Chapter 15 Acids and Bases

Chapter 15 Acids and Bases
Lecture Presentation Chapter 15 Acids and Bases

Acids and Bases 15.1 Heartburn 15.2 The Nature of Acids and Bases 15.3 Definitions of Acids and Bases 15.4 Acid Strength and the Acid Ionization Constant (Ka) 15.5 Autoionization of Water and pH 15.6 Finding the [H3O+] and pH of Strong and Weak Acid Solutions 15.7 Base solutions 15.8 The Acid-Base Properties of Ions and Salts 15.9 Polyprotic Acids Acid Strength and Molecular Structure Lewis Acids and Bases Acid Rain

Stomach Acid and Heartburn
The cells that line your stomach produce hydrochloric acid. To kill unwanted bacteria To help break down food To activate enzymes that break down food [H3O+] ~ 0.01 – 0.1 M If the stomach acid backs up into your esophagus, it irritates those tissues, resulting in heartburn. Acid reflux

Curing Heartburn Mild cases of heartburn can be cured by neutralizing the acid in the esophagus. Swallowing saliva, which contains bicarbonate ion Taking antacids that contain hydroxide ions and/or carbonate ions

GERD GERD (gastroesophageal reflux disease) is chronic leaking of stomach acid into the esophagus. In people with GERD, the muscles separating the stomach from the esophagus do not close tightly, allowing stomach acid to leak into the esophagus. Physicians diagnose GERD by attaching a pH sensor to the esophagus to measure the acidity levels of the fluids over time.

Properties of Acids Sour taste Ability to dissolve many metals
Ability to neutralize bases Change blue litmus paper to red

Common Acids

Structure of Acids HCl, HF
Binary acids have acid hydrogens attached to a nonmetal atom. HCl, HF

Structure of Acids Oxyacids have acid hydrogens attached to an oxygen atom. H2SO4, HNO3

Structure of Acids Carboxylic acids have COOH group.
HC2H3O2, H3C6H5O7 Only the first H in the formula is acidic. The H is on the COOH.

Properties of Bases Taste bitter Feel slippery
alkaloids = plant product that is alkaline often poisonous Feel slippery Ability to turn red litmus paper blue Ability to neutralize acids

Common Bases

Indicators Indicators are chemicals that change color depending on the solution’s acidity or basicity. Many vegetable dyes are indicators. Anthocyanins (花青素) Litmus From Spanish moss Red in acid, blue in base Phenolphthalein Found in laxatives Red in base, colorless in acid

Definitions of Acids and Bases
Arrhenius definition Based on H+ and OH- Brønsted–Lowry definition Based on reactions in which an H+ is transferred Lewis definition Based on reactions in which an electron pair is transferred

Arrhenius Theory Swedish Chemist Svante Arrhenius (1880s)
Acids: produce H+ ions in aqueous solution. HCl(aq) → H+(aq) + Cl−(aq)

Hydronium Ion H+ + H2O  H3O+
The H+ ions produced by the acid are so reactive they cannot exist in water. H+ ions are protons! Instead, they react with water molecules to produce complex ions, mainly hydronium ion, H3O+. H+ + H2O  H3O+ There are also minor amounts of H+ with multiple water molecules, H(H2O)n+.

Arrhenius Theory Bases: produce OH− ions in aqueous solution.
NaOH(aq) → Na+(aq) + OH(aq)

Arrhenius Acid–Base Reactions
The H+ from the acid combines with the OH− from the base to make a molecule of H2O. It is often helpful to think of H2O as H—OH. The cation from the base combines with the anion from the acid to make a salt. acid + base → salt + water HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Problems with Arrhenius Theory
It does not explain why molecular substances, such as NH3, dissolve in water to form basic solutions, even though they do not contain OH– ions. It does not explain how some ionic compounds, such as Na2CO3 or Na2O, dissolve in water to form basic solutions, even though they do not contain OH– ions. It does not explain why molecular substances, such as CO2, dissolve in water to form acidic solutions, even though they do not contain H+ ions. It does not explain acid–base reactions that take place outside aqueous solution.

Brønsted–Lowry Acid–Base Theory
It defines acids and bases based on what happens in a reaction. Any reaction involving H+ (proton) that transfers from one molecule to another is an acid–base reaction, regardless of whether it occurs in aqueous solution or if there is OH− present. All reactions that fit the Arrhenius definition also fit the Brønsted–Lowry definition.

