numerator: 30.0 g H2SO4 x mol H2SO4
  • denominator: 100 g soln y L soln
Convert the given numerator into the desired numerator.
30.0 g H2SO4 = 0.306_0__ The molar mass of H2SO4 is 98.08 g/mol, so
30.0 g H2SO4 ×
1 mol H2SO4
98.08 g H2SO4
= 0.306 mol H2SO4.
mol H2SO4
Next, convert the denominator of the given concentration into that of the desired concentration.
100 g solution = 0.08210_0_4_
100 g soln ×
1 mL soln
1.218 g soln
×
1 L soln
1000 mL soln
= 0.08210 L soln
L solution
Finally, divide the two to obtain the desired concentration units.
[H2SO4]
=
3.73_0__
0.306 mol H2SO4
0.08210 L soln
= 3.73 M
M

2.3 Dilutions

Introduction

Stock solutions are often fairly concentrated and must be diluted before they are used. In this section, we show how to make a solution of known volume and molarity from a more concentrated stock solution.

Objectives

2.3-1. Dilution Lecture

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2.3-2. Equations of Dilution

The amount of solute does not change with the addition of solvent, so we can write the following:
( 2.5 )
Ci × Vi = Cf × Vf
Dilution Equation
where C is any concentration that indicates the amount of solute per unit volume of solution. Therefore, C can be molarity or g/mL, but not molality or mass percent. We solve Equation 2.5 for the final concentration, Cf to obtain the following
( 2.6 )
Cf = Ci ×
Vi
Vf
Dilution Factor
where the ratio Vi/Vf is called the dilution factor. Thus, the final concentration equals the initial concentration times a dilution factor. If a solution is diluted successively several times, then the final concentration equals the initial concentration times the product of all of the successive dilution factors as shown in Equation 2.7.
( 2.7 )
Cf = C1 ×
V1
V2
×
V3
V4
×
V5
V6
Successive Dilutions

2.3-3. Molarity after Dilution Exercise

Exercise 2.12:

What is the molarity of sulfuric acid in a solution prepared by adding 30.0 mL of 18.0 M H2SO4 to enough water to make 500. mL of solution?
concentration = 1.08_0__
Cf = 18.0 M ×
30.0 mL
500. mL
= 1.08 M
M H2SO4

2.3-4. Volume Required for Dilution Exercise

Exercise 2.13:

How many mL of 18.0 M H2SO4 are required to prepare 300. mL of 1.50 M H2SO4?
volume of 18.0 M H2SO4 = 25_0_3_ The question asks for the initial volume given that Ci = 18.0 M; Cf = 1.50 M; and Vf = 300. mL.
Vi = 300. mL ×
1.50 M
18.0 M
= 25.0 mL
mL H2SO4

2.3-5. Concentration After Successive Dilutions Exercise

Exercise 2.14:

Solution A is prepared by diluting 20.0 mL of a stock 0.100 M HCl solution to 50.0 mL. Solution B is prepared by diluting 10.0 mL of solution A to 75.0 mL. Solution C is made by diluting 20.0 mL of solution B to 250.0 mL. What is the concentration of HCl in solution C? Note that the concentration is expressed as millimolar.
[HCl]
in solution C =
0.427_0__
[HCl] = 100. mM ×
20.0 mL
50.0 mL
×
10.0 mL
75.0 mL
×
20.0 mL
250.0 mL
= 0.427 mM
mM

2.3-6. Concentration of Initial Solution Exercise

Exercise 2.15:

50.00 mL of an unknown solution A is diluted to 500.00 mL to make solution B. 25.00 mL of solution B is diluted to 750.00 mL to make solution C. 15.00 mL of solution C is diluted to 1000.00 mL. What is the concentration of solution A, if the concentration of the final solution is 1.47 µM?
product of all dilution factors = 5.000e-5_0_4_
DF =
50.00 mL
500.00 mL
×
25.00 mL
750.00 mL
×
15.00 mL
1000.00 mL
concentration of solution A = 0.0294_0__
CA =
Cf
DF
=
1.47e−06 M
5.000e−05
= 0.0294 M
M

2.4 Determining Concentrations

Introduction

Analyzing solutions to determine their concentration is an important part of analytical chemistry. In this section, we discuss two important methods used to carry out such an analysis: spectrometry and titration.

