Chapter 16 Acids and Bases

Chapter 16 Acids and Bases
Chemistry, The Central Science, 11th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten Chapter 16 Acids and Bases Review of © 2009, Prentice-Hall, Inc.

Some Definitions Brønsted-Lowry An acid is a proton donor.
A base is a proton acceptor. © 2009, Prentice-Hall, Inc.

A Brønsted-Lowry acid… …must have a removable (acidic) proton.
A Brønsted-Lowry base… …must have a pair of nonbonding electrons. © 2009, Prentice-Hall, Inc.

If it can be either… HCO3- HSO4- H2O …it is amphiprotic.
© 2009, Prentice-Hall, Inc.

What Happens When an Acid Dissolves in Water?
Water acts as a Brønsted-Lowry base and abstracts a proton (H+) from the acid. As a result, the conjugate base of the acid and a hydronium ion are formed. © 2009, Prentice-Hall, Inc.

Conjugate Acids and Bases
The term conjugate comes from the Latin word “conjugare,” meaning “to join together.” Reactions between acids and bases always yield their conjugate bases and acids. © 2009, Prentice-Hall, Inc.

Practice - Brønsted-Lowry acids and bases
Identify the conjugate acid-base pairs: HClO + H2O D ClO- + H3O+ HS- + H2O D H2S + OH- HPO H2O D H2PO OH- HPO H2O D PO H3O+ An amphoteric compound is able to act as either an acid or a base. Which compounds in the above equations are amphoteric? © 2009, Prentice-Hall, Inc.

Identify the conjugate acid-base pairs:
HClO + H2O D ClO- + H3O+ HS- + H2O D H2S + OH- HPO H2O D H2PO OH- HPO H2O D PO H3O+ An amphoteric compound is able to act as either an acid or a base. Which compounds in the above equations are amphoteric? © 2009, Prentice-Hall, Inc.

Practice - Brønsted-Lowry acids and bases
Each of the following can act as both a Brønsted acid and a Brønsted base EXCEPT HCO3- H2O SO42- HS- HSO3- © 2009, Prentice-Hall, Inc.

Sample Exercise 16.1 (p. 675) a) What is the conjugate base of each of the following acids: HClO4 ClO4- H2S HS- PH4+ PH3 HCO3- CO32- b) What is the conjugate acid of each of the following bases? CN- HCN SO42- HSO4- H2O H3O+ HCO3- H2CO3 © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (16.1) Consider the following equilibrium reaction: HSO4-(aq) + OH-(aq) D SO42-(aq) + H2O(l) Which substances are acting as acids in the reaction? HSO4- and OH- HSO4- and H2O OH- and SO42- SO42- and H2O © 2009, Prentice-Hall, Inc.

Sample Exercise 16.2 (p. 676) The hydrogen sulfite ion (HSO3-) is amphoteric. Write an equation for the reaction of HSO3- with water, in which the ion acts as an acid. Write an equation for the reaction of HSO3- with water, in which the ion acts as a base. In both cases identify the conjugate acid-base pairs. © 2009, Prentice-Hall, Inc.

Sample Exercise 16.2 (p. 676) Write an equation for the reaction of HSO3- with water, in which the ion acts as an acid. HSO3- + H2O D SO32- + H3O+ HSO3- (acid) and SO32- (conjugate base) H2O (base) and H3O+ (conjugate acid) Write an equation for the reaction of HSO3- with water, in which the ion acts as a base. HSO3- + H2O D H2SO3 + OH- HSO3- (base) and H2SO3 (conjugate acid) H2O (acid) and OH- (conjugate base) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (16.2) The dihydrogen phosphate ion, H2PO4-, is amphiprotic. In which of the following reactions is this ion serving as a base? H3O+(aq) + H2PO4- D H3PO4(aq) + H2O(l) H3O+(aq) + HPO42- D H2PO H2O(l) H3PO4(aq) + HPO42- D 2 H2PO H2O(l) (i) only (i) and (ii) (i) and (iii) (ii) and (iii) (i), (ii) and (iii) © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (16.2) When lithium oxide (Li2O) is dissolved in water, the solution turns basic from the reaction of the oxide ion (O2-) with water. Write the reaction that occurs, and identify the conjugate acid-base pairs. O2- + H2O D OH- + OH- O2- (conjugate base) and OH- (conjugate acid) H2O (conjugate acid) and OH- (conjugate base) © 2009, Prentice-Hall, Inc.