Brønsted–Lowry Theory
The acid is an H+ donor. The base is an H+ acceptor. Base structure must contain an atom with an unshared pair of electrons. In a Brønsted–Lowry acid–base reaction, the acid molecule donates an H+ to the base molecule. H–A + :B  :A– + H–B+

Brønsted–Lowry Acids HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) acid base
Brønsted–Lowry acids are H+ donors. Any material that has H can potentially be a Brønsted–Lowry acid. Because of the molecular structure, often one H in the molecule is easier to transfer than others. When HCl dissolves in water, the HCl is the acid because HCl transfers an H+ to H2O, forming H3O+ ions. Water acts as base, accepting H+. HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) acid base

Brønsted–Lowry Bases NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq) base acid
Brønsted–Lowry bases are H+ acceptors. Any material that has atoms with lone pairs can potentially be a Brønsted–Lowry base. Because of the molecular structure, often one atom in the molecule is more willing to accept H+ transfer than others. When NH3 dissolves in water, the NH3(aq) is the base because NH3 accepts an H+ from H2O, forming OH–(aq). Water acts as acid, donating H+. NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq) base acid

Amphoteric Substances
Amphoteric substances can act as either an acid or a base because they have both a transferable H and an atom with lone pair electrons. Water acts as base, accepting H+ from HCl. HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) Water acts as acid, donating H+ to NH3. NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)

Brønsted–Lowry Acid–Base Reactions
One of the advantages of Brønsted–Lowry theory is that it illustrates reversible reactions to be as follows: H–A + :B  :A– + H–B+ The original base has an extra H+ after the reaction, so it will act as an acid in the reverse process. And the original acid has a lone pair of electrons after the reaction, so it will act as a base in the reverse process: :A– + H–B+  H–A + :B

Conjugate Acid–Base Pairs
In a Brønsted-–Lowry acid–base reaction, the original base becomes an acid in the reverse reaction. the original acid becomes a base in the reverse process. Each reactant and the product it becomes is called a conjugate pair.

Conjugate Pairs A base accepts a proton and becomes a conjugate acid.
An acid donates a proton and becomes a conjugate base.

H2SO4(aq) + H2O(l) ⟶ HSO4−(aq) + H3O+(aq)
Example 15.1 Identifying Brønsted–Lowry Acids and Bases and Their Conjugates In each reaction, identify the Brønsted–Lowry acid, the Brønsted–Lowry base, the conjugate acid, and the conjugate base. a. b. Solution a. Because H2SO4 donates a proton to H2O in this reaction, it is the acid (proton donor). After H2SO4 donates the proton, it becomes HSO4−, the conjugate base. Because H2O accepts a proton, it is the base (proton acceptor) After H2O accepts the proton, it becomes H3O+, the conjugate acid. H2SO4(aq) + H2O(l) ⟶ HSO4−(aq) + H3O+(aq)

HCO3−(aq) + H2O(l) H2CO3(aq) + OH−(aq)
Example 15.1 Identifying Brønsted–Lowry Acids and Bases and Their Conjugates Continued Solution b. Because H2O donates a proton to HCO3− in this reaction, it is the acid (proton donor). After H2O donates the proton, it becomes OH−, the conjugate base. Because HCO3− accepts a proton, it is the base (proton acceptor) After HCO3− accepts the proton, it becomes H2CO3, the conjugate acid. HCO3−(aq) + H2O(l) H2CO3(aq) + OH−(aq)

Arrow Conventions  in these notes
Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions. A single arrow indicates that all the reactant molecules are converted to product molecules at the end. A double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products.  in these notes

Strong or Weak A strong acid is a strong electrolyte.
Practically all the acid molecules ionize. A strong base is a strong electrolyte. Practically all the base molecules form OH– ions, either through dissociation or reaction with water. A weak acid is a weak electrolyte. Only a small percentage of the molecules ionize, . A weak base is a weak electrolyte only a small percentage of the base molecules form OH– ions, either through dissociation or reaction with water, .

Strong Acids Strong acids donate practically all their H+.
100% ionized in water Strong electrolyte [H3O+] = [strong acid] [X] means the molarity of X

Examples of Strong Acids

Weak Acids Weak acids donate a small fraction of their H+.
Most of the weak acid molecules do not donate H+ to water. Much less than 1% ionized in water [H3O+] << [weak acid]

Examples of Weak Acids

Strengths of Acids and Bases
Commonly, acid or base strength is measured by determining the equilibrium constant of a substance’s reaction with water. HAcid + H2O  Acid− + H3O+ Base + H2O  HBase+ + OH− The farther the equilibrium position lies toward the products, the stronger the acid or base. The position of equilibrium depends on the strength of attraction between the base form and the H+. Stronger attraction means weaker acid or stronger base.

Ionic Attraction and Acid Strength
The stronger an acid is at donating H, the weaker the conjugate base is at accepting H.

