Models of Acids and Bases

Models of Acids and Bases
Arrhenius: Acids produce H+ ions in solution, bases produce OH- ions. Brønsted–Lowry: Acids are proton (H+) donors, bases are proton acceptors. HCl + H2O Cl- + H3O+ acid base Copyright © Cengage Learning. All rights reserved

Brønsted–Lowry Reaction
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Acid in Water HA(aq) H2O(l) H3O+(aq) A-(aq) Conjugate base is everything that remains of the acid molecule after a proton is lost. Conjugate acid is formed when the proton is transferred to the base. Copyright © Cengage Learning. All rights reserved

Acid in Water - General Reaction (Continued)
Equilibrium expression Ka is the acid dissociation constant Water can be omitted from the acid dissociation reaction Does not affect the equilibrium position

Interactive Example 14.1 - Acid Dissociation (Ionization) Reactions
Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids: Hydrochloric acid (HCl) Acetic acid (HC2H3O2) The anilinium ion (C6H5NH3+) The hydrated aluminum(III) ion [Al(H2O)6]3+

Interactive Example 14.1 - Solution

Interactive Example 14.1 - Solution (Continued)
Although this formula looks complicated, writing the reaction is simple if you concentrate on the meaning of Ka Removing a proton, which can come only from one of the water molecules, leaves one OH– and five H2O molecules attached to the Al3+ ion So the reaction is

Acid Ionization Equilibrium

Defined by the equilibrium position of the acid’s ionization reaction
Strength of an Acid Defined by the equilibrium position of the acid’s ionization reaction Strong acid: Equilibrium lies far to the right Yields a weak conjugate base Weak acid: Equilibrium lies far to the left Weaker the acid, stronger its conjugate base Acid Conjugate base Copyright © Cengage Learning. All rights reserved

Various Ways to Describe Acid Strength
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Diprotic acids: Composed of two acidic protons
Types of Acids Diprotic acids: Composed of two acidic protons Completely dissociate in water Example - Sulfuric acid [H2SO4 (aq)] Oxyacids: Acidic proton is attached to an oxygen atom Examples - Nitric acid [HNO3(aq)] and phosphoric acid (H3PO4)

Types of Acids (Continued 1)
Organic acids: Contain a carbon atom backbone Contain the carboxyl group Generally weak in nature Examples - Acetic acid (CH3COOH) and benzoic acid (C6H5COOH)

Types of Acids (Continued 2)
Hydrohalic acids (HX) Acidic proton is attached to an atom other than oxygen X - Halogen atom Monoprotic acids: Contain one acidic proton Examples - Hydrogen sulfate ion (HSO4–) and phenol (HOC6H5)

Table 14.2 - Values of Ka for Some Common Monoprotic Acids

Autoionization Reaction for Water
Equilibrium expression Ion-product constant (Kw) (or the dissociation constant for water) always refers to the autoionization of water Copyright © Cengage Learning. All rights reserved

Ion-Product Constant (Kw)
No matter what the solution contains, the product of [H+] and [OH–] must always equal 1.0×10–14 at 25°C Temperature dependent Copyright © Cengage Learning. All rights reserved

Three Possible Situations
[H+] = [OH–]; neutral solution [H+] > [OH–]; acidic solution [OH–] > [H+]; basic solution Copyright © Cengage Learning. All rights reserved

Self-Ionization of Water

Water as an Acid and a Base
Autoionization: Transfer of a proton from one water molecule to another to produce a hydroxide ion and a hydronium ion One water molecule acts as an acid by furnishing a proton, and the other acts as a base by accepting the proton Copyright © Cengage Learning. All rights reserved

HA(aq) + H2O(l) H3O+(aq) + A-(aq)
CONCEPT CHECK! HA(aq) + H2O(l) H3O+(aq) A-(aq) acid base conjugate conjugate acid base What is the equilibrium constant expression for an acid acting in water? Use the following equation to write K: HA(aq) + H2O  H3O+(aq) + A-(aq) Although you can also write this as HA(aq)  H+(aq) + A-(aq), have the students use the form above until they are used to the fact that water is acting like a base. K = [H3O+][A-] / HA Copyright © Cengage Learning. All rights reserved

Example 14.4 - Autoionization of Water
At 60°C, the value of Kw is 1×10–13 Using Le Châtelier’s principle, predict whether the following reaction is exothermic or endothermic: Calculate [H+] and [OH–] in a neutral solution at 60°C

Kw increases from 1×10–14 at 25°C to 1×10–13 at 60°C
Example Solution (a) Kw increases from 1×10–14 at 25°C to 1×10–13 at 60°C Le Châtelier’s principle states that if a system at equilibrium is heated, it will adjust to consume energy Since the value of Kw increases with temperature, we must think of energy as a reactant, and so the process must be endothermic