Acid and Base Strength Strong acids are completely dissociated in water. Their conjugate bases cannot pull protons away “abstract protons”, are very weak. Weak acids only dissociate partially in water. Their conjugate bases have a slight ability to pull protons away, they are weak bases. © 2009, Prentice-Hall, Inc.

Acid and Base Strength Substances with negligible acidity do not dissociate in water. Their conjugate bases are exceedingly strong – abstract protons from acids/water readily. © 2009, Prentice-Hall, Inc.

Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl (aq) + H2O (l)  H3O+ (aq) + Cl- (aq) H2O is a much stronger base than Cl-, so the equilibrium lies so far to the right that K is not measured (K>>1). © 2009, Prentice-Hall, Inc.

Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. CH3CO2H (aq) + H2O (l) H3O+ (aq) + CH3CO2- (aq) Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1). © 2009, Prentice-Hall, Inc.

Sample Exercise 16.3 (p. 677) For the following proton-transfer reaction, use the above figure to predict whether the equilibrium lies predominantly to the left or to the right HSO4-(aq) + CO32-(aq)D SO42-(aq) + HCO3-(aq) Which base is more able to pull protons off the acid? CO32- or HCO3-? © 2009, Prentice-Hall, Inc.

Sample Exercise 16.3 (p. 677) For the following proton-transfer reaction, use the above figure to predict whether the equilibrium lies predominantly to the left or to the right HSO4-(aq) + CO32-(aq)D SO42-(aq) + HCO3-(aq) Which base is more able to pull protons off the acid? CO32- or HCO3-?  equilibrium lies to the right © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (16.3) Based on the information in Figure 16.4, place the following equilibria in order from smallest to largest value of Kc: CH3COOH(aq) + HS-(aq) D CH3COO-(aq) + H2S(aq) F-(aq) + NH4+(aq) D HF(aq) + NH3(aq) H2CO3 (aq) + Cl-(aq) D HCO3-(aq) + HCl(aq) (i) < (ii) < (iii) (ii) < (i) < (iii) (iii) < (i) < (ii) (ii) < (iii) < (i) (iii) < (ii) < (i) © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (16.3) For each of the following reactions, use the above figure to predict whether the equilibrium lies predominantly to the left or to the right: HPO42-(aq)+ H2O(l) D H2PO4-(aq)+ OH-(aq) b) NH4+(aq) + OH-(aq) D NH3(aq) + H2O(l) © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (16.3) For each of the following reactions, use the above figure to predict whether the equilibrium lies predominantly to the left or to the right: HPO42-(aq)+ H2O(l) D H2PO4-(aq)+ OH-(aq) to the left, as OH- is a stronger base than HPO42- and grabs one of the H’s from H2PO4- to form H2O NH4+(aq) + OH-(aq) D NH3(aq) + H2O(l) to the right, as OH- is a stronger base than NH3 and grabs one of the H’s from NH4+ to form H2O © 2009, Prentice-Hall, Inc.

Autoionization of Water
As we have seen, water is amphoteric. In pure water, a few molecules act as bases and a few act as acids. This is referred to as autoionization. H2O (l) + H2O (l) H3O+ (aq) + OH- (aq) © 2009, Prentice-Hall, Inc.