Autoionization of Water
Water is amphoteric; it can act either as an acid or a base. Therefore, there must be a few ions present. About 2 out of every 1 billion water molecules form ions through a process called autoionization. H2O Û H+ + OH– H2O + H2O Û H3O+ + OH– All aqueous solutions contain both H3O+ and OH–. The concentration of H3O+ and OH– are equal in water. [H3O+] = [OH–] = 10−7 M at 25 °C

Ion Product of Water The product of the H3O+ and OH– concentrations is always the same number. The number is called the ion product constant of water and has the symbol Kw. Also know as the dissociation constant of water [H3O+] × [OH–] = Kw = 1.00 × 10−14 at 25 °C If you measure one of the concentrations, you can calculate the other. As [H3O+] increases the [OH–] must decrease so the product stays constant. Inversely proportional

Acid Ionization Constant, Ka
Acid strength is measured by the equilibrium constant when it reacts with H2O. The equilibrium constant for this reaction is called the acid ionization constant, Ka. larger Ka = stronger acid

Table 15.5

Acidic and Basic Solutions
All aqueous solutions contain both H3O+ and OH– ions. Neutral solutions have equal [H3O+] and [OH–]. [H3O+] = [OH–] = 1.00 × 10−7 Acidic solutions have a larger [H3O+] than [OH–]. [H3O+] > 1.00 × 10−7; [OH–] < 1.00 × 10−7 Basic solutions have a larger [OH–] than [H3O+]. [H3O+] < 1.00 × 10−7; [OH–] > 1.00 × 10−7

Measuring Acidity: pH The acidity or basicity of a solution is often expressed as pH. pH = −log10[H3O+] pHwater = −log[10−7] = 7 Need to know the [H3O+] concentration to find pH pH < 7 is acidic; pH > 7 is basic. pH = 7 is neutral. [H3O+] = 10−pH

Significant Figures and Logs
When you take the log of a number written in scientific notation, the digits before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number. log(2.0 x 106) = log(106) + log(2.0) = … = Because the part of the scientific notation number that determines the significant figures is the decimal part, the significant figures are the digits after the decimal point in the log. log(2.0 × 106) = 6.30

What Does the pH Number Imply?
The lower the pH, the more acidic the solution; the higher the pH, the more basic the solution. 1 pH unit corresponds to a factor of 10 difference in acidity. Normal range of pH is 0 to 14. pH 0 is [H3O+] = 1 M; pH 14 is [OH–] = 1 M. pH can be negative (very acidic) or larger than 14 (very alkaline).

pOH Another way of expressing the acidity/basicity of a solution is pOH. pOH = −log[OH], [OH] = 10−pOH pOHwater = −log[10−7] = 7 Need to know the [OH] concentration to find pOH pOH < 7 is basic; pOH > 7 is acidic; pOH = 7 is neutral. pH + pOH = 14.0

Relationship between pH and pOH
pH + pOH = at 25 °C. You can use pOH to find the pH of a solution.

Example 15.3 Calculating pH from [H3O+] or [OH−]
Calculate the pH of each solution at 25 °C and indicate whether the solution is acidic or basic. a. [H3O+] = 1.8 × 10−4 M b. [OH−] = 1.3 × 10−2 M Solution a. To calculate pH, substitute the given [H3O+] into the pH equation. Since pH < 7, this solution is acidic. b. First use Kw to find [H3O+] from [OH−]. Then substitute [H3O+] into the pH expression to find pH. Since pH > 7, this solution is basic.

pK A way of expressing the strength of an acid or base is pK.
pKa = −log(Ka), Ka = 10−pKa pKb = −log(Kb), Kb = 10−pKb The stronger the acid, the smaller the pKa. Larger Ka = smaller pKa Because it is the –log The stronger the base, the smaller the pKb. Larger Kb = smaller pKb

[H3O+] and [OH−] in a Strong Acid or Strong Base Solution
There are two sources of H3O+ in an aqueous solution of a strong acid—the acid and the water. There are two sources of OH− in an aqueous solution of a strong acid—the base and the water. For a strong acid or base, the contribution of the water to the total [H3O+] or [OH−] is negligible. The [H3O+]acid shifts the Kw equilibrium so far that [H3O+]water is too small to be significant. Except in very dilute solutions, generally < 1 × 10−5 M

Finding pH of a Strong Acid or Strong Base Solution
For a monoprotic strong acid [H3O+] = [HAcid]. For polyprotic acids, the other ionizations can generally be ignored. For H2SO4, the second ionization cannot be ignored. 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00 For a strong ionic base, [OH−] = (number OH−) × [Base]. For molecular bases with multiple lone pairs available, only one lone pair accepts an H; the other reactions can generally be ignored. 0.10 M Ca(OH)2 has [OH−] = 0.20 M and pH =