Example Solution (b) At 60°C, For a neutral solution,

CONCEPT CHECK! If the equilibrium lies to the right, the value for Ka is __________. large (or >1) If the equilibrium lies to the left, the value for Ka is ___________. small (or <1) Large (or >1); small (or < 1) The students should know this from Chapter 13, but it is worth re-emphasizing. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! HA(aq) + H2O(l) H3O+(aq) + A–(aq) If water is a better base than A–, do products or reactants dominate at equilibrium? Does this mean HA is a strong or weak acid? Is the value for Ka greater or less than 1? products; strong; greater than 1 One of the advantages in having students write out the weak acid reaction with water is that we can make comparisons between the relative strengths of the conjugate base and water acting as a base. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Consider a 1.0 M solution of HCl. Order the following from strongest to weakest base and explain: H2O(l) A–(aq) (from weak acid HA) Cl–(aq) The bases from strongest to weakest are: A-, H2O, Cl-. Some students will try to claim that the Kb value for water is 1.0 x (or some will say 1.0 x 10-7) if they’ve read ahead in the chapter. Water does not have a Kb value; we must use reactions (like in the first Concept Check problem) to think about relative base strengths. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… How good is Cl–(aq) as a base?
Is A–(aq) a good base? The bases from strongest to weakest are: A–, H2O, Cl– Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Acetic acid (HC2H3O2) and HCN are both weak acids. Acetic acid is a stronger acid than HCN. Arrange these bases from weakest to strongest and explain your answer: H2O Cl– CN– C2H3O2– Copyright © Cengage Learning. All rights reserved

Let’s Think About It… H2O(l) + H2O(l) H3O+(aq) + OH–(aq) acid base conjugate conjugate acid base At 25°C, Kw = 1.0 × 10–14 The bases from weakest to strongest are: Cl–, H2O, C2H3O2–, CN– Weakest to strongest: Cl-, H2O, C2H3O2-, CN-. If HCN is weaker than HC2H3O2, the conjugate base will be stronger. Have the students see this by looking at the general equation HA(aq) + H2O  H3O+(aq) + A-(aq) They can also use this to see that water is a better base than the chloride ion. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Discuss whether the value of K for the reaction: HCN(aq) + F–(aq) CN–(aq) + HF(aq) is >1 <1 =1 (Ka for HCN is 6.2×10–10; Ka for HF is 7.2×10–4.) Explain your answer. The value for K is less than 1. This reaction is neither a Kb nor a Ka reaction. To think about it, we need to know which is the stronger acid, HCN or HF. This can be decided by knowing the Ka values. HF is stronger, which means that it is more likely to donate a proton. Also, F- must be a weaker base than CN-. Thus, CN- has a greater affinity for a proton than F-. Because of all of this, equilibrium lies on the left, so K is less than 1. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Calculate the value for K for the reaction: HCN(aq) + F–(aq) CN–(aq) + HF(aq) (Ka for HCN is 6.2×10–10; Ka for HF is 7.2×10–4.) K = 8.6 × 10–7 K = 8.6 x 10-7 K = Ka(HCN)/Ka(HF) = [6.2 x 10-10]/[7.2 x 10-4] The students should know how to manipulate the Ka expressions to derive this. Copyright © Cengage Learning. All rights reserved

pH changes by 1 for every power of 10 change in [H+].
pH = –log[H+] pH changes by 1 for every power of 10 change in [H+]. A compact way to represent solution acidity. pH decreases as [H+] increases. Copyright © Cengage Learning. All rights reserved

Significant Figures Number of decimal places in the log is equal to the number of significant figures in the original number [H+] = 1.0×10–9 M pH = 9.00 2 significant figures 2 decimal places Copyright © Cengage Learning. All rights reserved

pH Range pH = 7; neutral pH > 7; basic Higher the pH, more basic.
pH < 7; acidic Lower the pH, more acidic. Copyright © Cengage Learning. All rights reserved

pH is a log scale based on 10
Log Scales pOH = –log [OH–] pK = –log K pH is a log scale based on 10 pH changes by 1 for every power of 10 change in [H+] Copyright © Cengage Learning. All rights reserved

Calculate the pH for each of the following solutions. 1.0 × 10–4 M H+
EXERCISE! Calculate the pH for each of the following solutions. 1.0 × 10–4 M H+ pH = 4.00 b) M OH– pH = 12.60 pH = –log[H+] a) pH = –log[H+] = –log(1.0 × 10–4 M) = 4.00 b) Kw = [H+][OH–] = 1.00 × 10–14 = [H+](0.040 M) = 2.5 × 10–13 M H+ pH = –log[H+] = –log(2.5 × 10–13 M) = 12.60 Copyright © Cengage Learning. All rights reserved

pH and pOH Recall: Kw = [H+][OH–] –log Kw = –log[H+] – log[OH–]
pKw = pH + pOH 14.00 = pH + pOH Copyright © Cengage Learning. All rights reserved

Calculate the pOH for each of the following solutions. 1.0 × 10–4 M H+
EXERCISE! Calculate the pOH for each of the following solutions. 1.0 × 10–4 M H+ pOH = 10.00 b) M OH– pOH = 1.40 pH = –log[H+] and pOH = –log[OH–] and = pH + pOH a) pH = –log[H+] = –log(1.0 × 10–4 M) = 4.00; So, = pOH; pOH = 10.00 b) pOH = –log[OH–] = –log(0.040 M) = 1.40 Copyright © Cengage Learning. All rights reserved

EXERCISE! The pH of a solution is What is the [OH–] for this solution? [OH–] = 7.1 × 10–9 M pH = –log[H+] and pOH = –log[OH–] and = pH + pOH 14.00 = pOH; pOH = 8.15 [OH–] = 10^–8.15 = 7.1 × 10–9 M Copyright © Cengage Learning. All rights reserved

Interactive Example 14.6 - Calculations Using pH
The pH of a sample of human blood was measured to be 7.41 at 25°C Calculate pOH, [H+], and [OH–] for the sample