Objectives

2.4-1. Complementary Colors

White light is the result of all visible colors. When white light shines on a colored substance, however, some of the colors are absorbed. Those colors that are not absorbed can be either reflected or transmitted to the eye. We perceive this reflected or transmitted portion as the color of the substance. Consequently, the characteristic color of a material is not the color of light that it absorbs; rather it is the mixture of the remaining, unabsorbed colors that are observed. In other words, it appears as its complementary color. The approximate relationship between observed and absorbed colors is summarized in a color wheel as shown in Figure 2.7. For example, a solution appears orange because it absorbs blue light. Thus, when white light shines on the solution, the blue portion of the spectrum is absorbed by the solution, which leaves only the orange portion to be detected by your eyes.
Blue and orange are complements, green and red are complements, and violet and yellow are complements.
Figure 2.7: Color Wheel
Complementary colors are opposite one another in a color wheel, so orange is the complement of blue.

2.4-2. Spectrometry

Spectrometry, measuring the amount of light that a solute absorbs at some wavelength, is a convenient way to determine the concentration of a solute in a solution. The amount of light that is absorbed by the sample is called the absorbance (A) of the sample. Consider the experiment shown in Figure 2.8 below. Orange light of intensity Io enters the cell, where some of it is absorbed by the blue solution, so the outgoing intensity I, is less than the initial Io.
Absorbance depends upon distance, concentration, and absorptivity.
Figure 2.8: Absorbance and Beer's Law
Orange light is absorbed by a blue solution. The amount of light that is absorbed by the solution is called the absorbance (A), which is defined as –log(I/Io). The absorbance depends upon the concentration of the absorbing substance (c), the distance the light travels through the sample (l), and the molar absorptivity of the solute at the wavelength of the light (ε) as given by Beer's law.
The absorbance, which is defined as A = –log(I/Io),depends upon three factors: The relationship between these factors is given by Beer's Law.
( 2.8 )
A = εlc
Beer's Law

2.4-3. Determining an Unknown Concentration

The following procedure is used to determine an unknown concentration in a solution: Many solutes absorb very strongly, and the prepared solutions frequently absorb so strongly that too little light passes through the cell. In these cases, the solutions must be diluted before they can be measured. Indeed, it is often the case that they must be diluted several times in order to get a good reading of the absorbance. The following exercise takes you through this process except that no blank is used. Follow the above steps and dilute the samples to obtain the good absorbance readings. Don't forget to take into account your dilutions when doing the calculations.

2.4-4. Determining Concentration Exercise

Thank you for your patience as this section undergoes re-construction.

2.4-5. Titrations

Spectrometry is an easy way to determine an unknown concentration only if the solute absorbs light at a convenient wavelength (usually one in the visible). The concentration of a solute can also be determined by measuring the stoichiometric amount of a solution of known concentration required to react with the unknown solute. This method is called titration. The known solution, whose volume is to be determined, is called the titrant, while the solute being analyzed is the analyte. The point at which the titrant and analyte are in stoichiometric amounts is called the equivalence point. The equivalence point is usually approximated by an end point, which is the point at which an indicator undergoes a color change.
The analyte is in a flask and the titrant is in the buret.
Figure 2.9: Titration Apparatus

2.4-6. Titration Method

Acid-base titrations are very common in the chemistry lab. The net chemical equation for the reaction of a strong acid and a strong base is:
H3O1+ + OH1− 2 H2O
The above chemical equation shows that the reactants react in a 1:1 ratio, so the equivalence point is reached when the number of (milli)moles of base added equals the number of (milli)moles of acid present initially. However, the number of (milli)moles of a solute equals the volume of solution in (milli)liters times the molarity of the solute. Thus, the equivalence point in an acid base reaction is that point where
( 2.9 )
MacidVacid = MbaseVbase
Acid Base Equivalence Point
If the acid is the unknown, then a known amount of acid, the analyte, would be added to the flask, so Vacid would be known. The concentration of the base (Mbase), the titrant, would also be known and the volume of base required to reach the equivalence point (Vbase) would be determined by the titration. Macid would then be determined from Equation 2.9.