Ion-Product Constant The equilibrium expression for this process is
Kc = [H3O+] [OH-] This special equilibrium constant is referred to as the ion-product constant for water, Kw. At 25C, Kw = 1.0  10-14 © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (16.4) In a certain acidic solution at 25oC, [H+] is 100 times greater than [OH-]. What is the value for [OH-] for the solution? 1.0 x 10-8 M 1.0 x 10-7 M 1.0 x 10-6 M 1.0 x 10-2 M 1.0 x 10-9 M © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (16.4) Indicate whether solutions with each of the following ion concentrations is neutral, acidic, or basic: a) [H+] = 4 x 10-9 M b) [OH-] = 1 x 10-7 M c) [OH-] = 7 x M © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (16.4) Indicate whether solutions with each of the following ion concentrations is neutral, acidic, or basic: a) [H+] = 4 x 10-9 M basic b) [OH-] = 1 x 10-7 M neutral c) [OH-] = 7 x M acidic © 2009, Prentice-Hall, Inc.

Sample Exercise 16.5 (p. 680) Calculate the concentration of H+(aq) in
a) a solution in which [OH-] is M (1.0 x M) b) a solution in which [OH-] is 1.8 x 10-9 M (5.0 x 10-6 M) Assume T = 25oC. © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (16.5) Calculate the concentration of OH-(aq) in a solution in which a) [H+] = 2 x 10-6 M 5 x 10-9 M b) [H+] = [OH-] x 10-7 M c) [H+] = 100 x [OH-] 1 x 10-6 M © 2009, Prentice-Hall, Inc.

pH In pure water, Kw = [H3O+] [OH-] = 1.0  10-14
Since in pure water [H3O+] = [OH-], [H3O+] =  = 1.0  10-7 © 2009, Prentice-Hall, Inc.

pH Therefore, in pure water, pH = -log (1.0  10-7) = 7.00
An acid has a higher [H3O+] than pure water, so its pH is <7. A base has a lower [H3O+] than pure water, so its pH is >7. © 2009, Prentice-Hall, Inc.

pH These are the pH values for several common substances.

Practice Exercise 2 (16.6) a) In a sample of lemon juice [H+] is 3.8 x 10-4 M. What is the pH? (3.42) b) A commonly available window-cleaning solution has a [OH-] of 1.9 x 10-6 M. What is the pH? (8.28) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (16.7) A solution at 25oC has pOH = Which of the following statements is or are true? The solution is acidic. The pH of the solution is – For this solution, [OH-] = M. Only one of the statements is true. Statements (i) and (ii) are true. Statements (i) and (iii) are true. Statements (ii) and (iii) are true. All three statements are true. © 2009, Prentice-Hall, Inc.

Other “p” Scales The “p” in pH tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions). Some similar examples are pOH: -log [OH-] pKw: -log Kw © 2009, Prentice-Hall, Inc.

-log [H3O+] + -log [OH-] = -log Kw = 14.00
Watch This! Because [H3O+] [OH-] = Kw = 1.0  10-14, we know that -log [H3O+] + -log [OH-] = -log Kw = 14.00 or, in other words, pH + pOH = pKw = 14.00 © 2009, Prentice-Hall, Inc.

How Do We Measure pH? For less accurate measurements, one can use
Litmus paper “Red” paper turns blue above ~pH = 8 “Blue” paper turns red below ~pH = 5 Or an indicator. © 2009, Prentice-Hall, Inc.

Strong Acids You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. For the monoprotic strong acids, [H3O+] = [acid]. © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (16.8) Order the following three solutions from smallest to largest pH. 0.20 M HClO3 M HNO3 1.50 M HCl (i) < (ii) < (iii) (ii) < (i) < (iii) (iii) < (i) < (ii) (ii) < (iii) < (i) (iii) < (ii) < (i) © 2009, Prentice-Hall, Inc.

Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+). Again, these substances dissociate completely in aqueous solution. © 2009, Prentice-Hall, Inc.

Sample Exercise 16.9 (p. 685) What is the pH of
a M solution of NaOH? (12.45) b) a M solution of Ca(OH)2? (11.34) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (16.9) Order the following three solutions from smallest to largest pH: 0.030 M Ba(OH)2 0.040 M KOH Pure water (i) < (ii) < (iii) (ii) < (i) < (iii) (iii) < (i) < (ii) (ii) < (iii) < (i) (iii) < (ii) < (i) © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (16.9) What is the concentration of a solution of
KOH for which the pH is 11.89? (7.8 x 10-3 M) b) Ca(OH)2 for which the pH is 11.68? (2.4 x 10-3 M) © 2009, Prentice-Hall, Inc.