Since pH + pOH = 14.00, pOH = – pH = – 7.41 = 6.59 To find [H+] we must go back to the definition of pH, which is pH = – log[H+] Thus, 7.41 = – log[H+] or log[H+] = – 7.41

Interactive Example 14.6 - Solution (Continued 1)
We need to know the antilog of –7.41 Taking the antilog is the same as exponentiation antilog (n) = 10n Since pH = –log[H+], –pH = log[H+] [H+] can be calculated by taking the antilog of –pH [H+] = antilog (–pH)

In the present case, [H+] = antilog (–pH) = antilog (–7.41) = 10–7.41 [H+] = 3.9×10–8 M Similarly, [OH–] = antilog (–pOH), And [OH–] = antilog (–6.59) = 10–6.59 = 2.6×10–7 M

Fill in the missing information in the following table:
Exercise Fill in the missing information in the following table: 7.12 1.3 × 10–7 M 7.6 × 10–8 M acidic 0.92 13.08 0.12 M acidic 10.89 1.3 × 10–11 M 7.8 × 10–4 M basic 7.00 7.00 1.0 × 10–7 M neutral Copyright © Cengage Learning. All rights reserved

Focus on the solution components and their chemistry
Solving for pH Focus on the solution components and their chemistry Determine the significant components List the major species, and focus on those that furnish H+ ions Major species: Solution components that are available in large amounts Copyright © Cengage Learning. All rights reserved

Interactive Example 14.7 - pH of Strong Acids
Calculate the pH of 0.10 M HNO3 Calculate the pH of 1.0×10–10 M HCl

Interactive Example 14.7 - Solution (a)
Since HNO3 is a strong acid, the major species in solution are H+, NO3–, and H2O The concentration of HNO3 is virtually zero, since the acid completely dissociates in water Also, [OH–] will be very small because the H+ ions from the acid will drive the equilibrium to the left

Interactive Example 14.7 - Solution (a) (Continued 1)
This is an acidic solution where [H+] >> [OH–], so [OH–] << 10–7 M and the sources of H+ are: H+ from HNO3 (0.10 M) H+ from H2O The number of H+ ions contributed by the autoionization of water will be very small compared with the 0.10 M contributed by the HNO3 and can be neglected

Interactive Example 14.7 - Solution (a) (Continued 2)
Since the dissolved HNO3 is the only important source of H+ ions in this solution, [H+] = 0.10 M and pH = –log(0.10) = 1.00

Interactive Example 14.7 - Solution (b)
Normally, in an aqueous solution of HCl the major species are H+, Cl–, and H2O However, in this case the amount of HCl in solution is so small that it has no effect The only major species is H2O Thus, the pH will be that of pure water, or pH = 7.00

Problem-Solving Strategy - Solving Weak Acid Equilibrium Problems
List the major species in the solution Choose the species that can produce H+ Write balanced equations for the reactions producing H+ Using the values of the equilibrium constants for the reactions, decide which equilibrium will dominate in producing H+ Copyright © Cengage Learning. All rights reserved

Write the equilibrium expression for the dominant equilibrium
Problem-Solving Strategy - Solving Weak Acid Equilibrium Problems (Continued 1) Write the equilibrium expression for the dominant equilibrium List the initial concentrations of the species participating in the dominant equilibrium Define the change needed to achieve equilibrium Define x Write the equilibrium concentrations in terms of x Copyright © Cengage Learning. All rights reserved

Use the 5% rule to verify whether the approximation is valid
Problem-Solving Strategy - Solving Weak Acid Equilibrium Problems (Continued 2) Substitute the equilibrium concentrations into the equilibrium expression Solve for x the easy way Assume that [HA]0 – x ≈ [HA]0 Use the 5% rule to verify whether the approximation is valid Calculate [H+] and pH Copyright © Cengage Learning. All rights reserved

Thinking About Acid–Base Problems
What are the major species in solution? What is the dominant reaction that will take place? Is it an equilibrium reaction or a reaction that will go essentially to completion? React all major species until you are left with an equilibrium reaction. Solve for the pH if needed. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Consider an aqueous solution of 2.0 × 10–3 M HCl. What are the major species in solution? H+, Cl–, H2O What is the pH? pH = 2.70 Major Species: H+, Cl-, H2O pH = –log[H+] = –log(2.0 × 10–3 M) = 2.70 pH = 2.70 Copyright © Cengage Learning. All rights reserved

Calculate the pH of a 1.5 × 10–11 M solution of HCl. pH = 7.00
CONCEPT CHECK! Calculate the pH of a 1.5 × 10–11 M solution of HCl. pH = 7.00 pH = 7.00 Watch for student to say pH = -log(1.5 x 10-11) = This answers neglects that water is a major species and it turns out to be the major contributor of H+. If students think about this problem and consider major species, they will answer this correctly. Major species are H+, Cl-, and H2O. Dominant reaction is: H2O + H2O  H3O+ (aq) + OH- (aq) Water is a major contributor. Thus , pH = 7.00. Copyright © Cengage Learning. All rights reserved

Calculate the pH of a 1.5 × 10–2 M solution of HNO3.
CONCEPT CHECK! Calculate the pH of a 1.5 × 10–2 M solution of HNO3. Major Species: H+, NO3-, H2O The reaction controlling the pH is: H2O + H2O  H3O+(aq) + OH-(aq) This is because it is the only equilibrium reaction (NO3- is not a base in water). However, the major source for H+ is from the nitric acid, so the pH = -log(1.5 x 10-2) = 1.82. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… Why is this reaction not likely?
NO3–(aq) + H2O(l) HNO3(aq) + OH–(aq) Because the K value for HNO3 is very large and HNO3 virtually completely dissociates in solution. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… What reaction controls the pH?
H2O(l) + H2O(l) H3O+(aq) + OH–(aq) In aqueous solutions, this reaction is always taking place. But is water the major contributor of H+ (H3O+)? pH = 1.82 Copyright © Cengage Learning. All rights reserved