2.4-7. Titration Exercise

Thank you for your patience as this section undergoes re-construction.

2.4-8. Precipitation Reactions

Solubility rules can help you determine the precipitate.
The concentration of a solution can also be determined from the amount of precipitate that forms when an excess of another reactant is added. The procedure is the following:
Rule 1 Compounds of NH41+ and group 1A metal ions are soluble.
Rule 2 Compounds of NO31–, ClO41–, ClO31– and C2H3O21– are soluble.
Rule 3 Compounds of Cl1–, Br1–, and I1– are soluble except those of Ag1+, Cu1+, Tl1+, Hg22+,and Pb2+.
Rule 4 Compounds of SO42– are soluble except those of Ca2+, Sr2+, Ba2+, and Pb2+
Rule 5 Most other ionic compounds are insoluble.
Table 2.4: Solubility Rules
Solubility rules for ionic compounds in water.

2.4-9. Precipitation Exercise

Exercise 2.18:

What is the concentration of Ag1+ ions in a solution if addition of excess PO43– ions to 25.00 mL of the solution produced 163.2 mg of Ag3PO4 (Mm = 418.6 g/mol)?
The net ionic equation is: 3 Ag1+ + PO43– Ag3PO4

The number of millimoles of Ag3PO4 produced = 0.3899_0__
163.2 mg Ag3PO4 ×
1 mmol Ag3PO4
418.6 mg Ag3PO4
= 0.3899 mmol Ag3PO4
mmol
The number of millimoles of Ag1+ required = 1.170_0_4_
0.3899 mmol Ag3PO4 ×
3 mmol Ag1+
1 mmol Ag3PO4
= 1.170 mmol Ag1+
mmol
The concentration of Ag1+ in the original solution = 0.04678_0__ 1.1696 mmol Ag1+ was in 25.00 mL of solution.
[Ag1+] =
1.1696 mmol Ag1+
25.0 mL solution
= 0.04678 M
M

2.5 Colligative Properties

Introduction

The properties of a solution depend upon the concentrations of the solutes. Indeed, some solution properties depend only upon the concentration of particles in solution and not upon the identity of those particles. These properties are called colligative properties. In this section, we discuss the following colligative properties.
  • vapor pressure
  • boiling point
  • freezing point
  • osmotic pressure

Objectives

2.5-1. van't Hoff Factor and Colligative Concentrations

Colligative properties depend upon the concentration of particles in solution, which is different than the solute concentration when the solute is ionic and dissociates in water. The solution concentration is converted to a concentration of particles with the van't Hoff factor (i), which is the number of moles of particles produced when one mole of solute dissolves. For example, i = 3 for CaCl2 because dissolving one mole of CaCl2 produces three moles of ions: CaCl2 Ca2+ + 2 Cl1–. The total particle concentration is called the colligative concentration. It is equal to the solute concentration times its van't Hoff factor. The following equations define the colligative molarity and colligative molality.
( 2.10 )
Mc = i × M
mc = i × m
Colligative Concentrations

2.5-2. van't Hoff Exercise

Exercise 2.19:

What is the van't Hoff factor for each of the following?
CaSO4: i = 2_0__ CaSO4 Ca2+ + SO42–
CH3OH: i = 1_0__ CH3OH is a nonelectrolyte.
NH4ClO: i = 2_0__ NH4ClO NH41+ + ClO1–
Ba(OH)2: i = 3_0__ Ba(OH)2 Ba2+ + 2 OH1–
HCl: i = 2_0__ HCl H1+ + Cl1–
SF4: i = 1_0__ SF4 is a nonelectrolyte.

