Pearson Chemistry 12 New South WalesIn this chapter, you will begin by examining how material science can create advanced materials by controlling matter on the atomic scale. Nanoparticles contain just a few hundred or thousands of atoms. You will then develop a detailed picture of the structure of atoms, which is the foundation for all chemistry.
A diamond is a pure substance made entirely from carbon atoms. This chapter will explain how the carbon atoms are arranged in diamonds and why this structure results in the properties of diamond such as hardness and stability at high temperatures.
Carbon is an unusual element because it exists in several other forms that are very different from diamond. This chapter will also describe the arrangements of atoms in these other forms of carbon.
You will learn how chemists measure and compare masses of isotopes and atoms using a mass spectrometer.
Outcomes • investigate the components of atoms through relative charges and position in the
periodic table
• investigate the components of atoms through calculation of relative isotopic mass
• investigate the components of atoms through representation of the symbol, atomic number and atomic mass (nucleon number)
Credit line to come xxxxxxxxxxxxxxx
Atomic structure and atomic mass
CHAPTER
Most biological processes, particularly within the human body, exist in narrow limits, including the balance of acidic and basic components. Saliva is mildly acidic to begin the digestive process, and the stomach produces acid to further aid digestion and kill pathogens and unhelpful bacteria. A number of amino acids have acidic or basic side chains and are of great importance physiologically. Many acids and bases used in industry and in household applications are very strong, and can cause serious injury if they are mishandled or misused. It is important then to understand the nature of such acids and bases and quantify their strengths and applications.
In this chapter you will investigate the distinction between strong and weak acids, and concentrated and dilute acids. You will learn how to measure acidity and basicity, particularly with reference to their component concentrations and respective p-functions.
Content
INQUIRY QUESTION
What is the role of water in solutions of acids and bases? By the end of this chapter, you will be able to:
• conduct an investigation to demonstrate the preparation and use of indicators as illustrators of the characteristics and properties of acids and bases and their reversible reactions (ACSCH101)
• conduct a practical investigation to measure the pH of a range of acids and bases
• calculate pH, pOH, hydrogen ion concentration ([H+]) and hydroxide ion concentration ([OH−]) for a range of solutions (ACSCH102) ICT N
• conduct an investigation to demonstrate the use of pH to indicate the differences between the strength of acids and bases (ACSCH102)
• construct models and/or animations to communicate the differences between strong, weak, concentrated and dilute acids and bases (ACSCH099) ICT
• calculate the pH of the resultant solution when solutions of acids and/or bases are diluted or mixed ICT N
Chemistry Stage 6 Syllabus © NSW Education Standards Authority for and on behalf of the Crown in right of the State of NSW, 2017.
Using the Brønsted–Lowry theory
CHAPTER
CHEMISTRY INQUIRY CCT
• lab coat
• disposable gloves
• safety glasses
• blender
• sieve
• white paper
• vinegar
• lemon juice
• laundry detergent
1 Pour 1.0 L of distilled water into the blender.
2 Add three large red cabbage leaves.
3 Blend the water and cabbage leaves until the mixture appears smooth.
4 Using the sieve, strain off the liquid and keep the filtrate.
5 Half-fill each of the test tubes with the filtrate.
6 Set aside one test tube without adding any other solution to it. This will provide a reference for the colour of neutral solutions.
7 To one of the test tubes, add approximately 5.0 mL of 0.100 mol L−1 HCl solution and shake the tube gently to mix. This will provide a reference for the colour of acidic solutions.
8 To another test tube, add approximately 5.0 mL of 0.100 mol L−1 NaOH solution and shake the tube gently to mix. This will provide a reference for the colour of basic solutions.
9 To the remaining test tubes, add approximately 5.0 mL of the samples and shake each tube gently to mix. Dissolve any solid sample first in a small amount of distilled water before adding it to the indicator filtrate.
RECORD THIS …
Describe the colour change of each solution tested and determine by the colour whether the solution is acidic, basic or neutral compared with the reference test tube.
Present your results in a table.
REFLECT ON THIS …
1 Which substances are acidic?
2 Which substances are basic?
3 Were there any results that surprised you? Explain your response.
4 What is the purpose of the blender?
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163CHAPTER 7 | USING THE BRØNSTED–LOWRY THEORY
The acid solutions in the two beakers shown in Figure 7.1.1 are of equal concentration, yet the acid in the beaker on the left reacts more vigorously with zinc than the acid on the right. The acid on the left is described as a stronger acid than the one on the right.
FIGURE 7.1.1 Zinc reacts more vigorously with a strong acid (left) than with a weak acid (right). The acid solutions are of equal concentration and volume.
In Chapter 6 you learnt that the Brønsted−Lowry theory defines acids as proton donors, and bases as proton acceptors. In this section you will investigate the differences between: • strong and weak acids • strong and weak bases.
ACID AND BASE STRENGTH Experiments show that different acid solutions of the same concentration do not have the same pH (a measure of the concentration of hydronium ions in solution). (pH will be looked at more closely in Section 7.2.)
Some acids can donate a proton more readily than others. The Brønsted−Lowry theory defines the strength of an acid as its ability to donate protons to a base. The strength of a base is a measure of its ability to accept protons from an acid.
Since aqueous solutions of acids and bases are most commonly used, it is convenient to use an acid’s tendency to donate a proton to water, or a base’s tendency to accept a proton from water, as a measure of its strength.
Table 7.1.1 gives the names and chemical formulae of some strong and weak acids and bases.
TABLE 7.1.1 Examples of common strong and weak acids and bases
Strong acids Weak acids Strong bases Weak bases
hydrochloric acid, HCl
potassium hydroxide, KOH
calcium hydroxide, Ca(OH)2
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MODULE 6 | ACID/BASE REACTIONS164
Strong acids As you saw previously, hydrogen chloride gas (HCl) dissociates completely when it is bubbled through water; virtually no HCl molecules remain in the solution (Figure 7.1.2a). Similarly, pure HNO3 and H2SO4 are covalent molecular compounds that also dissociate completely in water:
HCl(g) + H2O(l) → H3O +(aq) + Cl−(aq)
H2SO4(l) + H2O(l) → H3O +(aq) + HSO4
−(aq) HNO3(l) + H2O(l) → H3O
+(aq) + NO3 −(aq)
The single reaction arrow (→) in each equation above indicates that the dissociation reaction is complete.
