Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus

Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus
fruits contain citric acid. React with certain metals to produce hydrogen gas. 2HCl (aq) + Mg (s) MgCl2 (aq) + H2 (g) React with carbonates and bicarbonates to produce carbon dioxide gas 2HCl (aq) + CaCO3 (s) CaCl2 (aq) + CO2 (g) + H2O (l) Aqueous acid solutions conduct electricity.

Bases Have a bitter taste. Feel slippery. Many soaps contain bases.
Cause color changes in plant dyes. Aqueous base solutions conduct electricity.

Pg 129

Common Household acids and bases.

Types of Chemical Reactions
Acid-Base Reactions The Arrhenius Concept The Arrhenius concept defines acids as substances that produce hydrogen ions, H+, when dissolved in water. An example is nitric acid, HNO3, a molecular substance that dissolves in water to give H+ and NO3-.

Acid-Base Reactions The Arrhenius Concept The Arrhenius concept defines bases as substances that produce hydroxide ions, OH-, when dissolved in water. An example is sodium hydroxide, NaOH, an ionic substance that dissolves in water to give sodium ions and hydroxide ions.

Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water

Hydronium ion, hydrated proton, H3O+

Acid-Base Reactions The Brønsted-Lowry Concept The Brønsted-Lowry concept of acids and bases involves the transfer of a proton (H+) from the acid to the base. In this view, acid-base reactions are proton-transfer reactions.

A Brønsted acid is a proton donor A Brønsted base is a proton acceptor
A Brønsted acid must contain at least one ionizable proton!

Acid-Base Reactions The Brønsted-Lowry Concept The Brønsted-Lowry concept defines an acid as the species (molecule or ion) that donates a proton (H+) to another species in a proton-transfer reaction. A base is defined as the species (molecule or ion) that accepts the proton (H+) in a proton-transfer reaction.

Acid-Base Reactions The Brønsted-Lowry Concept In the reaction of ammonia with water, H+ the H2O molecule is the acid because it donates a proton. The NH3 molecule is a base, because it accepts a proton.

Acid-Base Reactions The Brønsted-Lowry Concept The H+(aq) ion associates itself with water to form H3O+(aq). This “mode of transportation” for the H+ ion is called the hydronium ion.

Acid-Base Reactions The Brønsted-Lowry Concept The dissolution of nitric acid, HNO3, in water is therefore a proton-transfer reaction where HNO3 is an acid (proton donor) and H2O is a base (proton acceptor). H+

Acid-Base Reactions In summary, the Arrhenius concept and the Brønsted-Lowry concept are essentially the same in aqueous solution. The Arrhenius concept acid: proton (H+) donor base: hydroxide ion (OH-) donor

Acid-Base Reactions In summary, the Arrhenius concept and the Brønsted-Lowry concept are essentially the same in aqueous solution. The Brønsted-Lowry concept acid: proton (H+) donor base: proton (H+) acceptor

CH3COO- (aq) + H+ (aq) CH3COOH (aq) Brønsted base
Identify each of the following species as a Brønsted acid, base, or both. (a) HI, (b) CH3COO-, (c) H2PO4- HI (aq) H+ (aq) + I- (aq) Brønsted acid CH3COO- (aq) + H+ (aq) CH3COOH (aq) Brønsted base H2PO4- (aq) H+ (aq) + HPO42- (aq) Brønsted acid H2PO4- (aq) + H+ (aq) H3PO4 (aq) Brønsted base

Brønsted-Lowry Concept of Acids and Bases
Some species can act as an acid or a base. An amphoteric species is a species that can act either as an acid or a base (it can gain or lose a proton). For example, HCO3- acts as a proton donor (an acid) in the presence of OH- H+ 2

Some species can act as an acid or a base. An amphoteric species is a species that can act either as an acid or a base (it can gain or lose a proton). Alternatively, HCO3 can act as a proton acceptor (a base) in the presence of HF. H+ 2

The amphoteric characteristic of water is important in the acid-base properties of aqueous solutions. Water reacts as an acid with the base NH3. H+ 2