2.5-3. Colligative Molality Exercise

Exercise 2.20:

Determine the colligative molarity of each of the following solutions.
Solution i Mc
0.080 M K3PO4    
4_0__ K3PO4 3 K1+ + PO43–
0.32_0__ Mc = iM
0.042 M C6H12O6    
1_0__ C6H12O6 is a nonelectrolyte.
0.042_0__ Mc = iM

2.5-4. Phase Diagram

Figure 2.10 shows the phase diagrams of a pure substance and of a solution in which the substance is the solvent. The two diagrams are identical except for the region in blue, which is either a solid or gas in the pure substance but is all liquid in the solvent. Thus, the phase diagram of the solvent differs from that of the pure substance in three ways: The impact of these three effects is to increase the temperature-pressure range of the liquid state. The amount by which the liquid state is extended (shown in blue in the figure) depends only on the concentration of the solute particles, so freezing point depression, boiling point elevation, and vapor pressure lowering are all colligative properties, and we now examine the relationship between particle concentrations and their effect on each of these colligative properties. We then define a fourth colligative property, the osmotic pressure.
Differences in the phase diagrams of water and an aqueous solution
Figure 2.10: Phase diagrams of a solvent and a solution of the solvent.
The solvent is liquid only in the green region, while the solution is liquid in both the green and blue regions; i.e., the blue region shows the extent to which the liquid state is expanded due to the presence of the solute particles.

2.5-5. Vapor Pressure Lowering

The vapor pressure of a solution is less than that of the pure solvent.
The liquid vapor equilibrium is a dynamic equilibrium that is established when the rate of evaporation equals the rate of condensation. The pressure of the vapor in equilibrium with the liquid at a given temperature is the vapor pressure of the liquid at that temperature. Evaporation occurs from the surface of the liquid, so the rate of evaporation depends upon the concentration of particles at the surface. Figure 2.11 compares a pure solvent with a solution in which the mole fraction of a nonvolatile solute is 0.2. A solute mole fraction of 0.2 means that 20% of the particles in solution are solute particles, so 20% of the sites on the surface are occupied by nonvolatile solute particles. Consequently, evaporation can occur from only 80% of the surface sites, which results in a 20% reduction in the rate of evaporation, which in turn, causes a 20% reduction in the vapor pressure. We conclude that ΔP, the amount by which the vapor pressure of the solvent at some temperature is lowered by the addition of a solute with a mole fraction of Xsolute, is
( 2.11 )
ΔP = XsoluteP°solvent
Vapor Pressure Lowering of a Solvent
ΔP is the vapor pressure lowering; it is always positive because the vapor pressure of a solution is always lower than that of the pure solvent. P° is the vapor pressure of the pure solvent at the temperature under consideration. The vapor pressure of the solution is lower than that of the solvent by ΔP, so P = P° – ΔP = P° – XsoluteP° = (1 – Xsolute)P°. However, 1 – Xsolute = Xsolvent, so we can rewrite Equation 2.11 as
( 2.12 )
Psoln = XsolventP°solvent
Vapor Pressure of a Solution
The above indicates that the vapor pressure of a solution is equal to the vapor pressure of the pure solvent times the fraction of the surface sites occupied by the solvent; i.e., the mole fraction of the solvent.
Figure 2.11: Vapor Pressure Lowering
(a) Solvent: All of the surface sites are occupied by solvent molecules, which produces ten molecules in the vapor. (b) Solution: The mole fraction of the solute in the solution is ~0.2, so nonvolatile solute particles (blue spheres) occupy ~20% of the surface sites, which reduces the vapor pressure by ~20% as shown by the presence of only eight molecules in the vapor.