Acids that readily donate a proton are called strong acids. Strong acids donate protons easily. Therefore solutions of strong acids contain ions, with virtually no unreacted acid molecules remaining. Hydrochloric acid, sulfuric acid and nitric acid are the most common strong acids.
Weak acids Vinegar is a solution of ethanoic (acetic) acid. Pure ethanoic acid is a polar covalent molecular compound that dissociates in water to produce hydrogen ions and ethanoate (acetate) ions. In a 1.0 mol L−1 solution of ethanoic acid (CH3COOH), only a small proportion of ethanoic acid molecules are dissociated at any one time (Figure 7.1.2b). A 1.0 mol L−1 solution of ethanoic acid contains a high proportion of CH3COOH molecules and only some hydronium ions and ethanoate ions. In a 1.0 mol L−1 solution of ethanoic acid at 25°C, the concentrations of CH3COO−(aq) and H3O
+ are only about 0.004 mol L−1. The partial dissociation of a weak acid is shown in an equation using reversible
(double) arrows: CH COOH(aq)
base CH COO H O (aq)3 2 3 3+ +− +
Therefore ethanoic acid is described as a weak acid in water (Figure 7.1.2b).
H2O(l)
H3O +(aq)
FIGURE 7.1.2 (a) In a 1.0 mol L–1 solution of hydrochloric acid, the acid molecules are completely dissociated in water. (b) However, in a 1.0 mol L–1 solution of ethanoic acid only a small proportion of the ethanoic acid molecules are dissociated.
Strong bases The ionic compound sodium oxide (Na2O) dissociates in water, releasing sodium ions (Na+) and oxide ions (O2−). The oxide ions react completely with the water, accepting a proton to form hydroxide ions (OH−):
O (aq) base
H O(l) acid
2OH (aq)2 2+ →− −
The oxide ion is an example of a strong base. Strong bases accept protons easily. Sodium hydroxide (NaOH) is often referred to as a strong base. However,
according to the Brønsted−Lowry definition of acids and bases, it is more correct to say that sodium hydroxide is an ionic compound that is a source of the strong base OH−.
(a) (b)
CHEMFILE WE
Acid and base spills Concentrated acids and bases can be very hazardous, particularly because of their corrosive nature. Liquid spilled onto a flat surface such as a floor or bench also creates a slipping hazard. You could consider neutralising the strong acid or base by using a strong counterpart, such as a sodium hydroxide solution added to a hydrochloric acid spill. However, adding a strong base to a strong acid is not advisable, because too much strong base could be added, increasing the hazard. You could add water to dilute the acid, but this would increase the amount of liquid in the spill, causing it to spread further.
For this reason laboratories are equipped with spill kits. Acid and base spill kits contain quantities of solid weak acids and bases which neutralise the strong solution spilled, and may also contain booms to contain the spill and pads to absorb the neutralised liquid. An example of a solid weak base in a spill kit is sodium hydrogen carbonate. An example of weak acid in a spill kit is citric acid monohydrate.
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165CHAPTER 7 | USING THE BRØNSTED–LOWRY THEORY
Weak bases Ammonia is a covalent molecular compound that dissociates in water by accepting a proton. This dissociation can be represented by the equation:
NH (aq) base
H O(l) acid
NH (aq) OH (aq)3 2 4+ ++ −
Ammonia behaves as a base because it gains a proton. Water has donated a proton and so it behaves as an acid.
Only a small proportion of ammonia molecules are dissociated at any instant, so a 1.0 mol L–1 solution of ammonia contains mostly ammonia molecules together with a smaller number of ammonium ions and hydroxide ions. This is shown by the double arrow in the equation. Ammonia is a weak base in water.
CHEMFILE N
Super acids Fluorosulfuric acid (HSO3F) is one of the strongest acids known. It has a similar geometry to that of the sulfuric acid molecule (Figure 7.1.3). The highly electronegative fluorine atom causes the oxygen−hydrogen bond in fluorosulfuric acid to be more polarised than the oxygen−hydrogen bond in sulfuric acid. The acidic proton is easily transferred to a base.
S
O
HO
HO O
FIGURE 7.1.3 Structures of sulfuric acid (left) and fluorosulfuric acid (right) molecules
Fluorosulfuric acid is classified as a super acid. Super acids are acids that have an acidity greater than the acidity of pure sulfuric acid.
Super acids such as fluorosulfuric acid and triflic acid (CF3SO3H) are about 1000 times stronger than sulfuric acid. Carborane acid, H(CHB11Cl11), is 1 million times stronger than sulfuric acid. The strongest known super acid is fluoroantimonic acid (H2FSbF6), which is 1016 times stronger than pure sulfuric acid.
Super acids are used in the production of plastics and high-octane petrol, in coal gasification and in research.
Relative strength of conjugate acid−base pairs You will recall from Chapter 6 that conjugate acids and bases differ by one proton (H+). In the reaction represented by the equation:
+ +− −HF(aq) OH (aq) H O(l) F (aq)2
HF is the conjugate acid of F− and OH− is the conjugate base of H2O. HF and F− are a conjugate acid−base pair. H2O and OH− are another conjugate acid−base pair in this reaction.
The stronger an acid is, the weaker its conjugate base is. Similarly, the stronger a base is, the weaker its conjugate acid is. The relative strength of some conjugate acid−base pairs is shown in Figure 7.1.4.
Strength versus concentration When referring to solutions of acids and bases, it is important not to confuse the terms ‘strong’ and ‘weak’ with ‘concentrated’ and ‘dilute’. Concentrated and dilute describe the amount of acid or base dissolved in a given volume of solution. Hydrochloric acid is a strong acid because it readily donates protons. A concentrated solution of hydrochloric acid can be prepared by bubbling a large amount of hydrogen chloride into a given volume of water. By using only a small amount of hydrogen chloride, a dilute solution of hydrochloric acid would be produced.