Finding the pH of a Weak Acid
There are also two sources of H3O+ in an aqueous solution of a weak acid—the acid and the water. However, finding the [H3O+] is complicated by the fact that the acid only undergoes partial ionization. Calculating the [H3O+] requires solving an equilibrium problem for the reaction that defines the acidity of the acid. HAcid + H2O  Acid + H3O+

Example 15.5 Finding the [H3O+] of a Weak Acid Solution
Find the [H3O+] of a M HCN solution. Procedure For… Finding the pH (or [H3O+]) of a Weak Acid Solution To solve these types of problems, follow the outlined procedure. Solution Step 1 Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing the given concentration of the weak acid as its initial concentration. Leave room in the table for the changes in concentrations and for the equilibrium concentrations. (Note that the [H3O+] is listed as approximately zero because the autoionization of water produces a negligibly small amount of H3O+).

Continued Step 2 Represent the change in the concentration of H3O+ with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x. Always keep in mind the stoichiometry of the reaction. Step 3 Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

Continued Step 4 Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant (Ka). In many cases, you can make the approximation that x is small (as discussed in Section 14.8). Substitute the value of the acid ionization constant (from Table 15.5) into the Ka expression and solve for x. Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Therefore the approximation is valid.

(pH was not asked for in this problem.)
Example 15.5 Finding the [H3O+] of a Weak Acid Solution Continued Step 5 Determine the [H3O+] from the calculated value of x and calculate the pH if necessary. [H3O+] = 7.0 × 10−6 M (pH was not asked for in this problem.) Step 6 Check your answer by substituting the computed equilibrium values into the acid ionization expression. The calculated value of Ka should match the given value of Ka. Note that rounding errors and the x is small approximation could result in a difference in the least significant digit when comparing values of Ka. Since the calculated value of Ka matches the given value, the answer is valid.

Example 15.6 Finding the pH of a Weak Acid Solution
Find the pH of a M HNO2 solution. Procedure For… Finding the pH (or [H3O+]) of a Weak Acid Solution To solve these types of problems, follow the outlined procedure. Solution Step 1 Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing the given concentration of the weak acid as its initial concentration. Leave room in the table for the changes in concentrations and for the equilibrium concentrations. (Note that the [H3O+] is listed as approximately zero because the autoionization of water produces a negligibly small amount of H3O+).

Continued Step 2 Represent the change in the concentration of H3O+ with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x. Always keep in mind the stoichiometry of the reaction. Step 3 Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

Continued Step 4 Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant (Ka). In many cases, you can make the approximation that x is small (as discussed in Section 14.8). Substitute the value of the acid ionization constant (from Table 15.5) into the Ka expression and solve for x. Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Therefore the approximation is valid (but barely so).

Continued Step 5 Determine the [H3O+] from the calculated value of x and calculate the pH if necessary. Step 6 Check your answer by substituting the computed equilibrium values into the acid ionization expression. The calculated value of Ka should match the given value of Ka. Note that rounding errors and the x is small approximation could result in a difference in the least significant digit when comparing values of Ka. Since the calculated value of Ka matches the given value, the answer is valid.

Example 15.7 Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work Find the pH of a M HClO2 solution. Solution Step 1 Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing the given concentration of the weak acid as its initial concentration. (Note that the H3O+ concentration is listed as approximately zero. Although a little H3O+ is present from the autoionization of water, this amount is negligibly small compared to the amount of HClO2 or H3O+ formed by the acid.)

Example 15.7 Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work Continued Step 2 Represent the change in [H3O+] with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x. Step 3 Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

Example 15.7 Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work Continued Step 4 Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant (Ka). Make the x is small approximation and substitute the value of the acid ionization constant (from Table 15.5) into the Ka expression. Solve for x.

Example 15.7 Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work Continued Check to see if the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Therefore, the x is small approximation is not valid.

Example 15.7 Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work Continued Step 4a If the x is small approximation is not valid, solve the quadratic equation explicitly or use the method of successive approximations to find x. In this case, we solve the quadratic equation. Since x represents the concentration of H3O+, and since concentrations cannot be negative, we reject the negative root. x = 0.028

Example 15.7 Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work Continued Step 5 Determine the H3O+ concentration from the calculated value of x and calculate the pH (if necessary). Step 6 Check your answer by substituting the calculated equilibrium values into the acid ionization expression. The calculated value of Ka should match the given value of Ka. Note that rounding errors could result in a difference in the least significant digit when comparing values of Ka. Since the calculated value of Ka matches the given value, the answer is valid.

Percent Ionization Another way to measure the strength of an acid is to determine the percentage of acid molecules that ionize when dissolved in water; this is called the percent ionization. The higher the percent ionization, the stronger the acid. Because [ionized acid]equil = [H3O+]equil

Relationship between [H3O+]equilibrium and [HA]initial
Increasing the initial concentration of acid results in increased [H3O+] at equilibrium. Increasing the initial concentration of acid results in decreased percent ionization. This means that the increase in [H3O+] concentration is slower than the increase in acid concentration.