The amphoteric characteristic of water is important in the acid-base properties of aqueous solutions. Water can also react as a base with the acid HF. H+ 2

Copyright © Cengage Learning. All rights reserved
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 Ionization Equilibrium
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Definition of Strong Acid and Weak Acid
Ionization equilibrium lies far to the right. Yields a weak conjugate base. Weak acid: Ionization equilibrium lies far to the left. Weaker the acid, stronger its conjugate base. Copyright © Cengage Learning. All rights reserved

Strong Acid Weak Acid

Various Ways to Describe Acid Strength
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Self-ionization of Water
Self-ionization is a reaction in which two like molecules react to give ions. In the case of water, the following equilibrium is established. The equilibrium-constant expression for this system is: 2

[ ] - Acid-Base Properties of Water autoionization of water
H2O (l) H+ (aq) + OH- (aq) autoionization of water O H + - [ ] conjugate acid base H2O + H2O H3O+ + OH- acid conjugate base

Self-ionization is a reaction in which two like molecules react to give ions. The concentration of ions is extremely small, so the concentration of H2O remains essentially constant. This gives: constant 2

Self-ionization is a reaction in which two like molecules react to give ions. We call the equilibrium value for the ion product [H3O+][OH-] the ion-product constant for water, which is written Kw. At 25 oC, the value of Kw is 1.0 x Like any equilibrium constant, Kw varies with temperature. 2

These ions are produced in equal numbers in pure water, so if we let x = [H+] = [OH-] Thus, the concentrations of H+ and OH- in pure water are both 1.0 x 10-7 M. If you add acid or base to water they are no longer equal but the Kw expression still holds. 2

Review: Water as an Acid and a Base
Water is amphoteric: Behaves either as an acid or as a base. At 25°C: Kw = [H+][OH–] = 1.0 × 10–14 No matter what the solution contains, the product of [H+] and [OH–] must always equal 1.0 × 10–14 at 25°C. Copyright © Cengage Learning. All rights reserved

Solutions of Strong Acid or Base
By dissolving substances in water, you can alter the concentrations of H+(aq) and OH-(aq). In a neutral solution, the concentrations of H+(aq) and OH-(aq) are equal, as they are in pure water. In an acidic solution, the concentration of H+(aq) is greater than that of OH-(aq). In a basic solution, the concentration of OH-(aq) is greater than that of H+(aq). 2

At 25 oC, you observe the following conditions. In an acidic solution, [H+] > 1.0 x 10-7 M. In a neutral solution, [H+] = 1.0 x 10-7 M. In a basic solution, [H+] < 1.0 x 10-7 M. 2

The pH of a Solution Although you can quantitatively describe the acidity of a solution by its [H+], it is often more convenient to give acidity in terms of pH. The pH of a solution is defined as the negative logarithm of the molar hydrogen-ion concentration. 2

The pH of a Solution For a solution in which the hydrogen-ion concentration is 1.0 x 10-3, the pH is: Note that the number of decimal places in the pH equals the number of significant figures in the hydrogen-ion concentration. 2

The pH of a Solution In a neutral solution, whose hydrogen-ion concentration is 1.0 x 10-7, the pH = 7.00. For acidic solutions, the hydrogen-ion concentration is greater than 1.0 x 10-7, so the pH is less than 7.00. Similarly, a basic solution has a pH greater than 7.00. 2

The pH Scale 2

A Problem to Consider A sample of orange juice has a hydrogen-ion concentration of 2.9 x 10-4 M. What is the pH? 2

A Problem to Consider The pH of human arterial blood is What is the hydrogen-ion concentration? 2

The pH of a Solution A measurement of the hydroxide ion concentration, similar to pH, is the pOH. The pOH of a solution is defined as the negative logarithm of the molar hydroxide-ion concentration. 2

The pH of a Solution A measurement of the hydroxide ion concentration, similar to pH, is the pOH. Then because Kw = [H+][OH-] = 1.0 x at 25 oC, you can show that 2

A Problem to Consider An ammonia solution has a hydroxide-ion concentration of 1.9 x 10-3 M. What is the pH of the solution? You first calculate the pOH: Then the pH is: 2