GO TO Section 6.1 page 000
Acid
ng
FIGURE 7.1.4 The relative strength of some conjugate acid−base pairs
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MODULE 6 | ACID/BASE REACTIONS166
However, in both cases, the hydrogen chloride is completely dissociated—it is a strong acid. Similarly a solution of ethanoic acid may be concentrated or dilute. However, as it is partially dissociated, it is a weak acid (Figure 7.1.5).
Weak, concentrated ethanoic acid
Weak, dilute ethanoic acid
Strong, concentrated hydrochloric acid
Strong, dilute hydrochloric acid
acid (CH3COOH, HCl) H+(aq)conjugate base (CH3COO–(aq), Cl–(aq))
FIGURE 7.1.5 The concentration of ions in an acid solution depends on both the concentration and strength of the acid.
Terms such as ‘weak’ and ‘strong’ for acids and bases, and ‘dilute’ or ‘concentrated’ for solutions, are qualitative (or descriptive) terms. Solutions can be given accurate, quantitative descriptions by stating concentrations in mol L−1 or g L−1. The extent of dissociation of an acid or base can be quantified by measuring its dissociation constant. This will be discussed in greater detail in Chapter 8.
7.1 Review
SUMMARY
• A concentrated acid or base contains more moles of solute per litre of solution than a dilute acid or base.
• In the context of acids and bases, the terms ‘strong’ and ‘weak’ refer to the relative tendency to accept or donate protons.
- A strong acid donates a proton more readily than a weak acid.
- A strong base accepts a proton more readily than a weak base.
• The stronger an acid is, the weaker is its conjugate base. The stronger a base is, the weaker is its conjugate acid.
KEY QUESTIONS
1 Write balanced equations to show that, in water: a HClO4 is a strong acid b HCN is a weak acid c CH3NH2 is a weak base.
2 Write balanced equations for the three dissociation stages of arsenic acid (H3AsO4).
3 Which one of the following equations represents the reaction of a strong acid with water? A HNO3(aq) + H2O(l) → H3O
+(aq) + NO3 −(aq)
B + ++ −HF(aq) H O(l) H O (aq) F (aq)2 3 C LiOH(s) → Li+(aq) + OH−(aq) D NH3(aq) + H2O(l) NH4
+(aq) + OH−(aq)
4 Perchloric acid is a stronger acid than ethanoic (acetic) acid. 1.0 mol L−1 solutions of which acid would you expect to be a better conductor of electricity? Explain your answer.Pa
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167CHAPTER 7 | USING THE BRØNSTED–LOWRY THEORY
7.2 Acidity and basicity of solutions The Brønsted−Lowry theory that defines an acid as a proton donor and a base as a proton acceptor. You have also seen that acids and bases can be classified as strong or weak, depending on how easily they donate or accept protons.
In this section, you will learn about the pH and pOH scales, which are a measure of acidity and basicity, respectively. You will also learn about the relationship between the concentration of hydronium and hydroxide ions in different solutions.
IONIC PRODUCT OF WATER Water molecules can react with each other, as represented by the equation:
+ ++ −H O(l) H O(l) H O (aq) OH (aq)2 2 3 The production of the H3O
+ ion and OH− ion in this reaction is shown visually in Figure 7.2.1.
H3O + OH–H2O H2O
FIGURE 7.2.1 The ionisation of water molecules
Pure water undergoes this self-ionisation to a very small extent. In this reaction, water behaves as both a very weak acid and a very weak base, producing one hydronium ion (H3O
+) for every one hydroxide ion (OH−). Water therefore displays amphiprotic properties.
The concentration of hydronium and hydroxide ions is very low. In pure water at 25°C the H3O
+ and OH− concentrations are both 10−7 mol L–1. For each H3O + ion
present in a glass of water, there are 560 million H2O molecules! Experimental evidence shows that all aqueous solutions contain both H3O
+ and OH− ions, and that the product of their molar concentrations, [H3O
+][OH−], is always 1.0 × 10−14 at 25°C. If either [H3O
+] or [OH−] in an aqueous solution increases, then the concentration of the other must decrease proportionally.
Remember that [H3O +] represents the concentration of hydrogen ions and
[OH−] represents the concentration of hydroxide ions. The expression [H3O +][OH−]
is known as the ionic product of water and is represented by the symbol Kw: Kw = [H3O
+][OH−] = 1.0 × 10−14 at 25°C
Acidic and basic solutions In solutions of acidic substances, H3O
+ ions are formed by reaction of the acid with water, as well as from self-ionisation of water. So the concentration of H3O
+ ions will be greater than 10−7 mol L–1 at 25°C. Since the product [H3O
+][OH−] remains constant, the concentration of OH− ions in an acidic solution at this temperature must be less than 10−7 mol L–1.
The opposite is true for basic solutions. The concentration of OH− ions in a basic solution is greater than 10−7 mol L–1 and the concentration of H3O
+ ions is less than 10−7 mol L–1.
In summary, at 25°C: • pure water and neutral solutions: [H3O
+] = [OH−] = 10−7 mol L–1
• acidic solutions: [H3O +] > 10−7 mol L–1 and [OH−] < 10−7 mol L–1
• basic solutions: [H3O +] < 10−7 mol L–1 and [OH−] > 10−7 mol L–1.
The higher the concentration of H3O + ions in a solution, the more acidic the
solution is. Conversely, the higher the concentration of OH− ions in a solution, the more
basic the solution is.
Square brackets [ ] are often used to represent molar concentration.
In acid–base chemistry, the terms ‘dissociation’ and ‘ionisation’ can be used interchangeably, and the ionic product is also called the ionisation constant.
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MODULE 6 | ACID/BASE REACTIONS168
Calculating the concentration of H3O+ in aqueous solutions The expression for Kw can be used to determine the concentrations of hydronium and hydroxide ions in solution, knowing that the value of Kw in solutions at 25°C is 1.0 × 10−14.