Example 15.9 Finding the Percent Ionization of a Weak Acid
Find the percent ionization of a 2.5 M HNO2 solution. Solution To find the percent ionization, you must find the equilibrium concentration of H3O+. Follow the procedure in Example 15.5, shown in condensed form here. Therefore, [H3O+] = M.

Example 15.9 Finding the Percent Ionization of a Weak Acid
Continued Use the definition of percent ionization to calculate it. (Since the percent ionization is less than 5%, the x is small approximation is valid.) Example 15.6

Why Doesn’t the Increase in H3O+ Keep Up with the Increase in HA?
The reaction for ionization of a weak acid is as follows: HA(aq) + H2O(l)  A−(aq) + H3O+(aq) According to Le Châtelier’s principle, if we reduce the concentrations of all the (aq) components, the equilibrium should shift to the right to increase the total number of dissolved particles. We can reduce the (aq) concentrations by using a more dilute initial acid concentration. This will result in a larger percent ionization.

Finding the pH of Mixtures of Acids
Generally, you can ignore the contribution of the weaker acid to the [H3O+]equil. For a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H3O+] to shift the weak acid equilibrium to the left so far that the weak acid’s added [H3O+] is negligible. For mixtures of weak acids, you generally only need to consider the stronger for the same reasons, as long as one is significantly stronger than the other and their concentrations are similar.

Example 15.10 Mixtures of Weak Acids
Find the pH of a mixture that is M in HF and M in HClO. Solution The three possible sources of H3O+ ions are HF, HClO, and H2O. Write the ionization equations for the three sources and their corresponding equilibrium constants. Since the equilibrium constant for the ionization of HF is about 12,000 times larger than that for the ionization of HClO, the contribution of HF to [H3O+] is by far the greatest. You can therefore just calculate the [H3O+] formed by HF and neglect the other two potential sources of H3O+. Write the balanced equation for the ionization of HF and use it as a guide to prepare an ICE table.

Continued Substitute the expressions for the equilibrium concentrations into the expression for the acid ionization constant (Ka). Since the equilibrium constant is small relative to the initial concentration of HF, you can make the x is small approximation. Substitute the value of the acid ionization constant (from Table 15.5) into the Ka expression and solve for x. Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Therefore, the approximation is valid (though barely so).

Continued Determine the H3O+ concentration from the calculated value of x and find the pH.

Strong Bases The stronger the base, the more willing it is to accept H. Use water as the standard acid. For ionic bases, practically all units are dissociated into OH– or accept H+. Strong electrolyte Multi-OH strong bases completely dissociated

Weak Bases In weak bases, only a small fraction of molecules accept H+. Weak electrolyte Most of the weak base molecules do not take H from water. Much less than 1% ionization in water [HO–] << [weak base] Finding the pH of a weak base solution is similar to finding the pH of a weak acid.

Base Ionization Constant, Kb
Base strength is measured by the size of the equilibrium constant when it reacts with H2O :Base + H2O  OH− + H:Base+ The equilibrium constant is called the base ionization constant, Kb. Larger Kb = stronger base 78

Common Weak Bases Table 15.8 page 721

Structure of Amines

Example 15.11 Finding the [OH−] and pH of a Strong Base Solution
What is the OH− concentration and pH in each solution? a M KOH b M Sr(OH)2 Solution a. Since KOH is a strong base, it completely dissociates into K+ and OH− in solution. The concentration of OH− will therefore be the same as the given concentration of KOH. Use this concentration and Kw to find [H3O+]. Then substitute [H3O+] into the pH expression to find the pH.

Example 15.11 Finding the [OH−] and pH of a Strong Base Solution
Continued b. Since Sr(OH)2 is a strong base, it completely dissociates into 1 mol of Sr2+ and 2 mol of OH− in solution. The concentration of OH− will therefore be twice the given concentration of Sr(OH)2. Use this concentration and Kw to find [H3O+]. Substitute [H3O+] into the pH expression to find the pH.

Example 15.12 Finding the [OH−] and pH of a Weak Base Solution
Find the [OH−] and pH of a M NH3 solution. Solution Step 1 Write the balanced equation for the ionization of water by the base and use it as a guide to prepare an ICE table showing the given concentration of the weak base as its initial concentration. Leave room in the table for the changes in concentrations and for the equilibrium concentrations. (Note that you should list the OH− concentration as approximately zero. Although a little OH is present from the autoionization of water, this amount is negligibly small compared to the amount of OH− formed by the base.)