2.5-6. Vapor Pressure of Water

We will have made frequent use of the vapor pressure of water. The vapor pressure at 5 °C increments can be found in Table 2.5 and in the Resources.
T (°C) P° (torr)      T (°C) P° (torr)
0 4.6 50 92.5
5 6.5 55 118.0
10 9.2 60 149.9
15 12.8 65 187.5
20 17.5 70 233.7
25 23.8 75 289.1
30 31.8 80 355.1
35 41.2 85 433.6
40 55.3 90 525.8
45 71.9 95 633.9
Table 2.5: Vapor Pressure of Water at Selected Temperatures

2.5-7. Vapor Pressure Lowering Exercise

Exercise 2.21:

What is the vapor pressure at 25 °C of a solution prepared by dissolving 10.0 g NaCl in 100. g of water?
1. moles of NaCl in solution: 0.171_0__
10.0 g NaCl ×
1 mol NaCl
58.5 g NaCl
= 0.171 mol NaCl
2. moles of solute particles in solution: 0.342_0_3_
0.171 mol NaCl ×
2 mol particles
1 mol NaCl
= 0.342 mol particles
3. moles of solvent particles in solution: 5.56_0__
100 g H2O ×
1 mol H2O
18.0 g H2O
= 5.56 mol H2O
4. moles of solute and solvent particles in solution: 5.90_0__2 0.342 mol solute particles + 5.56 mol H2O = 5.90 mol
5. mole fraction of solvent: 0.942_0__
Xsolvent ×
5.56 mol H2O
5.90 mol particles
= 0.942
6. vapor pressure of solvent at 25 °C (see Table 2.5): 23.8_0__ The vapor pressure of pure water is found to be 23.8 torr, from Table 2.5. torr
7. vapor pressure of solution at 25 °C: 22.4_0__ P = (0.942)(23.8 torr) = 22.4 torr. torr

2.5-8. Boiling Point Elevation

The normal boiling point of a liquid is the temperature at which its vapor pressure is 1 atm. Although this is the definition of the normal boiling point, it is common to refer to it as simply the boiling point. The boiling point of water is 100 °C, so its vapor pressure at 100 °C is 1 atm, but the vapor pressure of an aqueous solution is less than 1 atm at 100 °C due to vapor pressure lowering. Consequently, an aqueous solution must be heated to a higher temperature to achieve a vapor pressure of 1 atm, so the boiling point of an aqueous solution is always higher than that of pure water. This reasoning can be applied to any solution, so we conclude that the boiling point of a solution is always higher than the boiling point of the pure solvent. The amount by which the boiling point of the solvent is increased by the addition of a nonvolatile solute is known as the boiling point elevation, ΔTb, which can be determined from the following equation:
( 2.13 )
ΔTb = kbmc = ikbm
Boiling Point Elevation
kb is the boiling point elevation constant, which has units of °C·m–1 and depends on the solvent, and mc is the colligative molality of the solute. ΔTb is the amount by which the boiling point of the solvent (T°b) is raised, so the boiling point of the solution (Tb) is given as
Tb = T°b + ΔTb

2.5-9. Freezing Point Depression

Spreading salt on icy streets and sidewalks melts the ice because solute particles reduce the freezing point of a solvent in much the same way as they reduce its vapor pressure. That is, solute particles block sites on the solid where solvent molecules might otherwise freeze, thereby reducing the rate of freezing. The amount by which the freezing point is lowered is called the freezing point depression, ΔTf, and is shown in Equation 2.14:
( 2.14 )
ΔTf = kfmc = ikfm
Freezing Point Depression
kf is the freezing point depression constant, which has units of °C·m–1 and depends on the solvent, and mc is the colligative molality of the solute. ΔTf is the amount by which the freezing point of the solvent (T°f) is lowered, so the freezing point of the solution (Tf) is given as
Tf = ΔT°fΔTf