Worked example 7.2.1
CALCULATING THE CONCENTRATION OF HYDRONIUM AND HYDROXIDE IONS IN AN AQUEOUS SOLUTION
For a 0.10 mol L−1 HCl solution at 25°C, calculate the concentrations of H3O + and
OH− ions.
Thinking Working
Find the concentration of the hydronium (H3O
+) ions. HCl is a strong acid, so it will dissociate completely in solution. Each molecule of HCl donates one proton to water to form one H3O
+ ion:
HCl(aq) + H2O(l) → H3O +(aq) + Cl−(aq)
Because HCl is completely dissociated in water, 0.10 mol L−1 HCl will produce a solution with a concentration of H3O
+ ions of 0.10 mol L−1.
[H3O +] = 0.10 mol L−1
Use the expression for the ionisation constant of water to calculate the concentration of OH− ions.
Kw = [H3O +][OH−] = 1.0 × 10−14
− =
=
= ×
×
×
− −
−
+
−
Worked example: Try yourself 7.2.1
CALCULATING THE CONCENTRATION OF HYDRONIUM AND HYDROXIDE IONS IN AN AQUEOUS SOLUTION
For a 5.6 × 10−6 mol L−1 HNO3 solution at 25°C, calculate the concentration of H3O
+ and OH− ions.
pH AND pOH: A CONVENIENT WAY TO MEASURE ACIDITY AND BASICITY
Definition of pH The range of H3O
+ concentrations in solutions is so great that a convenient scale, called the pH scale, has been developed to measure acidity. The pH scale was first proposed by the Danish scientist Sören Sörenson in 1909 as a way of expressing levels of acidity. The pH of a solution is defined as:
pH = −log10[H3O +]
+] = 10−pH
The pH scale eliminates the need to write cumbersome powers of 10 when we describe hydrogen ion concentration. It is part of a series of transformations known as p-functions. The use of pH greatly simplifies the measurement and calculation of acidity. Since the scale is based on the negative logarithm of the hydrogen ion concentration, the pH of a solution decreases as the concentration of hydrogen ions increases.
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169CHAPTER 7 | USING THE BRØNSTED–LOWRY THEORY
Definition of pOH The pOH scale is a measure of basicity, in the same way that the pH scale is a measure of acidity. The pOH of a solution is defined as:
pOH = −log10[OH–] Alternatively, this expression can be rearranged to give:
[OH–] = 10−pOH
At 25°C, the pOH and pH of a solution are linked according to the following relationship:
pH + pOH = 14 This means that the pOH scale is the reverse relationship of the pH scale. That
is, as the pOH of a solution increases, the pH decreases.
pH and pOH of acidic and basic solutions Figure 7.2.2 shows a pH meter, which can be used to measure the pH of a solution accurately.
Acidic, basic and neutral solutions can be defined in terms of their pH at 25°C. • Neutral solutions have both the pH and pOH equal to 7. • Acidic solutions have a pH less than 7 and a pOH greater than 7. • Basic solutions have a pH greater than 7 and a pOH less than 7.
FIGURE 7.2.2 A pH meter is used to measure the acidity of a solution.
On the pH scale, the most acidic solutions have pH values slightly less than 0, and the most basic solutions have pH values of about 14. On the pOH scale, more basic solutions have a lower pOH. The pH and pOH values of some common substances are provided in Table 7.2.1.
TABLE 7.2.1 pH and pOH values of some common substances at 25°C
Solution pH pOH [H3O+] ( mol L–1) [OH−] (mol L–1) [H3O+][OH−]
1.0 mol L−1 HCl 0.0 14.0 1 10−14 10−14
lemon juice 3.0 11.0 10−3 10−11 10−14
vinegar 4.0 10.0 10−4 10−10 10−14
tomato juice 5.0 9.0 10−5 10−9 10−14
rain water 6.0 8.0 10−6 10−8 10−14
pure water 7.0 7.0 10−7 10−7 10−14
seawater 8.0 6.0 10−8 10−6 10−14
soap 9.0 5.0 10−9 10−5 10−14
oven cleaner 13.0 1.0 10−13 10−1 10−14
1.0 mol L−1 NaOH 14.0 0.0 10−14 1 10−14
A solution with a pH of 2 has 10 times the concentration of hydronium ions as one with a pH of 3.
A solution with a pH of 2 has one tenth the concentration of hydroxide ions as one with a pH of 3.
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MODULE 6 | ACID/BASE REACTIONS170
The reactions of acids and bases are important in a large variety of everyday applications, including biological functions. The high acidity of gastric juices is essential for protein digestion in the stomach. There is also a complex system of pH control in your blood, because even small deviations from the normal pH range of 7.35–7.45 for any length of time can lead to serious illness or death.
Indicators One of the characteristic properties of acids and bases is their ability to change the colour of certain substances, called indicators. Many plant extracts are indicators, including litmus, a purple dye obtained from lichens. In the presence of acids, litmus turns red. Indicators can also be obtained from rose petals, blackberries, red cabbage and many other plants, although most commercial indicators are made from other substances. Some common indicators and their pH ranges are shown in Figure 7.2.3.
methyl violet
thymol blue
methyl orange
methyl red
bromothymol blue
Colour changes and pH ranges
violet
violet
pinkcolourless
FIGURE 7.2.3 Common indicators and their pH ranges
Indicators themselves are weak acids or weak bases. The conjugate acid form of the indicator is one colour and the conjugate base form is another colour. Indicators that undergo a single colour change are also used for many analyses.
Universal indicator Universal indicator (Figure 7.2.4) is widely used to estimate the pH of a solution. It is a mixture of several indicators and changes through a range of colours, from red through yellow, green and blue, to violet. If a more accurate measurement of pH is needed, you can use a pH meter instead of universal indicator.
FIGURE 7.2.4 Universal indicator pH scale. When universal indicator is added to a solution, it changes colour depending on the solution’s pH. The tubes contain solutions of pH 0 to 14 from left to right. The green tube (seventh from right) is neutral, pH 7.