Continued Step 2 Represent the change in the concentration of OH− with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x. Step 3 Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

Continued Step 4 Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the base ionization constant. In many cases, you can make the approximation that x is small (as discussed in Chapter 14). Substitute the value of the base ionization constant (from Table 15.8) into the Kb expression and solve for x. Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Therefore, the approximation is valid.

Continued Step 5 Determine the OH− concentration from the calculated value of x. Use the expression for Kw to find [H3O+]. Substitute [H3O+] into the pH equation to find pH.

Acid–Base Properties of Ions and Salts
Salts are water-soluble ionic compounds. Salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic. NaHCO3 solutions are basic. Na+ is the cation of the strong base NaOH. HCO3− is the conjugate base of the weak acid H2CO3. Salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic. NH4Cl solutions are acidic. NH4+ is the conjugate acid of the weak base NH3. Cl− is the anion of the strong acid HCl.

Anions as Weak Bases Therefore, every anion can potentially be a base.
A−(aq) + H2O(l)  HA(aq) + OH−(aq) The stronger the acid, the weaker the conjugate base. An anion that is the conjugate base of a strong acid is pH neutral. Cl−(aq) + H2O(l)  HCl(aq) + OH−(aq) An anion that is the conjugate base of a weak acid is basic. F−(aq) + H2O(l)  HF(aq) + OH−(aq)

Example 15.13 Determining Whether an Anion Is Basic or pH-Neutral
Classify each anion as a weak base or pH-neutral: a. NO3− b. NO2− c. C2H3O2− Solution a. From Table 15.3, we can see that NO3− is the conjugate base of a strong acid (HNO3) and is therefore pH-neutral. b. From Table 15.5 (or from its absence in Table 15.3), we know that NO2− is the conjugate base of a weak acid (HNO2) and is therefore a weak base.

Example 15.13 Determining Whether an Anion Is Basic or pH-Neutral
Continued c. From Table 15.5 (or from its absence in Table 15.3), we know that C2H3O2− is the conjugate base of a weak acid (HC2H3O2) and is therefore a weak base.

Relationship between Ka of an Acid and Kb of Its Conjugate Base
Many reference books only give tables of Ka values because Kb values can be found from them. When you add equations, you multiply the K’s.

Example 15.14 Determining the pH of a Solution Containing an Anion Acting as a Base
Find the pH of a M NaCHO2 solution. The salt completely dissociates into Na+(aq) and CHO2−(aq), and the Na+ ion has no acid or base properties. Solution Step 1 Since the Na+ ion does not have any acidic or basic properties, you can ignore it. Write the balanced equation for the ionization of water by the basic anion and use it as a guide to prepare an ICE table showing the given concentration of the weak base as its initial concentration.

Continued Step 2 Represent the change in the concentration of OH− with the variable x. Define the changes in the concentrations of the other reactants and products in terms of x. Step 3 Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

Continued Therefore, the approximation is valid.

Continued Step 5 Determine the OH− concentration from the calculated value of x. Use the expression for Kw to find [H3O+]. Substitute [H3O+] into the pH equation to find pH.

NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
Cations as Weak Acids Some cations can be thought of as the conjugate acid of a weak base. Others are the counterions of a strong base. Therefore, some cations can potentially be acidic. MH+(aq) + H2O(l)  MOH(aq) + H3O+(aq) The stronger the base, the weaker the conjugate acid. A cation that is the counterion of a strong base is pH neutral. A cation that is the conjugate acid of a weak base is acidic. NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq) Because NH3 is a weak base, the position of this equilibrium favors the right.

Metal Cations as Weak Acids
Cations of small, highly charged metals are weakly acidic. Alkali metal cations and alkali earth metal cations are pH neutral. Cations are hydrated. Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+ (aq) + H3O+(aq)

Classifying Salt Solutions as Acidic, Basic, or Neutral
If the salt cation is the counterion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solution. NaCl Ca(NO3)2 KBr If the salt cation is the counterion of a strong base and the anion is the conjugate base of a weak acid, it will form a basic solution. NaF Ca(C2H3O2) KNO2

If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution. NH4Cl If the salt cation is a highly charged metal ion and the anion is the conjugate base of a strong acid, it will form an acidic solution. Al(NO3)3

If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base. NH4F because HF is a stronger acid than NH4+, Ka of NH4+ is larger than Kb of the F−; therefore, the solution will be acidic. (NH4Cl) (NH4F) (Al(NO3)3) (AlF3) (NaCl) (NaF)

Example 15.16 Determining the Overall Acidity or Basicity of Salt Solutions
Determine if the solution formed by each salt is acidic, basic, or neutral. a. SrCl b. AlBr c. CH3NH3NO3 d. NaCHO e. NH4F Solution a. The Sr2+ cation is the counterion of a strong base (Sr(OH)2) and is pH-neutral. The Cl− anion is the conjugate base of a strong acid (HCl) and is pH-neutral as well. The SrCl2 solution is therefore pH-neutral (neither acidic nor basic). b. The Al3+ cation is a small, highly charged metal ion (that is not an alkali metal or an alkaline earth metal) and is a weak acid. The Br− anion is the conjugate base of a strong acid (HBr) and is pH-neutral. The AlBr3 solution is therefore acidic.