The pH of rainwater collected in a certain region of the northeastern United States on a particular day was What is the H+ ion concentration of the rainwater? pH = -log [H+] [H+] = 10-pH = = 1.5 x 10-5 M The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood? pH + pOH = 14.00 pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60 pH = – pOH = – 6.60 = 7.40

The pH of a Solution The pH of a solution can accurately be measured using a pH meter 2

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! 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

What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation. Start 0.002 M 0.0 M 0.0 M HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq) End 0.0 M 0.002 M 0.002 M pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7 What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution? Ba(OH)2 is a strong base – 100% dissociation. Start 0.018 M 0.0 M 0.0 M Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq) End 0.0 M 0.018 M 0.036 M pH = – pOH = log(0.036) = 12.6

As an example, calculate the concentration of OH- ion in 0.10 M HCl. Because you started with 0.10 M HCl (a strong acid) the reaction will produce 0.10 M H+(aq). Substituting [H+]=0.10 into the ion-product expression, we get: 2

Similarly, in a solution of a strong base you can normally ignore the self-ionization of water as a source of OH-(aq). The OH-(aq) concentration is usually determined by the strong base concentration. However, the self-ionization still exists and is responsible for a small concentration of H+ ion. 2

As an example, calculate the concentration of H+ ion in M NaOH. Because you started with M NaOH (a strong base) the reaction will produce M OH-(aq). Substituting [OH-]=0.010 into the ion-product expression, we get: 2

Strong Electrolyte – 100% dissociation
NaCl (s) Na+ (aq) + Cl- (aq) H2O Weak Electrolyte – not completely dissociated CH3COOH CH3COO- (aq) + H+ (aq) Strong Acids are strong electrolytes HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq) HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq) HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq) H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)

Weak Acids are weak electrolytes
HF (aq) + H2O (l) H3O+ (aq) + F- (aq) HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq) HSO4- (aq) + H2O (l) H3O+ (aq) + SO42- (aq) H2O (l) + H2O (l) H3O+ (aq) + OH- (aq) Strong Bases are strong electrolytes NaOH (s) Na+ (aq) + OH- (aq) H2O KOH (s) K+ (aq) + OH- (aq) H2O Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq) H2O

Weak Bases are weak electrolytes
F- (aq) + H2O (l) OH- (aq) + HF (aq) NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq) Conjugate acid-base pairs: The conjugate base of a strong acid has no measurable strength. H3O+ is the strongest acid that can exist in aqueous solution. The OH- ion is the strongest base that can exist in aqueous solution.

Weak Acids (HA) and Acid Ionization Constants
HA (aq) + H2O (l) H3O+ (aq) + A- (aq) HA (aq) H+ (aq) + A- (aq) Ka = [H+][A-] [HA] Ka is the acid ionization constant weak acid strength Ka

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. CAUTION!!! Use the 5% rule to verify whether the approximation is valid. CAUTION!!! Calculate [H+] and pH. Copyright © Cengage Learning. All rights reserved

What is the pH of a 0.5 M HF solution (at 250C)?
Ka = [H+][F-] [HF] = 7.1 x 10-4 HF (aq) H+ (aq) + F- (aq) HF (aq) H+ (aq) + F- (aq) Initial (M) 0.50 0.00 0.00 Change (M) -x +x +x Equilibrium (M) x x x Ka = x2 x = 7.1 x 10-4 Ka << 1 0.50 – x  0.50 Ka  x2 0.50 = 7.1 x 10-4 x2 = 3.55 x 10-4 x = M [H+] = [F-] = M pH = -log [H+] = 1.72 [HF] = 0.50 – x = 0.48 M

When can I use the approximation?
Ka << 1 0.50 – x  0.50 When x is less than 5% of the value from which it is subtracted. 0.019 M 0.50 M x 100% = 3.8% Less than 5% Approximation ok. x = 0.019 What is the pH of a 0.05 M HF solution (at 250C)? Ka  x2 0.05 = 7.1 x 10-4 x = M 0.006 M 0.05 M x 100% = 12% More than 5% Approximation not ok. Must solve for x exactly using quadratic equation or method of successive approximation.