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SKILLBUILDER WE
Using litmus paper Litmus paper is commonly used to test whether a substance is acidic or basic. Blue litmus paper will remain blue in the presence of a base, but will turn red in the presence of an acid (Figure 7.2.5). Red litmus paper will remain red in the presence of an acid, but will turn blue in the presence of a base (Figure 7.2.6). Purple litmus paper turns red in the presence of an acid, blue in the presence of a base, and remains purple if the substance is neutral.
Litmus paper does not tell you the pH value of a substance, but pH indicator papers can be used for this. Packets of pH inidcator papers include a colour chart that enables you to determine the pH of any substance (Figure 7.2.7).
FIGURE 7.2.5 Blue litmus paper turns red when exposed to the citric acid in an orange.
FIGURE 7.2.6 Red litmus paper turns blue when exposed to basic substances, such as wet alkaline soil.
FIGURE 7.2.7 A packet of pH indicator papers with a colour chart
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MODULE 6 | ACID/BASE REACTIONS172
Calculations involving pH and pOH The pH of an aqueous solution can be calculated using the formula pH = −log10[H3O
+], and similarly pOH = −log10[OH–]. Your scientific calculator has a logarithm function that will simplify pH and pOH calculations.
Worked example 7.2.2
CALCULATING THE pH AND pOH OF AN AQUEOUS SOLUTION FROM [H3O+]
Calculate the pH and pOH of a solution in which the concentration of [H3O +] is
0.14 mol L−1. Give your answer correct to 2 decimal places.
Thinking Working
ions in the solution. [H3O
+] = 0.14 mol L−1
pH = −log10[H3O +]
Use the logarithm function on your calculator to determine the answer (ensure to use log to the base 10 and not the natural logarithm, which may appear as ln).
pH = −log10[H3O +]
pOH = 14 – pH
pOH = 14 – pH
Worked example: Try yourself 7.2.2
CALCULATING THE pH AND pOH OF AN AQUEOUS SOLUTION FROM [H3O+]
Calculate the pH and pOH of a solution in which the concentration of [H3O +] is
6 × 10−9 mol L−1. Give your answer correct to 2 significant figures.
Worked example 7.2.3
CALCULATING THE pH AND pOH IN A SOLUTION OF A BASE
What is the pH and pOH of a 0.005 mol L−1 solution of Ba(OH)2 at 25°C?
Thinking Working
Write down the equation for the dissociation of Ba(OH)2 in water.
In water, 1 mol of Ba(OH)2 completely dissociates to release 2 mol of OH− ions.
Ba(OH)2(aq) → Ba2+(aq) + 2OH−(aq)
Determine the concentration of [OH−] ions.
[OH−] = 2 × [Ba(OH)2]
= 0.01 mol L−1
To calculate the pH, determine the [H3O
+] in the solution by substituting the [OH−] into the ionic product of water:
Kw = [H3O +][OH−] = 1.0 × 10−14
Kw = [H3O +][OH−] = 1.0 × 10−14
=
=
+
×
−
−
1.0 10 0.01
pH = −log10[H3O +]
pH = −log10[H3O +]
= −log10(1 × 10−12)
Determine the pOH by using the pOH formula:
pOH = –log10[OH−]
pOH = −log10[OH−]
pH = 14 − pOH
pH = 14 – pOH
CALCULATING pH AND pOH IN A SOLUTION OF A BASE
What is the pH and pOH of a 0.01 mol L−1 solution of Ba(OH)2 at 25°C?
Worked example 7.2.4
CALCULATING pH AND pOH IN A SOLUTION WHERE THE SOLUTE MOLAR CONCENTRATION IS NOT GIVEN
Calculate the pH and pOH of a solution at 25°C that contains 1.0 g NaOH in 100 mL of solution.
Thinking Working
=n(NaOH) m M
NaOH(aq) → Na+(aq) + OH−(aq)
NaOH is completely dissociated in water.
Determine the number of moles of OH− based on the dissociation equation.
n(OH−) = n(NaOH)
= 0.025 mol
Use the formula for determining concentration given number of moles and volume:
=c n V
n = 0.025 mol
V = 0.100 L =
pOH = –log10[OH−]
pH = 14 − pOH pH = 14 – pOH
= 14 – 0.60
Worked example: Try yourself 7.2.4
CALCULATING PH AND POH IN A SOLUTION WHERE THE SOLUTE MOLAR CONCENTRATION IS NOT GIVEN
Calculate the pH and pOH of a solution at 25°C that contains 0.50 g KOH in 500 mL of solution.
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Worked example 7.2.5
CALCULATING [H3O+] AND [OH−] IN A SOLUTION OF A GIVEN pH
Calculate [H3O +] and [OH−] in a solution of pH 5.0 at 25°C.
Thinking Working
Decide which form of the relationship between pH and [H3O
+] should be used:
pH = −log10[H3O +]
+], use:
[H3O +] = 10−pH
Substitute the value of pH into the relationship expression and use a calculator to determine the answer.
[H3O +] = 10−pH
Determine the [OH−] in the solution by substituting the [H3O
+] into the ionic product of water:
Kw = [H3O +][OH−] = 1.0 × 10−14
Kw = [H3O +][OH−] = 1.0 × 10−14
− =
= × ×
+
−
−
Worked example: Try yourself 7.2.5
CALCULATING [H3O+] AND [OH−] IN A SOLUTION OF A GIVEN pH
Calculate [H3O +] and [OH−] in a solution of pH 10.4 at 25°C.
+ ADDITIONAL N
Effect of temperature on pH and pOH At the start of this section the ionic product of water was defined as:
Kw = [H3O +] [OH−] = 1.0 × 10−14 at 25°C
You can use this relationship to calculate either [H3O +] or
[OH−] at 25°C in different solutions. But what happens if the temperature is not 25°C?
From experimental data it is known that the value of Kw increases as the temperature increases, as shown in Table 7.2.2.