Continued c. The CH3NH3+ ion is the conjugate acid of a weak base (CH3NH2) and is acidic. The NO3− anion is the conjugate base of a strong acid (HNO3) and is pH-neutral. The CH3NH3NO3 solution is therefore acidic. d. The Na+ cation is the counterion of a strong base and is pH-neutral. The CHO2− anion is the conjugate base of a weak acid and is basic. The NaCHO2 solution is therefore basic.

Continued e. The NH4+ ion is the conjugate acid of a weak base (NH3) and is acidic. The F− ion is the conjugate base of a weak acid and is basic. To determine the overall acidity or basicity of the solution, compare the values of Ka for the acidic cation and Kb for the basic anion. Obtain each value of K from the conjugate by using Ka × Kb = Kw. Since Ka is greater than Kb, the solution is acidic.

Ionization in Polyprotic Acids
Because polyprotic acids ionize in steps, each H has a separate Ka. Ka1 > Ka2 > Ka3 Generally, the difference in Ka values is great enough so that the second ionization does not happen to a large enough extent to affect the pH. Most pH problems just do first ionization. Except H2SO4  uses [H2SO4] as the [H3O+] for the second ionization. [A2−] = Ka2 as long as the second ionization is negligible.

HSO4 + H2O  SO42 + H3O+ Ka2 = 1.2 × 10−2
Ionization in H2SO4 The ionization constants for H2SO4 are as follows: H2SO4 + H2O  HSO4 + H3O+ strong HSO4 + H2O  SO42 + H3O+ Ka2 = 1.2 × 10−2 For most sulfuric acid solutions, the second ionization is significant and must be accounted for. Because the first ionization is complete, use the given [H2SO4] = [HSO4−]initial = [H3O+]initial.

Example 15.18 Dilute H2SO4 Solutions
Find the pH of a M sulfuric acid (H2SO4) solution. Solution Sulfuric acid is strong in its first ionization step and weak in its second. Begin by writing the equations for the two steps. As the concentration of an H2SO4 solution becomes smaller, the second ionization step becomes more significant because the percent ionization increases (as discussed in Section 15.6). Therefore, for a concentration of M, you can’t neglect the H3O+ contribution from the second step, as you can for other polyprotic acids. You must calculate the H3O+ contributions from both steps. The [H3O+] that results from the first ionization step is M (because the first step is strong). To determine the [H3O+] formed by the second step, prepare an ICE table for the second step in which the initial concentration of H3O+ is M. The initial concentration of HSO4− must also be M (due to the stoichiometry of the ionization reaction).

Continued Substitute the expressions for the equilibrium concentrations (from the table just shown) into the expression for Ka2. In this case, you cannot make the x is small approximation because the equilibrium constant (0.012) is not small relative to the initial concentration (0.0100). Substitute the value of Ka2 and multiply out the expression to arrive at the standard quadratic form.

Continued Solve the quadratic equation using the quadratic formula. Since x represents a concentration, and since concentrations cannot be negative, we reject the negative root. x = Determine the H3O+ concentration from the calculated value of x and calculate the pH. Notice that the second step produces almost half as much H3O+ as the first step—an amount that must not be neglected. This will always be the case with dilute H2SO4 solutions.

Example 15.17 Finding the pH of a Polyprotic Acid Solution
Find the pH of a M ascorbic acid (H2C6H6O6) solution. Solution To find the pH, you must find the equilibrium concentration of H3O+. Treat the problem as a weak acid pH problem with a single ionizable proton. The second proton contributes a negligible amount to the concentration of H3O+ and can be ignored. Follow the procedure from Example 15.6, shown in condensed form here. Use Ka1 for ascorbic acid from Table Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Calculate the pH from H3O+ concentration.

Example 15.17 Finding the pH of a Polyprotic Acid Solution
Continued The approximation is valid. Therefore,

Example 15.19 Finding the Concentration of the Anions for a Weak Diprotic Acid Solution
Find the [C6H6O62−] of the M ascorbic acid (H2C6H6O6) solution in Example Solution To find the [C6H6O62−], use the concentrations of [HC6H6O6−] and H3O+ produced by the first ionization step (as calculated in Example 15.17) as the initial concentrations for the second step. Because of the 1:1 stoichiometry, [HC6H6O6−] = [H3O+]. Then solve an equilibrium problem for the second step similar to that of Example 15.6, shown in condensed form here. Use Ka2 for ascorbic acid from Table

Example 15.19 Finding the Concentration of the Anions for a Weak Diprotic Acid Solution
Continued Since x is much smaller than 2.8 × 10−3, the x is small approximation is valid. Therefore, [C6H6O62−] = 1.6 × 10−12 M.