Solving Weak Acid Equilibrium Problems
List the major species in the solution. Choose the species that can produce H+, and write balanced equations for the reactions producing H+. Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing H+. Write the equilibrium expression for the dominant equilibrium. Copyright © Cengage Learning. All rights reserved

List the initial concentrations of the species participating in the dominant equilibrium. Define the change needed to achieve equilibrium; that is, define x. Write the equilibrium concentrations in terms of x. Substitute the equilibrium concentrations into the equilibrium expression. Copyright © Cengage Learning. All rights reserved

Solve for x the “easy” way, that is, by assuming that [HA]0 – x about equals [HA]0. Use the 5% rule to verify whether the approximation is valid. Calculate [H+] and pH. Copyright © Cengage Learning. All rights reserved

Interactive Example 14.8 - The pH of Weak Acids
The hypochlorite ion (OCl–) is a strong oxidizing agent often found in household bleaches and disinfectants It is also the active ingredient that forms when swimming-pool water is treated with chlorine In addition to its oxidizing abilities, the hypochlorite ion has a relatively high affinity for protons (it is a much stronger base than Cl–, for example)

Interactive Example 14.8 - The pH of Weak Acids (Continued)
It also forms the weakly acidic hypochlorous acid (HOCl, Ka = 3.5×10–8) Calculate the pH of a M aqueous solution of hypochlorous acid

We list the major species Since HOCl is a weak acid and remains mostly undissociated, the major species in a M HOCl solution are HOCl and H2O Both species can produce H+

Since HOCl is a significantly stronger acid than H2O, it will dominate in the production of H+ We therefore use the following equilibrium expression:

The initial concentrations appropriate for this equilibrium are [HOCl]0 = M [OCl–]0 = 0 [H+]0 ≈ 0 (We neglect the contribution from H2O)

Since the system will reach equilibrium by the dissociation of HOCl, let x be the amount of HOCl (in mol/L) that dissociates in reaching equilibrium The equilibrium concentrations in terms of x are [HOCl] = [HOCl]0 – x = – x [OCl–] = [OCl–]0 + x = 0 + x = x [H+] = [H+]0 + x ≈ 0 + x = x

Substituting these concentrations into the equilibrium expression gives Since Ka is so small, we can expect a small value for x Thus, we make the approximation [HA]0 – x ≈[HA]0, or – x ≈ 0.100

This leads to the following expression: Solving for x gives x = 5.9×10–5 The approximation – x ≈ must be validated

To do this, we compare x to [HOCl]0: Since this value is much less than 5%, the approximation is considered valid We calculate [H+] and pH [H+] = x =5.9×10–5 M and pH = 4.23

What are the major species in solution? HCN, H2O
CONCEPT CHECK! Consider a 0.80 M aqueous solution of the weak acid HCN (Ka = 6.2 × 10–10). What are the major species in solution? HCN, H2O Major Species: HCN, H2O. This is because HCN is a weak acid. Thus, while H+ and CN- are in solution, the concentrations are very small. Primary reaction: HCN(aq) + H2O  H3O+(aq) + CN-(aq) The primary reaction is chosen because the K value is significantly larger. Thus, it is the primary source of H3O+ (or H+). Copyright © Cengage Learning. All rights reserved

Let’s Think About It… Why aren’t H+ or CN– major species?
See notes on slide 41. Copyright © Cengage Learning. All rights reserved

Consider This HCN(aq) + H2O(l) H3O+(aq) + CN–(aq) Ka = 6.2 × 10-10
H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Kw = 1.0 × 10-14 Which reaction controls the pH? Explain. See notes on slide 1. Copyright © Cengage Learning. All rights reserved

Calculate the pH of a 0.50 M aqueous solution of the weak acid HF.
EXERCISE! Calculate the pH of a 0.50 M aqueous solution of the weak acid HF. (Ka = 7.2 × 10–4) Major Species: HF, H2O Possibilities for primary reaction: HF(aq) + H2O  H3O+(aq) + F-(aq) H2O + H2O  H3O+(aq) + OH-(aq) The first reaction is the primary reaction because the K value is significantly larger. Thus, it is the primary source of H3O+ (or H+). Use an ICE table, K expression, and pH = –log[H+] to solve for pH. pH = 1.72 Copyright © Cengage Learning. All rights reserved

Let’s Think About It… What are the major species in solution? HF, H2O
Why aren’t H+ and F– major species? See Slide 4. Copyright © Cengage Learning. All rights reserved

HF(aq) + H2O(l) H3O+(aq) + F–(aq)
Let’s Think About It… What are the possibilities for the dominant reaction? HF(aq) + H2O(l) H3O+(aq) + F–(aq) Ka=7.2 × 10-4 H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Kw=1.0 × 10-14 Which reaction controls the pH? Why? See Slide 4. Copyright © Cengage Learning. All rights reserved