Overview of Chapter 16 New material: On your own:
Application of equilibrium to weak acids and bases: Ka, Kb Calculate Ka from pH and vice versa Calculate percent ionization of weak acids Polyprotic acids and their successive Ka’s Types of weak bases Calculate concentrations of all species of weak acids and weak bases at equilibrium, using Ka, Kb, pH or pOH Relationship between Ka and Kb, derived from Kw – S2 Application of Ka/Kb: acid-base properties of salt solutions – S2 Factors that affect acid strength – binary acids and oxyacids – S2 Carboxylic acids – S2 Lewis acids and bases – S2 On your own: Arrhenius acids and bases Brønsted-Lowry acids and bases Conjugate acid-base pairs Relative strengths of acids and bases Autoionization of water and Kw [H+] > pH [OH-] > pOH pH + pOH = 14.00 Determine pH and/or pOH of strong acids and bases © 2009, Prentice-Hall, Inc.

Application of Equilibrium to Weak Acids and Weak Bases

Dissociation Constants
For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant is called the acid-dissociation constant, Ka. Note that H2O is omitted from Ka because it is a pure liquid HA (aq) + H2O (l) A- (aq) + H3O+ (aq) [H3O+] [A-] [HA] Kc = © 2009, Prentice-Hall, Inc.

Ionization Constants and Equations WS

Calculating Ka from the pH
pH  [H+] at equilibrium Equilibrium [H+]  RICE table Use RICE table to find the [other species] © 2009, Prentice-Hall, Inc.

Calculating Ka from the pH Sample Exercise 16.10 (p. 688)
The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is Calculate Ka for formic acid at this temperature. We know that [H3O+] [HCOO-] [HCOOH] Ka = © 2009, Prentice-Hall, Inc.

The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is Calculate Ka for formic acid at this temperature. To calculate Ka, we need the equilibrium concentrations of all three things. We can find [H3O+], which is the same as [HCOO-], from the pH. © 2009, Prentice-Hall, Inc.

Calculating Ka from pH Now we can set up a table… [HCOOH], M [H3O+], M
[HCOO-], M Initially 0.10 Change - 4.2  10-3 + 4.2  10-3 At Equilibrium  10-3 = = 0.10 4.2  10-3 © 2009, Prentice-Hall, Inc.

Calculating Ka from pH [4.2  10-3] [4.2  10-3] Ka = [0.10]
= 1.8  10-4 © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (16.10) Niacin, one of the B vitamins, has the following molecular structure: A M solution of niacin has a pH of 3.26. What is the acid-dissociation constant, Ka, for niacin? (1.6 x 10-5) © 2009, Prentice-Hall, Inc.

Calculating Percent Ionization Sample Exercise 16.11 (p. 689)
A 0.10 M solution of formic acid (HCOOH) contains 4.2 x 10-3 M H+(aq). Calculate the percentage of the acid that is ionized. (4.2%) © 2009, Prentice-Hall, Inc.

Calculating Percent Ionization Sample Exercise 16.11 (p. 689)
[H3O+]eq [HA]initial Percent Ionization =  100 In this example [H3O+]eq = 4.2  10-3 M [HCOOH]initial = 0.10 M 4.2  10-3 0.10 Percent Ionization =  100 = 4.2% © 2009, Prentice-Hall, Inc.

Using Ka to Calculate pH
• Using Ka, we can calculate the concentration of H+ (and hence the pH). Example: Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25C. Ka for acetic acid at 25C is 1.8  10-5. © 2009, Prentice-Hall, Inc.

Substitute into the equilibrium constant expression and solve. Ka = 1.8 x 10-5 = [H+][C2H3O2-] = (x)(x) [HC2H3O2] – x Note that Ka is very small (1.8 x 10-5) relative to [HC2H3O2] (0.30 M). © 2009, Prentice-Hall, Inc.