Strengths of Binary Acids
The more d+ H─X d− polarized the bond, the more acidic the bond. The stronger the H─X bond, the weaker the acid. Binary acid strength increases to the right across a period. Acidity: H─C < H─N < H─O < H─F Binary acid strength increases down the column. Acidity: H─F < H─Cl < H─Br < H─I

Relationship between Bond Strength and Acidity

Strengths of Oxyacids, H–O–Y
The more electronegative the Y atom, the stronger the oxyacid. HClO > HIO Acidity of oxyacids decreases down a group. Weaken the H–O bond. The larger the oxidation number of the central atom, the stronger the oxyacid. H2CO3 > H3BO3 Acidity of oxyacids increases to the right across a period. The more oxygens attached to Y, the stronger the oxyacid. Further weakens and polarizes the H–O bond HClO3 > HClO2

Relationship between Electronegativity and Acidity

Relationship between Number of Oxygens on the Central Atom and Acidity

Lewis Acid–Base Theory
Lewis acid–base theory focuses on transferring an electron pair. Lone pair  bond Bond  lone pair Does NOT require H atoms The electron donor is called the Lewis base. Electron rich; therefore nucleophile The electron acceptor is called the Lewis acid. Electron deficient; therefore electrophile

Lewis Bases The Lewis base has electrons it is willing to give away to or share with another atom. The Lewis base must have a lone pair of electrons on it that it can donate. Anions are better Lewis bases than neutral atoms or molecules. N: < N:− Generally, the more electronegative an atom, the less willing it is to be a Lewis base. O: < S: 122

Lewis Acids They are electron deficient, either from being attached to electronegative atom(s) or not having an octet. They must have an empty orbital willing to accept the electron pair. H+ has empty 1s orbital. B in BF3 has empty 2p orbital and an incomplete octet. Many small, highly charged metal cations have empty orbitals they can use to accept electrons. Atoms that are attached to highly electronegative atoms and have multiple bonds can be Lewis acids.

Lewis Acid–Base Reactions
The base donates a pair of electrons to the acid. It generally results in a covalent bond forming H3N: + BF3  H3N─BF3 The product that forms is called an adduct. Arrhenius and Brønsted–Lowry acid–base reactions are also Lewis.

Examples of Lewis Acid–Base Reactions

Examples of Lewis Acid–Base Reactions
Ag+(aq) :NH3(aq)  Ag(NH3)2+(aq) Lewis Acid Base Adduct

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
U.S. Fuel Consumption Over 85% of the energy use in the United States comes from the combustion of fossil fuels. Oil, natural gas, coal Combustion of fossil fuels produces CO2. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) Natural fossil fuels also contain small amounts of S that burn to produce SO2(g). S(s) + O2(g) → SO2(g) The high temperatures of combustion allow N2(g) in the air to combine with O2(g) to form oxides of nitrogen. N2(g) + 2 O2(g) → 2 NO2(g)

pH of Rain in Different Regions
Figure pg 740

What Is Acid Rain? Natural rain water has a pH of 5.6.
Naturally slightly acidic due mainly to CO2 Rain water with a pH lower than 5.6 is called acid rain. Acid rain is linked to damage in ecosystems and structures.

What Causes Acid Rain? Many natural and pollutant gases dissolved in the air are nonmetal oxides. CO2, SO2, NO2 Nonmetal oxides are acidic. CO2(g) + H2O(l)  H2CO3(aq) 2 SO2(g) + O2(g) + 2 H2O(l)  2 H2SO4(aq) 4 NO2(g) + O2(g) + 2 H2O(l)  4 HNO3(aq) Processes that produce nonmetal oxide gases as waste increase the acidity of the rain. Natural – volcanoes and some bacterial action Man made – combustion of fuel

Damage from Acid Rain Acids react with metals and materials that contain carbonates. Acid rain damages bridges, cars, and other metallic structures. Acid rain damages buildings and other structures made of limestone or cement. Acidifying lakes affects aquatic life. Soil acidity causes more dissolving of minerals and leaching more minerals from soil, making it difficult for trees.

Acid Rain Legislation 1990 Clean Air Act attacks acid rain.
Forces utilities to reduce SO2 The result is acid rain in the Northeast stabilized and beginning to be reduced.

Chapter 15 Acids and Bases

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