Steps Toward Solving for pH
HF(aq) + H2O H3O+(aq) + F–(aq) Initial 0.50 M ~ 0 Change –x +x Equilibrium 0.50–x x See Slide 4. Ka = 7.2 × 10–4 pH = 1.72 Copyright © Cengage Learning. All rights reserved

Percent Dissociation (Ionization)
For a given weak acid, the percent dissociation increases as the acid becomes more dilute. Copyright © Cengage Learning. All rights reserved

A solution of 8.00 M formic acid (HCHO2) is 0.47% ionized in water.
EXERCISE! A solution of 8.00 M formic acid (HCHO2) is 0.47% ionized in water. Calculate the Ka value for formic acid. Ka = 1.8 × 10–4 Ka = 1.8 x 10-4 If 8.00 M of the acid is 0.47% ionized, then M dissociates. HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq) I C E Copyright © Cengage Learning. All rights reserved

EXERCISE! Calculate the pH of an 8.00 M solution of formic acid. Use the data from the previous slide to help you solve this problem. pH = 1.42 pH = –log[H+] = –log[0.038 M] = 1.42 Major species: HCHO2, H2O Dominant reaction: HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq) Copyright © Cengage Learning. All rights reserved

EXERCISE! The value of Ka for a 4.00 M formic acid solution should be: higher than lower than the same as the value of Ka of an 8.00 M formic acid solution. Explain. the same; Students should understand that Ka is not a function of initial concentration. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! The percent ionization of a 4.00 M formic acid solution should be: higher than lower than the same as the percent ionization of an 8.00 M formic acid solution. Explain. higher; To make 4.00 M solution from 8.00 M solution, we must add water (dilute). With a larger volume, Le Chatelier's principle tells us that equilibrium will shift to the right. Thus, percent ionization will increase. However, this increase if offset by the dilution factor, and the solution has an overall lower concentration of H+. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! The pH of a 4.00 M formic acid solution should be: higher than lower than the same as the pH of an 8.00 M formic acid solution. Explain. higher; They should also understand that a lower concentration of the same weak acid will be less acidic (and thus the pH will be higher). Copyright © Cengage Learning. All rights reserved

EXERCISE! Calculate the percent ionization of a 4.00 M formic acid solution in water. % Ionization = 0.67% % dissociation = 0.67% Major species: HCHO2, H2O Dominant reaction: HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq) HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq) I C -x x x E x x x Ka = 1.8 x 10-4 = (x)2/(4.00-x) x = 0.027; (0.027/4.00) x 100 = 0.67% Copyright © Cengage Learning. All rights reserved

Calculate the pH of a 4.00 M solution of formic acid.
EXERCISE! Calculate the pH of a 4.00 M solution of formic acid. pH = 1.57 pH = 1.57 Major species: HCHO2, H2O Dominant reaction: HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq) From preview exercise, x = [H3O+] = [H+] = 0.027; pH = 1.57 Copyright © Cengage Learning. All rights reserved

Interactive Example 14.9 - The pH of Weak Acid Mixtures
Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2×10–10) and 5.00 M HNO2 (Ka = 4.0×10–4) Also calculate the concentration of cyanide ion (CN–) in this solution at equilibrium

Since HCN and HNO2 are both weak acids and are largely undissociated, the major species in the solution are HCN, HNO2, and H2O All three of these components produce H+

A mixture of three acids might lead to a very complicated problem However, the situation is greatly simplified by the fact that even though HNO2 is a weak acid, it is much stronger than the other two acids present (as revealed by the K values) Thus, HNO2 can be assumed to be the dominant producer of H+, and we will focus on its equilibrium expression

The initial concentrations, the definition of x, and the equilibrium concentrations are as follows:

It is convenient to represent these concentrations in the following shorthand form (called an ICE table): Substitute the equilibrium concentrations in the equilibrium expression

Make the approximation that 5.00 – x = 5.00 We solve for x: x = 4.5×10–2 Using the 5% rule, we show that the approximation is valid

Therefore, [H+] = x = 4.5×10–2 M and pH = 1.35 We also want to calculate the equilibrium concentration of cyanide ion in this solution The CN– ions in this solution come from the dissociation of HCN

Although the position of this equilibrium lies far to the left and does not contribute significantly to [H+], HCN is the only source of CN– Thus, we must consider the extent of the dissociation of HCN to calculate [CN–] The equilibrium expression for the preceding reaction is

We know [H+] for this solution from the results of the first part of the problem It is important to understand that there is only one kind of H+ in this solution It does not matter from which acid the H+ ions originate The equilibrium [H+] we need to insert into the HCN equilibrium expression is 4.5×10–2 M, even though the H+ was contributed almost entirely from the dissociation of HNO2

What is [HCN] at equilibrium? We know [HCN]0 = 1.00 M, and since Ka for HCN is so small, a negligible amount of HCN will dissociate Thus, [HCN] = [HCN]0 – amount of HCN dissociated ≈ [HCN]0 = 1.00 M

Since [H+] and [HCN] are known, we can find [CN–] from the equilibrium expression:

Note the significance of this result Since [CN–] = 1.4×10–8 M and HCN is the only source of CN–, this means that only 1.4×10–8 mol/L of HCN dissociated This is a very small amount compared with the initial concentration of HCN, which is exactly what we would expect from its very small Ka value, and [HCN] = 1.00 M as assumed

For a solution of any weak acid HA:
Percent Dissociation Definition For a solution of any weak acid HA: [H+] decreases as [HA]0 decreases Percent dissociation increases as [HA]0 decreases Copyright © Cengage Learning. All rights reserved