Substitute into the equilibrium constant expression and solve. Keep this x x2 = 1.8 x 10-5 0.30 – x Neglect x in the denominator since it is extremely small relative to 0.30. (Use ballpark figure of 103 X difference for neglecting x in the denominator ) © 2009, Prentice-Hall, Inc.

What do we do if we are faced with having to solve a quadratic equation in order to determine the value of x? Often this cannot be avoided. However, if the Ka value is quite small, we find that we can make a simplifying assumption. • Assume that x is negligible compared to the initial concentration of the acid. • This will simplify the calculation. • It is always necessary to check the validity of any assumption. • Once we have the value of x, check to see how large it is compared to the initial concentration. • If x is <5% of the initial concentration, the assumption is probably a good one. • If x>5% of the initial concentration, then it may be best to solve the quadratic equation or use successive approximations. © 2009, Prentice-Hall, Inc.

Polyprotic Acids… …have more than one acidic proton
If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation. © 2009, Prentice-Hall, Inc.

Weak Bases The equilibrium constant expression for this reaction is
[HB] [OH-] [B-] Kb = where Kb is the base-dissociation constant. The larger Kb, the stronger the base. © 2009, Prentice-Hall, Inc.

Weak Bases The equilibrium constant expression for this reaction is
[NH4+] [OH-] [NH3] Kb = where Kb is the base-dissociation constant. © 2009, Prentice-Hall, Inc.

Kb can be used to find [OH-] and, through it, pH.
Weak Bases Kb can be used to find [OH-] and, through it, pH. © 2009, Prentice-Hall, Inc.

pH of Basic Solutions Sample Exercise 16.15 (p. 697)
Calculate the concentration of OH- in a 0.15M solution of NH3. (1.6 x 10-3 M) © 2009, Prentice-Hall, Inc.

Practice Problem 2 (16.15) (more of a conceptual question)
Which of the following compounds should produce the highest pH as a 0.05 M solution: pyridine, methylamine, or nitrous acid? (methylamine) © 2009, Prentice-Hall, Inc.

Types of Weak Bases Weak bases generally fall into one of two categories. 1. Neutral substances with a lone pair of electrons that can accept protons. Most neutral weak bases contain nitrogen. Amines are related to ammonia and have one or more N–H bonds replaced with N–C bonds (e.g., CH3NH2 is methylamine). © 2009, Prentice-Hall, Inc.

Types of Weak Bases 2. Anions of weak acids are also weak bases.
e.g.: ClO– is the conjugate base of HClO (weak acid): ClO–(aq) + H2O(l) D HClO(aq) + OH–(aq) Kb = 3.3 x 10–7 © 2009, Prentice-Hall, Inc.

Sample Exercise (p. 698) A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a pH of How many moles of NaClO were added to the water? Kb = 3.3 x 10-7 (See info immediately above.) (0.60 mol) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (16.16) The benzoate ion, C6H5COO-, is a weak base with Kb = 1.6 x How many moles of sodium benzoate are present in 0.50 L of a solution of NaC6H5COO if the pH is 9.04? 0.38 0.66 0.76 1.5 2.9 © 2009, Prentice-Hall, Inc.

Ka and Kb Note: Conjugate acid has one more H+ than its conjugate base. Ka and Kb are related in this way: Ka  Kb = Kw Therefore, if you know one of them, you can calculate the other. Alternatively, pKa + pKb = pKw = (at 25oC) © 2009, Prentice-Hall, Inc.