What is the pH of a 0.122 M monoprotic acid whose
Ka is 5.7 x 10-4? HA (aq) H+ (aq) + A- (aq) Initial (M) 0.122 0.00 0.00 Change (M) -x +x +x Equilibrium (M) x x x Ka = x2 x = 5.7 x 10-4 Ka << 1 0.122 – x  0.122 Ka  x2 0.122 = 5.7 x 10-4 x2 = 6.95 x 10-5 x = M M 0.122 M x 100% = 6.8% More than 5% Approximation not ok.

Ka = x2 x = 5.7 x 10-4 x x – 6.95 x 10-5 = 0 -b ± b2 – 4ac  2a x = ax2 + bx + c =0 x = x = HA (aq) H+ (aq) + A- (aq) Initial (M) Change (M) Equilibrium (M) 0.122 0.00 -x +x x x [H+] = x = M pH = -log[H+] = 2.09

Review: Solving weak acid ionization problems:
Identify the major species that can affect the pH. In most cases, you can ignore the autoionization of water. Ignore [OH-] because it is determined by [H+]. Use ICE to express the equilibrium concentrations in terms of single unknown x. Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. Calculate concentrations of all species and/or pH of the solution.

A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A M solution of nicotinic acid has a pH of 3.39 at 25 oC. Calculate the acid-ionization constant for this acid at 25 oC. It is important to realize that the solution was made M in nicotinic acid, however, some molecules ionize making the equilibrium concentration of nicotinic acid less than M. We will abbreviate the formula for nicotinic acid as HNic. 2

A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A M solution of nicotinic acid has a pH of 3.39 at 25 oC. Calculate the acid-ionization constant for this acid at 25 oC. Let x be the moles per liter of product formed. Initial 0.012 Change -x +x Equilibrium 0.012-x x 2

A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A M solution of nicotinic acid has a pH of 3.39 at 25 oC. Calculate the acid-ionization constant for this acid at 25 oC. The equilibrium-constant expression is: 2

A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A M solution of nicotinic acid has a pH of 3.39 at 25 oC. Calculate the acid-ionization constant for this acid at 25 oC. Substituting the expressions for the equilibrium concentrations, we get 2

A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A M solution of nicotinic acid has a pH of 3.39 at 25 oC. Calculate the acid-ionization constant for this acid at 25 oC. We can obtain the value of x from the given pH. 2

A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A M solution of nicotinic acid has a pH of 3.39 at 25 oC. Calculate the acid-ionization constant for this acid at 25 oC. Substitute this value of x in our equilibrium expression. Note first, however, that the concentration of unionized acid remains virtually unchanged. 2

A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A M solution of nicotinic acid has a pH of 3.39 at 25 oC. Calculate the acid-ionization constant for this acid at 25 oC. Substitute this value of x in our equilibrium expression. 2

A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A M solution of nicotinic acid has a pH of 3.39 at 25 oC. Calculate the acid-ionization constant for this acid at 25 oC. To obtain the degree of dissociation: The percent ionization is obtained by multiplying by 100, which gives 3.4%. 2

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

Calculations With Ka How do you know when you can use this simplifying assumption (non-quadratic formula)? It can be shown that if the acid concentration, Ca, divided by the Ka exceeds 100, that is, then this simplifying assumption of ignoring the subtracted x gives an acceptable error of less than 5%. 2

Calculations With Ka How do you know when you can use this simplifying assumption? If the simplifying assumption is not valid, you can solve the equilibrium equation exactly by using the quadratic equation. The next example illustrates this with a solution of aspirin (acetylsalicylic acid), HC9H7O4, a common headache remedy. 2

A Problem To Consider What is the pH at 25 oC of a solution obtained by dissolving g of acetylsalicylic acid (aspirin), HC9H7O4, in L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25 oC. The molar mass of HC9H7O4 is g. From this we find that the sample contained mol of the acid. 2