Interactive Example 14.10 - Calculating Percent Dissociation
Calculate the percent dissociation of acetic acid (Ka = 1.8×10–5) in 1.00 M of HC2H3O2

Since acetic acid is a weak acid, the major species in this solution are HC2H3O2 and H2O Both species are weak acids, but acetic acid is a much stronger acid than water Thus, the dominant equilibrium will be The equilibrium expression is

The initial concentrations, definition of x, and equilibrium concentrations are: Insert the equilibrium concentrations into the equilibrium expression and make the usual approximation that x is small compared with [HA]0

Thus, x2 ≈ 1.8×10–5 and x ≈ 4.2×10–3 The approximation 1.00 – x ≈ 1.00 is valid by the 5% rule, so [H+] = x = 4.2×10–3 M

The percent dissociation is

Exercise Use Le Châtelier’s principle to explain why percent dissociation increases as the concentration of a weak acid decreases Dilution shifts equilibrium to the side with the greater number of particles (% dissociation increases) Copyright © Cengage Learning. All rights reserved

Interactive Example 14.11 - Calculating Ka from Percent Dissociation
Lactic acid (HC3H5O3) is a chemical that accumulates in muscle tissue during exertion In a M aqueous solution, lactic acid is 3.7% dissociated Calculate the value of Ka for this acid

From the small value for the percent dissociation, it is clear that HC3H5O3 is a weak acid Thus, the major species in the solution are the undissociated acid and water HC3H5O3 and H2O However, even though HC3H5O3 is a weak acid, it is a much stronger acid than water and will be the dominant source of H+ in the solution

The dissociation reaction is The equilibrium expression is

The initial and equilibrium concentrations are as follows: The change needed to reach equilibrium can be obtained from the equation of percent dissociation

For this acid,

Now we can calculate the equilibrium concentrations: [HC3H5O3] = 0.10 – x = 0.10 M (to the correct number of significant figures) [C3H5O3–] = [H+] = x = 3.7×10–3 M These concentrations can now be used to calculate the value of Ka for lactic acid

Arrhenius: bases produce OH– ions.
Brønsted–Lowry: bases are proton acceptors. Have at least one unshared pair of electrons that is capable of forming a bond with a proton In a basic solution at 25°C, pH > 7. Ionic compounds containing OH- are generally considered strong bases (completely dissociate in water). LiOH, NaOH, KOH, Ca(OH)2 – What do these have in common? pOH = –log[OH–] pH = – pOH Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Calculate the pH of a 1.0 × 10–3 M solution of calcium hydroxide. pH = 11.30 pH = 11.30 Students need to see that the concentration of the hydroxide ion is twice that of the calcium hydroxide. Copyright © Cengage Learning. All rights reserved

CN–(aq) + H2O(l) HCN(aq) + OH–(aq)
Equilibrium expression for weak bases uses Kb. Kb will be very small for weak bases. CN–(aq) + H2O(l) HCN(aq) + OH–(aq) Copyright © Cengage Learning. All rights reserved

pH calculations for solutions of weak bases are very similar to those for weak acids.
Kw = [H+][OH–] = 1.0 × 10–14 pOH = –log[OH–] pH = – pOH Copyright © Cengage Learning. All rights reserved

Interactive Example 14.14 - The pH of Weak Bases II
Calculate the pH of a 1.0-M solution of methylamine (Kb = 4.38×10–4)

Since methylamine (CH3NH2) is a weak base, the major species in solution are CH3NH2 and H2O Both are bases; however, water can be neglected as a source of OH–, so the dominant equilibrium is

The ICE table is: Substitute the equilibrium concentrations in the equilibrium expression and make the usual approximation

The approximation is valid by the 5% rule, so [OH–] = x =2.1×10–2 M pOH = 1.68 pH = – 1.68 = 12.32

Acids that can furnish more than one proton.
Always dissociates in a stepwise manner, one proton at a time. The conjugate base of the first dissociation equilibrium becomes the acid in the second step. For a typical weak polyprotic acid: Ka1 > Ka2 > Ka3 For a typical polyprotic acid in water, only the first dissociation step is important to pH. Copyright © Cengage Learning. All rights reserved

Which proton is most readily lost?

Interactive Example 14.15 - The pH of a Polyprotic Acid
Calculate the pH of a 5.0-M H3PO4 solution and the equilibrium concentrations of the species H3PO4, H2PO4–, HPO42–, and PO43–

The major species in solution are H3PO4 and H2O None of the dissociation products of H3PO4 is written, since the Ka values are all so small that they will be minor species The dominant equilibrium is the dissociation of H3PO4

The ICE table is: Substitute the equilibrium concentrations into the expression for Ka1 and make the usual approximation

Thus, x ≈ 1.9×10–1 Since 1.9×10–1 is less than 5% of 5.0, the approximation is acceptable [H+] = x = 0.19 M pH = 0.72

So far we have determined that [H+] = [H2PO4–] = 0.19 M [H3PO4] = 5.0 – x = 4.8 M The concentration of HPO42– can be obtained by using the expression for Ka2

Thus, [HPO42–] = Ka2 = 6.2×10–8 M To calculate [PO43–], we use the expression for Ka3 and the values of [H+] and [HPO42–] calculated previously:

These results show that the second and third dissociation steps do not make an important contribution to [H+] This is apparent from the fact that [HPO42–] is 6.2×10–8 M, which means that only 6.2×10–8 mol/L H2PO4– has dissociated The value of [PO43–] shows that the dissociation of HPO42– is even smaller We must use the second and third dissociation steps to calculate [HPO42–] and [PO43–], since these steps are the only sources of these ions 136

Calculate the pH of a 1.00 M solution of H3PO4.
EXERCISE! Calculate the pH of a 1.00 M solution of H3PO4. Ka1 = 7.5 × 10-3 Ka2 = 6.2 × 10-8 Ka3 = 4.8 × 10-13 pH = 1.08 pH = 1.08 (if the students neglect "x", they will get 1.06) This is a good problem to discuss how to treat polyprotic acids. The students should understand why only the first dissociation reaction is important. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Calculate the equilibrium concentration of PO43- in a 1.00 M solution of H3PO4. Ka1 = 7.5 × 10-3 Ka2 = 6.2 × 10-8 Ka3 = 4.8 × [PO43-] = 3.6 × M [PO43-] = 3.6 x M From the previous exercise, the students should see that [H+] = [H2PO4-] = M (from the first dissociation reaction). Using the Ka2 expression, students can solve for [HPO42-] = 6.2 x 10-8 M. H2PO4-(aq) + H2O  H3O+(aq) + HPO42-(aq) I C -x x x E x x x Many students will use “0” as the initial value of [H3O+]. Note that [HPO42-] is equal to Ka2, which should make sense since [H+] = [H2PO4-] (“x” is negligible). Using the Ka3 expression, students can solve for the phosphate ion concentration. HPO42-(aq) + H2O  H3O+(aq) + PO43-(aq) I 6.2 x C -x x +x E 6.2 x x x x Again, students may forget to include the initial concentrations from the previous Ka table. Since “x” is negligible, the values are the same. Also, make sure the students understand that [H3O+] can only have one value; that is, there are not different [H3O+]’s for the different dissociations. Copyright © Cengage Learning. All rights reserved

Another name for ionic compounds
Salts Another name for ionic compounds Disintegrate into ions when dissolved in water Have independent movement in dilute solutions Behave as acids or bases under certain conditions Copyright © Cengage Learning. All rights reserved 139 139

Salts That Produce Neutral Solutions
When dissolved in water, salts that consist of cations of strong bases and anions of strong acids have no effect on [H+] Gives a neutral solution (pH = 7) Examples - KCl, NaCl, NaNO3, and KNO3 Copyright © Cengage Learning. All rights reserved 140 140

Salts A basic solution is formed if the anion of the salt is the conjugate base of a weak acid. NaF, KC2H3O2 Kw = Ka × Kb Use Kb when starting with base. Copyright © Cengage Learning. All rights reserved

Salts That Produce Basic Solutions
A basic solution is formed if the anion of the salt is the conjugate base of a weak acid Kb value for the anion can be obtained from the relationship Kb = Kw/Ka Use Kb when starting with base. Example - Sodium acetate (NaC2H3O2), NaF Copyright © Cengage Learning. All rights reserved 142 142

Interactive Example 14.18 - Salts as Weak Bases
Calculate the pH of a 0.30-M NaF solution The Ka value for HF is 7.2×10–4

The major species in solution are Na+, F–, and H2O Since HF is a weak acid, the F– ion must have a significant affinity for protons, and the dominant reaction will be This yields the following Kb expression:

Value of Kb can be calculated from Kw and the Ka value for HF: The corresponding ICE table is:

Thus, The approximation is valid by the 5% rule

[OH–] = x = 2.0×10–6 M pOH = 5.69 pH = – 5.69 = 8.31 As expected, the solution is basic

Base Strength in Aqueous Solutions (Continued)
CN– appears to be a weak base when cyanide ion reacts with water Competes with the hydroxide ion for H+ instead of competing with water Relative base strengths OH– > CN– > H2O Copyright © Cengage Learning. All rights reserved 149

Salts That Produce Acidic Solutions
Salts in which the anion is not a base and the cation is the conjugate acid of a weak base Example - Ammonium chloride (NH4Cl) Salts that contain highly charged metal ions Example - Solid aluminum chloride (AlCl3) Higher the charge on the metal ion, stronger the acidity of the hydrated ion Use Ka when starting with acid. Copyright © Cengage Learning. All rights reserved 150 150

Interactive Example 14.19 - Salts as Weak Acids I
Calculate the pH of a 0.10-M NH4Cl solution The Kb value for NH3 is 1.8×10–5

The major species in solution are NH4+, Cl–, and H2O Note that both NH4+ and H2O can produce H+ The dissociation reaction for the NH4+ ion is

Note that although the Kb value for NH3 is given, the reaction corresponding to Kb is not appropriate here, since NH3 is not a major species in the solution Instead, the given value of Kb is used to calculate Ka for NH4+ from the relationship Ka×Kb = Kw

Thus, Although NH4+ is a very weak acid, as indicated by its Ka value, it is stronger than H2O and will dominate in the production of H+ Thus, we will focus on the dissociation reaction of NH4+ to calculate the pH in this solution

Thus, 155

The approximation is valid by the 5% rule, so [H+] = x = 7.5×10–6 M pH = 5.13 156

Interactive Example 14.21 - The Acid–Base Properties of Salts
Predict whether an aqueous solution of each of the following salts will be acidic, basic, or neutral NH4C2H3O2 NH4CN