Sample Exercise 16.17 (p. 701) Calculate
the base-dissociation constant, Kb, for the fluoride ion (F-); (1.5 x 10-11) the acid-dissociation constant, Ka, for the ammonium ion (NH4+). (5.6 x 10-10) © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (16.17) (Use Appendix D for Ka)
By using information from Appendix D, put the following three substances in order of weakest to strongest base: (CH3)3N HCOO- BrO- (i) < (ii) < (iii) (ii) < (i) < (iii) (iii) < (i) < (ii) (ii) < (iii) < (i) (iii) < (ii) < (i) © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (16.17) (Use Appendix D for Ka)
a) Which of the following anions has the largest base-dissociation constant: NO2-, PO43-, or N3-? (PO43-, Kb = 2.4 x 10-2) b) The base quinoline has the following structure: Its conjugate acid is listed in handbooks as having a pKa of 4.90. What is the base-dissociation constant for quinoline? (7.9 x 10-10) © 2009, Prentice-Hall, Inc.

Acid/Base Properties of Salt Solutions

Reactions of Anions with Water
Anions are bases. As such, they can react with water in a hydrolysis reaction to form OH- and the conjugate acid: X- (aq) + H2O (l) HX (aq) + OH- (aq) © 2009, Prentice-Hall, Inc.

Reactions of Cations with Water
Cations with acidic protons (like NH4+) will lower the pH of a solution. Most metal cations that are hydrated in solution also lower the pH of the solution. © 2009, Prentice-Hall, Inc.

Reactions of Cations with Water
Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water. This makes the O-H bond more polar and the water more acidic. Greater charge and smaller size make a cation more acidic. © 2009, Prentice-Hall, Inc.

Predicting the pH of a solution (qualitatively)
Salts derived from a strong acid and strong base are neutral. e.g. NaCl, Ca(NO3)2 Salts derived from a strong base and weak acid are basic. e.g. NaClO, Ba(C2H3O2)2 Salts derived from a weak base and strong acid are acidic. e.g. NH4Cl © 2009, Prentice-Hall, Inc.

Predicting the pH of a solution (qualitatively)
Salts derived from a weak acid and weak base can be either acidic or basic. We need to compare Ka and Kb for hydrolysis of the anion and the cation. e.g. NH4CN Both ions undergo significant hydrolysis. Is the salt solution acidic or basic? Ka of NH4+ < Kb of CN-  solution should be basic. © 2009, Prentice-Hall, Inc.

Sample Exercise (p. 704) Determine whether aqueous solutions of each of the following salts will be acidic, basic, or neutral: Hint: Analyze each compound – derived from strong or weak acids and bases? Ba(CH3COO)2, NH4Cl CH3NH3Br KNO3 Al(ClO4)3 © 2009, Prentice-Hall, Inc.

Sample Exercise (p. 704) Determine whether aqueous solutions of each of the following salts will be acidic, basic, or neutral: Ba(CH3COO)2, anion of weak acid  basic NH4Cl cation of weak base  acidic CH3NH3Br KNO3 neutral – derived from strong acid + strong base Al(ClO4)3 cation of weak base (Al(OH)3) acidic, Al3+ - increase the polarity of O-H in H2O © 2009, Prentice-Hall, Inc.

Practice Exercise 1 (16.18) Order the following solutions from lowest to highest pH: 0.10 M NaClO 0.10 M KBr 0.10 M NH4ClO4 (i) < (ii) < (iii) (ii) < (i) < (iii) (iii) < (i) < (ii) (ii) < (iii) < (i) (iii) < (ii) < (i) © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (16.18) In each of the following, indicate which salt in each of the following pairs will form the more acidic (or less basic) M solution: a) NaNO3 or Fe(NO3)3 b) KBr, or KBrO c) CH3NH3Cl or BaCl2 d) NH4NO2 or NH4NO3 © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (16.18) In each of the following, indicate which salt in each of the following pairs will form the more acidic (or less basic) M solution: a) NaNO3 (neutral) or Fe(NO3)3 (acidic) b) KBr (neutral), or KBrO (basic) c) CH3NH3Cl (acidic) or BaCl2 (neutral) d) NH4NO2 or NH4NO3 © 2009, Prentice-Hall, Inc.

Practice Exercise 2 (16.19) Predict whether the dipotassium salt of citric acid (K2HC6H5O7) will form an acidic or basic solution in water. (see Table 16.3 for data) (acidic) © 2009, Prentice-Hall, Inc.