A Problem To Consider What is the pH at 25 oC of a solution obtained by dissolving g of acetylsalicylic acid (aspirin), HC9H7O4, in L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25 oC. The molar mass of HC9H7O4 is g. Hence, the concentration of the acetylsalicylic acid is mol/0.500 L = M (Retain two significant figures, the same number of significant figures in Ka). 2

A Problem To Consider What is the pH at 25 oC of a solution obtained by dissolving g of acetylsalicylic acid (aspirin), HC9H7O4, in L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25 oC. Note that which is less than 100, so we must solve the equilibrium equation exactly. 2

A Problem To Consider What is the pH at 25 oC of a solution obtained by dissolving g of acetylsalicylic acid (aspirin), HC9H7O4, in L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25 oC. We will abbreviate the formula for acetylsalicylic acid as HAcs and let x be the amount of H3O+ formed per liter. The amount of acetylsalicylate ion is also x mol; the amount of nonionized acetylsalicylic acid is ( x) mol. 2

A Problem To Consider What is the pH at 25 oC of a solution obtained by dissolving g of acetylsalicylic acid (aspirin), HC9H7O4, in L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25 oC. These data are summarized below. Initial 0.0036 Change -x +x Equilibrium x x 2

A Problem To Consider What is the pH at 25 oC of a solution obtained by dissolving g of acetylsalicylic acid (aspirin), HC9H7O4, in L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25 oC. The equilibrium constant expression is 2

A Problem To Consider What is the pH at 25 oC of a solution obtained by dissolving g of acetylsalicylic acid (aspirin), HC9H7O4, in L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25 oC. If we substitute the equilibrium concentrations and the Ka into the equilibrium constant expression, we get 2

A Problem To Consider What is the pH at 25 oC of a solution obtained by dissolving g of acetylsalicylic acid (aspirin), HC9H7O4, in L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25 oC. You can solve this equation exactly by using the quadratic formula. Rearranging the preceding equation to put it in the form ax2 + bx + c = 0, we get 2

A Problem To Consider What is the pH at 25 oC of a solution obtained by dissolving g of acetylsalicylic acid (aspirin), HC9H7O4, in L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25 oC. Now substitute into the quadratic formula. 2

A Problem To Consider What is the pH at 25 oC of a solution obtained by dissolving g of acetylsalicylic acid (aspirin), HC9H7O4, in L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25 oC. Now substitute into the quadratic formula. The lower sign in ± gives a negative root which we can ignore 2

A Problem To Consider What is the pH at 25 oC of a solution obtained by dissolving g of acetylsalicylic acid (aspirin), HC9H7O4, in L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25 oC. Taking the upper sign, we get Now we can calculate the pH. 2

Weak Bases and Base Ionization Constants
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) Kb = [NH4+][OH-] [NH3] Kb is the base ionization constant weak base strength Kb Solve weak base problems like weak acids except solve for [OH-] instead of [H+].

Ionization Constants of Conjugate Acid-Base Pairs
HA (aq) H+ (aq) + A- (aq) Ka A- (aq) + H2O (l) OH- (aq) + HA (aq) Kb H2O (l) H+ (aq) + OH- (aq) Kw KaKb = Kw Weak Acid and Its Conjugate Base Ka = Kw Kb Kb = Kw Ka

Base-Ionization Equilibria
Equilibria involving weak bases are treated similarly to those for weak acids. Ammonia, for example, ionizes in water as follows. The corresponding equilibrium constant is: 2

Equilibria involving weak bases are treated similarly to those for weak acids. Ammonia, for example, ionizes in water as follows. The concentration of water is nearly constant. 2

Equilibria involving weak bases are treated similarly to those for weak acids. In general, a weak base B with the base ionization has a base ionization constant equal to 2

A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. As before, we will follow the three steps in solving an equilibrium. Write the equation and make a table of concentrations. Set up the equilibrium constant expression. Solve for x = [OH-]. 2

A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. Pyridine ionizes by picking up a proton from water (as ammonia does). Initial 0.20 Change -x +x Equilibrium 0.20-x x 2