TABLE 7.2.2 The effect of temperature on pH and pOH of pure water
Temperature (°C) Kw pH = pOH
0 1.1 × 10−15 7.47
5 1.9 × 10−15 7.37
15 4.5 × 10−15 7.17
25 1.0 × 10−14 7.00
35 2.1 × 10−14 6.83
45 4.0 × 10−14 6.70
55 7.3 × 10−14 6.57
In Chapter 4 it was noted that a change in temperature changes the value of the equilibrium constant, and that endothermic reactions are favoured at higher temperatures. The self-ionisation of water is an endothermic process. This means that the value of Kw will be larger at higher temperatures, and therefore the concentration of both the hydronium and hydroxide ions will be higher. This is why it is important that the temperature is given when the value of Kw is stated.
The pH and pOH of pure water are both 7.00 at 25°C. At other temperatures, even though the pH and pOH are not equal to 7.00, pure water can still be described as neutral because the concentrations of H3O
+ and OH− ions are equal. It is important to note, then, that as you move away from 25°C the pH + pOH = 14 relationship no longer applies.
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7.2 Review
• Water self-ionises according to the equation:
+ ++ −H O(l) H O(l) H O (aq) OH (aq)2 2 3
• The ionic product for water is:
Kw = [H3O +][OH−] = 1.0 × 10−14 at 25°C
• The concentration of H3O + is measured using the
pH scale:
• The concentration of OH− is measured using the pOH scale:
pOH = −log10[OH−]
• At 25°C the pH and pOH of a neutral solution are both 7. The pH of an acidic solution is less than 7 and the pH of a basic solution is greater than 7. The pOH of a basic solution is less than 7 and the pOH of an acidic solution is greater than 7.
• The relationship between pH and pOH at 25°C can be expressed as:
pH + pOH = 14
KEY QUESTIONS
1 Calculate [OH−] at 25°C in an aqueous solution with [H3O
+] = 0.001 mol L−1.
2 What is [OH−] in a solution at 25°C with [H3O
+] = 5.70 × 10−9 mol L−1?
3 Calculate [H3O +] at 25°C in an aqueous solution in
which [OH−] = 1.0 × 10−5 mol L−1.
4 What is the pH and pOH of a solution in which [H3O
+] = 0.01 mol L−1?
5 Calculate the pH and pOH of a 0.001 mol L−1 solution of nitric acid (HNO3).
6 The pH of water in a lake is 6.0. Calculate [H3O +] and
[OH−] in the lake.
7 Determine the pH and pOH of a 200 mL solution that contains 0.365 g of dissolved HCl.
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MODULE 6 | ACID/BASE REACTIONS176
7.3 Dilution of acids and bases Although acids are frequently purchased as concentrated solutions, you will often need to use them in a more diluted form. For example, a bricklayer needs a 10% solution of hydrochloric acid to remove mortar splashes from bricks used to build a house. The brick-cleaning solution is prepared by diluting concentrated hydrochloric acid by a factor of 10.
In this section you will learn how to calculate the concentration and pH of acids and bases once they have been diluted.
CONCENTRATION OF ACIDS AND BASES The concentration of acids and bases is usually expressed in units of mol L−1. This is also referred to as molar concentration or molarity.
REVISION
You will recall from Year 11 that the molar concentration of a solution, in mol L−1, is given by the expression:
=c n V
n is the amount of solute in mol
V is the volume of the solution in L.
The most convenient way of preparing a solution of a dilute acid is by mixing concentrated acid with water, as shown in Figure 7.3.1. This is known as a dilution.
The amount of solute in a solution does not change when a solution is diluted; the volume of the solution increases and the concentration decreases. The change in concentration or volume can be calculated using the formula:
c1V1 = c2V2
where c1 and V1 are the initial concentration and volume
c2 and V2 are the concentration and volume after dilution.
You can calculate the concentration of a dilute acid if you know the:
• volume of the concentrated solution
• concentration of the concentrated solution
• total volume of water added.
GO TO Year 11 Chapter 8
Measure out 10.0 mL of 10.0 mol L–1 HCl with a pipette.
Transfer to a 100.0 mL volumetric flask partly filled with water.
Fill the flask with distilled water to the calibration mark and shake thoroughly.
FIGURE 7.3.1 Preparing a 1.00 mol L–1 HCl solution by diluting a 10.0 mol L–1 solution. Heat is released when a concentrated acid is mixed with water, so the volumetric flask is partly filled with water before the acid is added. (Extra safety precautions would be required for diluting concentrated sulfuric acid.)
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177CHAPTER 7 | USING THE BRØNSTED–LOWRY THEORY
The molar concentration of some concentrated acids are shown in Table 7.3.1.
TABLE 7.3.1 Molar concentrations of some concentrated acids
Concentrated acid (% by mass) Formula Molarity ( mol L–1)
Ethanoic (acetic) acid (99.5%) CH3COOH 17
Hydrochloric acid (36%) HCl 12
Nitric acid (70%) HNO3 16
Phosphoric acid (85%) H3PO4 15
Sulfuric acid (98%) H2SO4 18
In the laboratory you can prepare solutions of a base of a required concentration by: • diluting a more concentrated solution, or • dissolving a weighed amount of the base in a measured volume of water, as
shown in Figure 7.3.2.
N g C F
T
(a) Accurately weigh out a mass of the base. (b) Transfer the base to a volumetric flask. (c) Ensure complete transfer of the base by washing with water. (d) Dissolve the base in water. (e) Add water to make the solution up to the calibration mark and shake thoroughly.
FIGURE 7.3.2 Preparing a solution by dissolving a weighed amount of base in a measured volume of water.
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CALCULATING MOLAR CONCENTRATION AFTER DILUTION
Calculate the molar concentration of hydrochloric acid when 10.0 mL of water is added to 5.0 mL of 1.2 mol L−1 HCl.
Thinking Working
The number of moles of solute does not change during a dilution.
So c1V1 = c2V2, where c is the concentration in mol L−1 and V is the volume of the solution. (Each of the volume units must be the same, although not necessarily litres.)
c1V1 = c2V2
Identify given values for concentrations and volumes before and after dilution.