2.5-10. Solvent Constants

The solvent constants relevant for colligative properties are shown in Table 2.6 and in the Resources.
Solvent Freezing Point (°C) kf (°C·m–1) Boiling Point (°C) kb (°C·m–1)
Acetic acid 16.6 3.90 117.9 3.07
Benzene 5.5 4.90 80.1 2.53
Chloroform –63.5 4.70 61.7 3.63
Cyclohexane 6.6 20.0 80.7 2.79
Ethanol –117.3 1.99 78.5 1.22
para-Xylene 11.3 4.30
Water 0.0 1.86 100.0 0.512
Table 2.6: Freezing Point Depression and Boiling Point Elevation Data
for Some Solvents

2.5-11. Antifreeze Exercise

Exercise 2.22:

Antifreeze is ethylene glycol (C2H6O2, Mm = 62.1 g/mol, d = 1.11 g/mL). What are the boiling and freezing points of an aqueous solution that is 50.% ethylene glycol by volume? Assume a density of 1.0 g/mL for water.
First, we determine the molality of the solution by assuming that water is the solvent and ethylene glycol is the solute. Thus, we need the number of kg of water and the number of moles of C2H6O2. To determine the amounts, we must first choose an amount of solution. The easiest amount is exactly 1 L of solution.
The mass of H2O in 1 L of solution = 0.50_0_2_ The solution is 50% water by volume, so 1 L solution contains 500 mL water. The density of water is 1.0 g/mL.
500 mL H2O ×
1.0 g H2O
1.0 mL H2O
×
1 kg H2O
1e+03 g H2O
= 0.50 kg H2O
kg
The number of moles of C2H6O2 in 1 L of solution = 8.94_0_3_ The solution is 50% C2H6O2, so 1 L solution contains 500 mL C2H6O2. The density of C2H6O2 is given as 1.11 g/mL.
500 mL C2H6O2 ×
1.11 g C2H6O2
1 mL C2H6O2
×
1 mol C2H6O2
62.1 g C2H6O2
= 8.94 mol C2H6O2
mol
The molality of C2H6O2 = 18_0__
8.94 mol C2H6O2
0.50 kg H2O
= 18 m
m
The van't Hoff factor for C2H6O2 = 1_0__ C2H6O2 is a nonelectrolyte, so i = 1
The colligative molality of the solution = 18_0_2_ mc = im = (1)(18 m) = 18 m m

Use Table 2.6 to determine the following for water.
1. freezing point = 0_0__ The freezing point of water is 0 °C. °C

2. boiling point = 100_0__ The boiling point of water is 100 °C. °C

3. freezing point depression constant = 1.86_0__ The freezing point depression constant for water is found to be 1.86, from Table 2.6. °C/m

4. boiling point elevation constant = 0.512_0__ The boiling point elevation constant for water is found to be 0.512, from Table 2.6. °C/m

The freezing point depression of the solution = 33_0__
ΔTf = kfmc = (1.86)(18) = 33 °C
°C
The boiling point elevation of the solution = 9.2_0__
ΔTb = kbmc = (0.512)(18) = 9.2 °C
°C
The freezing point of the solution = -33_0__
Tf = 0.0 − 33 = −33 °C
°C
The boiling point of the solution = 109.2_0__
Tb = 100.0 + 9.2 = 109.2 °C
°C

2.5-12. Salt Exercise

Exercise 2.23:

The salt that is commonly used to melt ice on roads and sidewalks is CaCl2 (Mm = 111 g/mol). Consider a solution prepared by dissolving 62.3 g CaCl2 in 100. g of water.
molality of CaCl2 = 5.61_0__ Mm = 111 g/mol, m = 62.3 g
62.3 g CaCl2 ×
1 mol CaCl2
111 g CaCl2
= 0.561 mol CaCl2

mc =
0.561 mol CaCl2
0.100 kg H20
= 5.61 m
m

colligative molality of the solution = 16.8_0__ i = 3 for CaCl2, mc = 3(5.61) = 16.8 m m