Interactive Example 14.21 - Solution (a)
The ions in solution are NH4+ and C2H3O2– Ka for NH4+ is 5.6×10–10 Kb for C2H3O2– is 5.6×10–10 Thus, Ka for NH4+ is equal to Kb for C2H3O2–, and the solution will be neutral (pH = 7)

Interactive Example 14.21 - Solution (b)
The solution will contain NH4+ and CN– ions The Ka value for NH4+ is 5.6×10–10 and Since Kb for CN– is much larger than Ka for NH4+, CN– is a much stronger base than NH4+ is an acid The solution will be basic

Qualitative Prediction of pH of Salt Solutions (from Weak Parents)

EXERCISE! HC2H3O2 Ka = × 10-5 HCN Ka = × 10-10 Calculate the Kb values for: C2H3O2− and CN− Kb (C2H3O2-) = 5.6 × 10-10 Kb (CN-) = 1.6 × 10-5 Kb (C2H3O2-) = 5.6 x 10-10 Kb (CN-) = 1.6 x 10-5 Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Arrange the following 1.0 M solutions from lowest to highest pH. HBr NaOH NH4Cl NaCN NH3 HCN NaCl HF Justify your answer. HBr, HF, HCN, NH4Cl, NaCl, NaCN, NH3, NaOH The order is: HBr (strong acid), HF (Ka = 7.2 x 10-4), HCN (Ka = 6.2 x 10-10), NH4Cl (Ka = 5.6 x 10-10), NaCl (neutral), NaCN (Kb = 1.6 x 10-5), NH3 (Kb = 1.8 x 10-5), NaOH (strong base). Have the students use the Ka and Kb values to decide. They need not calculate the pH values to answer this question. Copyright © Cengage Learning. All rights reserved

Let’s Think About It… The possibilities for the dominant reactions are: F–(aq) + H2O(l) HF(aq) + OH–(aq) H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Na+(aq) + H2O(l) NaOH + H+(aq) Na+(aq) + F–(aq) NaF Copyright © Cengage Learning. All rights reserved

Let’s Think About It… How do we decide which reaction controls the pH?
F–(aq) + H2O(l) HF(aq) + OH–(aq) H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Determine the equilibrium constant for each reaction. The primary reaction is the first one, and the Kb value is 1.39 x This is over 1000 times larger than the K value for the second reaction (Kw). Copyright © Cengage Learning. All rights reserved

Calculate the pH of a 0.75 M aqueous solution of NaCN.
EXERCISE! Calculate the pH of a 0.75 M aqueous solution of NaCN. Ka for HCN is 6.2 × 10–10. Major Species: Na+, CN-, H2O Possibilities: CN-(aq) + H2O  HCN(aq) + OH-(aq) H2O + H2O  H3O+(aq) + OH-(aq) The primary reaction is the first one, and the Kb value is 1.6 x This is many times larger than the K value for the second reaction (Kw). pH = 11.54 Copyright © Cengage Learning. All rights reserved

Let’s Think About It… What are all possibilities for the dominant reaction? The possibilities for the dominant reaction are: CN–(aq) + H2O(l) HCN(aq) + OH–(aq) H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Na+(aq) + H2O(l) NaOH + H+(aq) Na+(aq) + CN–(aq) NaCN Which of these reactions really occur? Copyright © Cengage Learning. All rights reserved

Steps Toward Solving for pH
CN–(aq) + H2O HCN(aq) + OH–(aq) Initial 0.75 M ~ 0 Change –x +x Equilibrium 0.75–x x x can be safely neglected here. Kb = 1.6 × 10–5 pH = 11.54

Bond Strengths and Acid Strengths for Hydrogen Halides

Oxyacids Contains the group H–O–X.
For a given series the acid strength increases with an increase in the number of oxygen atoms attached to the central atom. The greater the ability of X to draw electrons toward itself, the greater the acidity of the molecule. Weaken and polarize the O—H bond Copyright © Cengage Learning. All rights reserved

Several Series of Oxyacids and Their Ka Values

Comparison of Electronegativity of X and Ka Value

Acidic Strength in Oxides
A compound containing the H—O—X group will produce: An acidic solution in water if the O—X bond is strong and covalent A basic solution in water if the O—X bond is ionic High electronegativity of X will lead to covalent and strong O—X bond Low electronegativity will lead to ionic and weak O—X bond Copyright © Cengage Learning. All rights reserved 179 179

Formed when covalent oxides are dissolved in water
Acidic Oxides Formed when covalent oxides are dissolved in water Leads to the formation of an acidic solution O—X bond remains intact When H—O—X grouping is dissolved in water, the O—X bond will remain intact. It will be the polar and relatively weak H—O bond that will tend to break, releasing a proton. Examples - SO2, CO2, and NO2 Copyright © Cengage Learning. All rights reserved 180 180

Ionic oxides dissolve in water to form basic solutions
Basic Oxides Ionic oxides dissolve in water to form basic solutions If X has a very low electronegativity, the O—X bond will be ionic and subject to being broken in polar water, producing a basic solution. Examples - K2O and CaO Oxide ion has a high affinity for protons and reacts with water to produce hydroxide ions Copyright © Cengage Learning. All rights reserved 181 181

Three Models for Acids and Bases

When analyzing an acid-base equilibrium problem:
Ask this question: What are the major species in the solution and what is their chemical behavior? What major species are present? Does a reaction occur that can be assumed to go to completion? What equilibrium dominates the solution? Let the problem guide you. Be patient. Copyright © Cengage Learning. All rights reserved

Models of Acids and Bases

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