A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. Note that which is much greater than 100, so we may use the simplifying assumption that (0.20-x)  (0.20). 2

A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. The equilibrium expression is 2

A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. If we substitute the equilibrium concentrations and the Kb into the equilibrium constant expression, we get 2

A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. Using our simplifying assumption that the x in the denominator is negligible, we get 2

A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. Solving for x we get 2

A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. Solving for pOH Since pH + pOH = 14.00 2

Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. Sulfuric acid, for example, can lose two protons in aqueous solution. The first proton is lost completely followed by a weak ionization of the hydrogen sulfate ion, HSO4-. 2

Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. For a weak diprotic acid like carbonic acid, H2CO3, two simultaneous equilibria must be considered. 2

Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. Each equilibrium has an associated acid-ionization constant. For the loss of the first proton 2

Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. Each equilibrium has an associated acid-ionization constant. For the loss of the second proton 2

Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. In general, the second ionization constant, Ka2, for a polyprotic acid is smaller than the first ionization constant, Ka1. In the case of a triprotic acid, such as H3PO4, the third ionization constant, Ka3, is smaller than the second one, Ka2. 2

Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions. However, reasonable assumptions can be made that simplify these calculations as we show in the next example. 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x For diprotic acids, Ka2 is so much smaller than Ka1 that the smaller amount of hydronium ion produced in the second reaction can be neglected. 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x The pH can be determined by simply solving the equilibrium problem posed by the first ionization. 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x If we abbreviate the formula for ascorbic acid as H2Asc, then the first ionization is: 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x Initial 0.10 Change -x +x Equilibrium 0.10-x x 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x The equilibrium constant expression is 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x Substituting into the equilibrium expression 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x Assuming that x is much smaller than 0.10, you get 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x The hydronium ion concentration is M, so 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x The ascorbate ion, Asc2-, which we will call y, is produced only in the second ionization of H2Asc. 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x Assume the starting concentrations for HAsc- and H3O+ to be those from the first equilibrium. 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x Initial 0.0028 Change -y +y Equilibrium x y y 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x The equilibrium constant expression is 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x Substituting into the equilibrium expression 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x Assuming y is much smaller than , the equation simplifies to 2

A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x Hence, The concentration of the ascorbate ion equals Ka2. 2

Review: Polyprotic Acids
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

Acid-Base Properties of Salts
Neutral Solutions: Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strong acid (e.g. Cl-, Br-, and NO3-). NaCl (s) Na+ (aq) + Cl- (aq) H2O Basic Solutions: Salts derived from a strong base and a weak acid. NaCH3COOH (s) Na+ (aq) + CH3COO- (aq) H2O CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq)

Acid-Base Properties of Salts
Acid Solutions: Salts derived from a strong acid and a weak base. NH4Cl (s) NH4+ (aq) + Cl- (aq) H2O NH4+ (aq) NH3 (aq) + H+ (aq) Salts with small, highly charged metal cations (e.g. Al3+, Cr3+, and Be2+) and the conjugate base of a strong acid. Al(H2O)6 (aq) Al(OH)(H2O)5 (aq) + H+ (aq) 3+ 2+

Solutions in which both the cation and the anion hydrolyze:
Acid-Base Properties of Salts Solutions in which both the cation and the anion hydrolyze: Kb for the anion > Ka for the cation, solution will be basic Kb for the anion < Ka for the cation, solution will be acidic Kb for the anion  Ka for the cation, solution will be neutral

Acid-Base Properties of a Salt Solution
One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. Consider a solution of sodium cyanide, NaCN. A 0.1 M solution has a pH of 11.1 and is therefore fairly basic. 2

One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. Sodium ion, Na+, is unreactive with water, but the cyanide ion, CN-, reacts to produce HCN and OH-. 2

One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. From the Brønsted-Lowry point of view, the CN- ion acts as a base, because it accepts a proton from H2O. 2

One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. You can also see that OH- ion is a product, so you would expect the solution to have a basic pH. This explains why NaCN solutions are basic. The reaction of the CN- ion with water is referred to as the hydrolysis of CN-. 2