freezing point of the solution = -31.3_0__ kf = 1.86 °C/m for water, so ΔTf = (1.86)(16.8) = 31.3 °C
Tf = 0.0 – 31.3 = –31.3 °C
°C

boiling point of the solution = 108.6_0__ kb = 0.512 °C/m for water, so ΔTb = (0.512)(16.8) = 8.62 °C
Tb = 100.0 + 8.62 = 108.6 °C
°C

mole fraction of solvent = 0.767_0_3_
100 g H2O ×
1 mol H2O
18.0 g H2O
= 5.556 mol H2O

ntotal = 5.556 mol H2O + 3(0.561 mol ions) = 7.24 mol

Xsolvent =
5.556 mol H2O
7.24 mol particles
= 0.767

vapor pressure of solvent at 20 °C = 17.5_0__ The vapor pressure of water at 20 °C is found to be 17.5 torr, from Table 2.5. torr

vapor pressure of solution at 20 °C = 13.4_0__
P = XsolventP° = (0.767)(17.5 torr) = 13.4 torr.
torr

2.5-13. Molar Mass from Freezing Point Exercise

Exercise 2.24:

A common laboratory in general chemistry is the determination of a molar mass from the freezing point depression that it causes. In this example, we outline the procedure.
1.00 g of a non-dissociating solute is dissolved in 10.0 g of para-xylene. What is the molar mass of the solute, if the freezing point of the solution was 1.95 °C lower than that of the pure solvent? kf = 4.30 °C/m for para-xylene.
Use Equation 2.14
ΔTf = kfmc = ikfm
to determine the molality of the solution:
0.453_0__
m =
ΔTf
kf
=
1.95 °C
4.30 °C/m
= 0.453 m
m


Use the molality and the mass of the solvent to determine the number of moles of solute in the solution:
4.53e-3_0__
0.010 kg p-xylene ×
0.453 mol solute
1 kg p-xylene
= 4.53e−03 mol solute
mol


Use the mass and number of moles of the solute to determine its molar mass:
221_0__ 1.00 g solute is 4.53e-03 mol, so
Mm =
1.00 g
4.53e−03 mol
= 221 g/mol
g/mol

2.5-14. Osmosis

Solvent molecules flow through a semipermeable membrane from dilute solutions to concentrated solutions.
A semipermeable membrane allows solvent molecules to pass through but denies passage of solute particles. The rate of passage of solvent molecules through the membrane is proportional to the concentration of the solvent molecules, so the rate is greater for a pure solvent than for a solution. When a solvent and a solution are separated by a semipermeable membrane, more solvent molecules move from the solvent to the solution than in the reverse direction. The net movement of the solvent molecules through the membrane from the solvent or a more dilute solution into a more concentrated one is called osmosis.
osmosis
Figure 2.12: Osmosis
(a) Rate of movement through the membrane is the same in both directions because the solvent concentration is the same on both sides. (b) A solute is added to side II, so the concentration of solvent is less in side II than in side I. Consequently, the rate of solvent motion I II is greater than II I.

2.5-15. Osmotic Pressure

Osmotic pressure depends on colligative molarity.
Osmosis of the solvent through the membrane from a solvent to a solution decreases the concentration of particles in the solution, but it also increases its volume while decreasing the volume of the solvent. These volume changes result in a height difference in the two columns as represented by h in Figure 2.13. The height difference results in a pressure differential at the membrane that increases the flow of solvent molecules from the solution side. Eventually, the combination of the increased pressure coupled with the decrease in concentration of the solute particles increases the flow of solvent particles from the solution back into the solvent to the point where the flow of solvent molecules through the membrane is the same in both directions. The pressure at which equilibrium is attained is called the osmotic pressure (Π) of the solution. The osmotic pressure that is developed is given in Equation 2.15
( 2.15 )
Π = Mc × R × T = i × M × R × T
Osmotic Pressure
where R is the ideal gas law constant (R = 0.0821 L·atm/K·mol).
osmotic pressure
Figure 2.13: Osmotic Pressure
Solvent molecules penetrate the semipermeable membrane, but solute molecules do not. Thus, there is a net flow of water molecules into the side with more solute.