Identify the unknown.
10.0 mL is added to 5.0 mL, so the final volume is 15.0 mL. (In practice, small volume changes can occur when solutions are mixed, however assume no volume changes occur.)
c1 = 1.2 mol L−1
V1 = 5.0 mL
V2 = 15.0 mL
You are required to calculate, c2, the concentration after dilution.
Transpose the equation and substitute the known values into the equation to find the required value.
= ×c c V V2
Worked example: Try yourself 7.3.1
CALCULATING MOLAR CONCENTRATION AFTER DILUTION
Calculate the molar concentration of nitric acid when 80.0 mL of water is added to 20.0 mL of 5.00 mol L−1 HNO3.
Worked example 7.3.2
CALCULATING THE VOLUME OF WATER TO BE ADDED IN A DILUTION
How much water must be added to 30.0 mL of 2.50 mol L−1 HCl to dilute the solution to 1.00 mol L−1?
Thinking Working
The number of moles of solute does not change during a dilution.
So c1V1 = c2V2, where c is the concentration in mol L−1 and V is the volume of the solution. (Each of the volume units must be the same, although not necessarily litres.)
c1V1 = c2V2
Identify given values for concentrations and volumes before and after dilution.
Identify the unknown.
V1 = 30.0 mL
c2 = 1.00 mol L−1
You are required to calculate V2, the volume of the diluted solution.
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179CHAPTER 7 | USING THE BRØNSTED–LOWRY THEORY
Transpose the equation and substitute the known values into the equation to find the required value.
= ×V c V c2
= 75.0 mL
The keyword in the question is added, therefore we calculate the volume of water to be added by finding the difference between the two volumes.
Volume of dilute solution = 75.0 mL
Initial volume of acid = 30.0 mL
So 75.0 − 30.0 = 45.0 mL of water must be added.
Worked example: Try yourself 7.3.2
CALCULATING THE VOLUME OF WATER TO BE ADDED IN A DILUTION
How much water must be added to 15.0 mL of 10.0 mol L−1 NaOH to dilute the solution to 2.00 mol L−1?
EFFECT OF DILUTION ON THE pH OF STRONG ACIDS AND BASES Consider a 0.1 mol L–1 solution of hydrochloric acid. Since pH = −log10[H3O
+], the pH of this solution is 1.0.
If this 1.0 mL solution is diluted by a factor of 10 to 10.0 mL by the addition of 9.0 mL of water, the concentration of H3O
+ ions decreases to 0.01 mol L–1 and the pH increases to 2.0. A further dilution by a factor of 10 to 100 mL will increase pH to 3.0. However, note that when acids are repeatedly diluted, the pH cannot increase above 7.
Similarly, the progressive dilution of a 0.10 mL NaOH solution will cause the pH to decrease until it reaches very close to 7.
You will now look at how to calculate the pH of solutions of strong acids and bases after dilution.
Worked example 7.3.3
CALCULATING THE pH OF A DILUTED ACID
5.0 mL of 0.010 mol L−1 HNO3 is diluted to 100.0 mL. Calculate the pH of the diluted solution.
Thinking Working
Identify given values for concentrations and volumes before and after dilution.
c1 = 0.010 mol L−1
V1 = 5.0 mL
V2 = 100.0 mL
+ after dilution, by transposing the formula:
c1V1 = c2V2
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CALCULATING THE PH OF A DILUTED ACID
10.0 mL of 0.1 mol L−1 HCl is diluted to 30.0 mL. Calculate the pH of the diluted solution.
Calculation of the pH of a base after dilution The following steps show the sequence used to calculate the pH of a base after it has been diluted. Remember that pH is a measure of the hydronium ion concentration. 1 Calculate [OH−] in the diluted solution. 2 Calculate pOH using pOH = –log10[OH–]. 3 Calculate the pH of the solution using pH = 14 – pOH.
Worked example 7.3.4
CALCULATING THE pH AND pOH OF A DILUTED BASE
10.0 mL of 0.1 mol L−1 NaOH is diluted to 100.0 mL. Calculate the pH and pOH of the diluted solution.
Thinking Working
Identify given values for concentrations and volumes before and after dilution.
c1 = 0.1 mol L−1
V1 = 10.0 mL
V2 = 100.0 mL
c2 = ?
Calculate c2, which is [OH−] after dilution, by transposing the formula:
c1V1 = c2V2
pOH = –log10[OH−]
pOH = –log10[OH−]
pH = 14 – pOH
pH = 14 − pOH
CALCULATING THE pH AND pOH OF A DILUTED BASE
15.0 mL of 0.02 mol L−1 KOH is diluted to 60.0 mL. Calculate the pH and pOH of the diluted solution.Pa ge
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CHEMISTRY IN ACTION PSC
Colourful chemistry The flower colour of the popular garden plant Hydrangea macrophylla depends on the pH of the soil (Figure 7.3.3). In acidic soil hydrangeas are blue, but in basic soil they are pink. You can make the soil more acidic with garden sulfur, which has the added benefit of preventing rot fungi from germinating when handling cuttings of succulents. More basic conditions can be achieved by adding garden lime, which is mainly calcium carbonate (CaCO3), a mild base. Garden lime has the added benefit of providing a source of calcium for plants. FIGURE 7.3.3 The different colours displayed by these hydrangeas is
caused by the acidity or basicity of the soil.
7.3 Review SUMMARY
• The amount of an acid or base in solution does not change during a dilution. The volume of the solution increases and the concentration decreases.
• Solutions of acids or bases of a required concentration can be prepared by diluting more concentrated solutions, using the formula:
c1V1 = c2V2
where c1 and V1 are the initial volume and concentration, and c2 and V2 are the final volume and concentration after dilution.
• The pH increases when a solution of an acid is diluted.
• The pH of a diluted acid can be determined by calculating the concentration of hydronium ions in the diluted solution.