Factors Affecting Acid Strength (Sec. 16.10-16.11)
The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound. So acidity increases from left to right across a row and from top to bottom down a group. © 2009, Prentice-Hall, Inc.

Factors Affecting Acid Strength
In oxyacids, in which an -OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid. © 2009, Prentice-Hall, Inc.

Sample Exercise (p. 708) Arrange the compounds in each of the following series in order of increasing acid strength: a) AsH3, HI, NaH, H2O NaH < AsH3 < H2O < HI b) H2SO4, H2SeO3, H2SeO4 H2SeO3 < H2SeO4 < H2SO4 © 2009, Prentice-Hall, Inc.

Practice 1 (16.20) Arrange the following substances in order from weakest to strongest acid: HClO3, HOI, HBrO2, HClO2, HIO2 HIO2 < HOI < HClO3 < HBrO2 < HClO2 HOI < HIO2 < HBrO2 < HClO2 < HClO3 HBrO2 < HIO2 < HClO2 < HOI < HClO3 HClO3 < HClO2 < HBrO2 < HIO2 < HOI HOI < HClO2 < HBrO2 < HIO2 < HClO3 © 2009, Prentice-Hall, Inc.

Practice 2 (16.20) In each of the following pairs choose the compound that leads to the more acidic (or less basic) solution: a) HBr, HF; b) PH3, H2S; c) HNO2, HNO3; d) H2SO3, H2SeO3. © 2009, Prentice-Hall, Inc.

Comparison of Different Types of Acids and Bases

Arrhenius (traditional) acids and bases (C19th)
Acid: compound containing H that ionizes to yield H+ in solution Base: compound containing OH that ionizes to yield OH- in solution (Note: does not describe acid/base behavior in solvents other than water) Note: Every Arrhenius acid/base is also a Brønsted-Lowry acid/base. © 2009, Prentice-Hall, Inc.

Bronsted-Lowry Acids and Bases (1923)
Acid: H+ (proton) donor Base: H+ (proton) acceptor NH H2O D NH OH- ammonia water ammonium ion hydroxide ion (B-L base) (B-L acid) (B-L acid) (B-L base) Note: Every Brønsted-Lowry acid/base is also a Lewis acid/base. © 2009, Prentice-Hall, Inc.

Lewis Acids and Bases (1920’s) (not required for AP Chem exam)
Acid: accepts pair of e-‘s Base: donates pair of e-‘s H [:O:H]-  H:O:H Lewis acid Lewis base .. .. .. .. © 2009, Prentice-Hall, Inc.

Lewis Acids Lewis acids are defined as electron-pair acceptors.
Atoms with an empty valence orbital can be Lewis acids. © 2009, Prentice-Hall, Inc.

Lewis Bases Lewis bases are defined as electron-pair donors.
Anything that could be a Brønsted-Lowry base is a Lewis base. Lewis bases can interact with things other than protons, however. © 2009, Prentice-Hall, Inc.

Hydrolysis of Metal Ions
The Lewis concept may be used to explain the acid properties of many metal ions. Metal ions are positively charged and attract water molecules (via the lone pairs on the oxygen atom of water). Hydrated metal ions act as acids. • For example: Fe(H2O)63+(aq) D Fe(H2O)5(OH)2+(aq) + H+(aq) Ka = 2 x 10–3 © 2009, Prentice-Hall, Inc.

Hydrolysis of Metal Ions
In general: • The higher the charge, the stronger the M–OH2 interaction. • Ka values generally increase with increasing charge • The smaller the metal ion, the more acidic the ion. • Ka values generally decrease with decreasing ionic radius • Thus the pH of an aqueous solution increases as the size of the ion increases (e.g., Ca2+ vs. Zn2+) and as the charge increases (e.g., Na+ vs. Ca2+ and Zn2+ vs. Al3+). © 2009, Prentice-Hall, Inc.

Chapter 16 Acids and Bases

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Chapter 16 Acids and Bases

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