2.5-16. Osmotic Pressure of Sea Water Exercise

Exercise 2.25:

What is the osmotic pressure developed by seawater at 25 °C. Assume that seawater is a 0.53 M solution of NaCl.
Π = 26_0__ The van't Hoff factor for NaCl = 2, the temperature is 298 K, R = 0.0821 L·atm/K·mol
Π = iMRT = (2)(0.53)(0.0821)(298) = 26 atm
atm

2.5-17. Molar Mass from Osmotic Pressure Exercise

Exercise 2.26:

1.00 L of an aqueous solution containing 0.40 g of a peptide has an osmotic pressure of 3.74 torr at 27 °C. What is the molar mass of the peptide?
Π = 4.92e-3_0__
3.74 torr ×
1 atm
760 torr
= 4.92e−03 atm
atm

molar concentration of the peptide = 2.00e-4_0_3_ Use Equation 2.15 and Π = 4.92e-03 atm, R = 0.0821 L·atm/K·mol, T = 300K, and i = 1 (peptides are nonelectrolytes).
M =
Π
iRT
=
4.92e−03 atm
(1)(0.0821 L\pre{·}atm/K\pre{·}mol)(300 K)
= 2.00e−04 M
M

number of moles of peptide in the solution = 2.00e-4_0_3_
n = M × V = (2.0e−04 M)(1.00 L) = 2.00e−04 mol.
The peptide is in a 1.00-L solution.
mol

molar mass of peptide = 2000_0_2_
Mm =
0.40 g peptide
2.00e−04 mol peptide
= 2.0e+03 g/mol
g/mol

2.5-18. Applications of Osmosis

Purification: Solvents can be purified by a process called reverse osmosis. If a pressure that exceeds the osmotic pressure is applied to a solution, solvent can be forced from the solution into the pure solvent. This process has been used to purify seawater in the Middle East. The challenge to purifying seawater lies in finding membranes that can withstand the large pressure (over 26 atm) required. Biology: Cell membranes are semipermeable. When the solution around a cell has the same colligative concentration as within the cell, the cell maintains its size. However, if the cell is placed in water, water passes into the cell and can rupture it. Placing the cell in a solution with a greater colligative molarity causes water to leave the cell. This is why drinking seawater does not quench your thirst. Water transport in plants: Water enters a tree through the membranes in the roots, but it evaporates from the leaves resulting in a substantial concentration difference between the roots and leaves. The large concentration difference can develop osmotic pressures of up to ~20 atm in tall trees.

2.6 Colloids

Introduction

Sometimes, rather than dissolving, one material will be suspended as small aggregates in another. Thus, sand will stay suspended in water as long as the water is stirred; but soon after the stirring has stopped, the sand settles to the bottom of the container. Because these mixtures do not meet the rigorous criterion of a solution, they are called dispersions or suspensions.

Objectives

2.6-1. Types of Colloids

Suspensions in which the particle size is very small (1 nm to 1 µm) are called colloidal suspensions or simply colloids. White paint is a colloidal suspension of SiO2 and TiO2 particles, which are used to make the paint opaque and white, respectively. Colloidal suspensions, which can be stable (will not settle) for years, are classified according to their composition. Whipped cream is a foam: a gas suspended in a liquid. Jellies and starch solutions are sols: solids suspended in a liquid. Milk is an emulsion: a liquid suspended in a liquid. Aerosols can be liquids suspended in a gas (hair sprays) or solids suspended in a gas (smoke). Fog is also an aerosol (water in air) and smog, the combination of smoke and fog, is also a colloidal suspension.

2.7 Exercises and Solutions

Select the links to view either the end-of-chapter exercises or the solutions to the odd exercises.