• The pH decreases when a solution of a base is diluted. The pH of a diluted base can be determined by:
- calculating the concentration of hydroxide ions in the diluted base, and
- using the ionic product of water to calculate the concentration of hydronium ions in the diluted base, or alternatively
- calculating pOH based on the concentration of hydroxide ions in the diluted base, then converting to pH using the expression:
pH = 14 – pOH
KEY QUESTIONS
1 1.0 L of water is added to 3.0 L of 0.10 mol L−1 HCl. What is the concentration of the diluted acid?
2 How much water must be added to 10 mL of a 2.0 mol L−1 sulfuric acid solution to dilute it to 0.50 mol L−1?
3 What volume of water must be added to dilute a 20.0 mL volume of 0.600 mol L−1 HCl to 0.100 mol L−1?
4 Describe the effect on the pH of a monoprotic acid solution of pH 1.0 when it is diluted by a factor of 10.
5 Calculate the pH and pOH of the solution at 25°C that is formed by the dilution of a 20.0 mL solution of 0.100 mol L−1 NaOH to 50.0 mL.
6 For each of the solutions a−e (all at 25°C), calculate: i the concentration of H3O
+ ions ii the concentration of OH− ions iii the pH iv the pOH. a 0.001 mol L−1 HNO3(aq) b 0.03 mol L−1 HCl(aq) c 0.01 mol L−1 NaOH(aq) d 10−4.5 mol L−1 HCl(aq) e 0.005 mol L−1 Ba(OH)2(aq).
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acidic solution amphiprotic basic solution concentrated solution concentration dilution dissociation constant
indicator ionic product of water molarity neutral solution p-function pH scale pOH scale
self-ionisation strong acid strong base super acid weak acid weak base
REVIEW QUESTIONS
1 Write an equation to show that perchloric acid (HClO4) acts as a strong acid in water.
2 Write an equation to show that hypochlorous acid (HClO3) acts as a weak acid in water.
3 Write an equation to show that ammonia (NH3) acts as a weak base in water.
4 Write a balanced ionic equation showing the H2PO4 −
ion acting as a weak base in water.
5 Write an equation to show that dimethylamine ((CH3)2NH) acts as a weak base in water.
6 a Write an equation showing the dissociation of calcium hydroxide (Ca(OH)2) in water.
b Explain why calcium hydroxide acts as a strong base.
7 Calculate [OH−] at 25°C in aqueous solutions with [H3O
+] equal to: a 0.001 mol L−1
b 10−5 mol L−1
c 5.7 × 10−9 mol L−1
d 3.4 × 10−12 mol L−1
e 6.5 × 10−2 mol L−1
f 2.23 × 10−13 mol L−1
8 Calculate the pOH from the [OH−] values calculated in Question 7.
9 State the concentration of the following ions in solutions at 25°C, with the given pH values. a hydronium ions b hydroxide ions. i pH 1 ii pH 3 iii pH 7 iv pH 11.7
10 The pH of human blood is tightly regulated so that it is between 7.35 and 7.45 at 37°C. What are the minimum and maximum concentrations of hydronium ions in blood within this pH range?
11 The pH of a cola drink is 3 and of black coffee is 5. How many more times acidic is the cola than black coffee?
12 Calculate the concentration of H3O + and OH− ions in
solutions with the following pH values at 25°C: a 3.0 b 10.0 c 8.5 d 5.8 e 9.6 f 13.5
13 The pH of a particular brand of tomato juice is 5.3 at 25°C. What is the concentration of hydroxide ions in this tomato juice?
14 A solution of hydrochloric acid has a pH of 2. a What is the molar concentration of hydrogen ions in
this solution? b What amount of hydrogen ions, in mol, would be
present in 500 mL of this solution? c What is the pOH of the hydrochloric acid solution?
15 Calculate the pH and pOH of each of the following mixtures at 25°C. a 10 mL of 0.025 mol L−1 HCl is diluted to 50 mL of
solution. b 20 mL of 0.0050 mol L−1 KOH is diluted to 500 mL of
solution. c 10 mL of 0.15 mol L−1 HCl is diluted to 1.5 L of
solution.
16 The molarity of concentrated sulfuric acid is 18.0 mol L−1. What volume of concentrated sulfuric acid is required to prepare 1.00 L of 2.00 mol L−1 H2SO4 solution?
17 When a 10.0 mL solution of hydrochloric acid was diluted, the pH changed from 2.00 to 4.00. What volume of water was added to the acid solution?
18 40.0 mL of 0.10 mol L−1 HNO3 is diluted to 500.0 mL. Will the pH increase or decrease?
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183CHAPTER 7 | USING THE BRØNSTED–LOWRY THEORY
19 A laboratory assistant forgot to label 0.10 mol L−1 solutions of sodium hydroxide (NaOH), hydrochloric acid (HCl), glucose (C6H12O6), ammonia (NH3) and ethanoic acid (CH3COOH). In order to identify them, temporary labels A−E were placed on the bottles and the electrical conductivity and pH of each solution was measured. The results are shown in the table below. Identify each solution and briefly explain your reasoning.
Solution Electrical conductivity pH
A poor 11
B zero 7
C good 13
D good 1
E poor 3
20 A standard laboratory usually dilutes concentrated stock solutions of acids or bases to make less concentrated solutions for use. a Write an equation to represent the dissociation of
sodium hydroxide in water. b Is sodium hydroxide regarded as a strong or weak
base? Explain your response.
c What mass of sodium hydroxide (NaOH) pellets would be needed to be added to a 500 mL volumetric flask in order to produce a 12 mol L−1 stock solution?
d i Calculate the volume of the stock solution required to make a 250.0 mL solution of 0.020 mol L−1 solution of NaOH.
ii Suggest a reason why it would not be suitable to make this dilution directly, and suggest an alternative method.
e What is the resultant pOH of the diluted solution? f Hence calculate the pH of the diluted solution and
the concentration of hydronium ions.
21 Reflect on the Inquiry task on page 000. You have been given two solutions, one contains a strong acid and the other contains a weak acid. Both acids are dilute and have the same concentration. How could you use the red cabbage indicator to tell the two solutions apart?
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