The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. The CN- ion hydrolyzes to give the conjugate acid and hydroxide. 2

The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. The hydrolysis reaction for CN- has the form of a base ionization so you writ the Kb expression for it. 2

The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. The NH4+ ion hydrolyzes to the conjugate base (NH3) and hydronium ion. 2

The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. This equation has the form of an acid ionization so you write the Ka expression for it. 2

Predicting Whether a Salt is Acidic, Basic, or Neutral
How can you predict whether a particular salt will be acidic, basic, or neutral? The Brønsted-Lowry concept illustrates the inverse relationship in the strengths of conjugate acid-base pairs. Consequently, the anions of weak acids (poor proton donors) are good proton acceptors. Anions of weak acids therefore, are basic. 2

How can you predict whether a particular salt will be acidic, basic, or neutral? One the other hand, the anions of strong acids (good proton donors) have virtually no basic character, that is, they do not hydrolyze. For example, the Cl- ion, which is conjugate to the strong acid HCl, shows no appreciable reaction with water. 2

How can you predict whether a particular salt will be acidic, basic, or neutral? Conversely, the cations of weak bases are acidic. One the other hand, the cations of strong bases have virtually no acidic character, that is, they do not hydrolyze. For example, 2

To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt. Consider potassium acetate, KC2H3O2. The potassium ion is the cation of a strong base (KOH) and does not hydrolyze. 2

To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt. Consider potassium acetate, KC2H3O2. The acetate ion, however, is the anion of a weak acid (HC2H3O2) and is basic. A solution of potassium acetate is predicted to be basic. 2

These rules apply to normal salts (those in which the anion has no acidic hydrogen) A salt of a strong base and a strong acid. The salt has no hydrolyzable ions and so gives a neutral aqueous solution. An example is NaCl. 2

These rules apply to normal salts (those in which the anion has no acidic hydrogen) A salt of a strong base and a weak acid. The anion of the salt is the conjugate of the weak acid. It hydrolyzes to give a basic solution. An example is NaCN. 2

These rules apply to normal salts (those in which the anion has no acidic hydrogen) A salt of a weak base and a strong acid. The cation of the salt is the conjugate of the weak base. It hydrolyzes to give an acidic solution. An example is NH4Cl. 2

These rules apply to normal salts (those in which the anion has no acidic hydrogen) A salt of a weak base and a weak acid. Both ions hydrolyze. You must compare the Ka of the cation with the Kb of the anion. If the Ka of the cation is larger the solution is acidic. If the Kb of the anion is larger, the solution is basic. 2

The pH of a Salt Solution
To calculate the pH of a salt solution would require the Ka of the acidic cation or the Kb of the basic anion. The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species. Thus the Kb for CN- is related to the Ka for HCN. 2

To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-. Ka When these two reactions are added you get the ionization of water. Kw Kb 2

To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-. Ka Kb When two reactions are added, their equilibrium constants are multiplied. Kw 2

To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-. Ka Kb Therefore, Kw 2

For a solution of a salt in which only one ion hydrolyzes, the calculation of equilibrium composition follows that of weak acids and bases. The only difference is first obtaining the Ka or Kb for the ion that hydrolyzes. The next example illustrates the reasoning and calculations involved. 2

A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 oC? The Ka for HCN is 4.9 x Sodium cyanide gives Na+ ions and CN- ions in solution. Only the CN- ion hydrolyzes. 2

A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 oC? The Ka for HCN is 4.9 x The CN- ion is acting as a base, so first, we must calculate the Kb for CN-. Now we can proceed with the equilibrium calculation. 2

A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 oC? The Ka for HCN is 4.9 x Let x = [OH-] = [HCN], then substitute into the equilibrium expression. 2

A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 oC? The Ka for HCN is 4.9 x This gives 2

A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 oC? The Ka for HCN is 4.9 x Solving the equation, you find that Hence, As expected, the solution has a pH greater than 7.0. 2

Molecular Structure and Acid Strength
H X H+ + X- The stronger the bond The weaker the acid HF << HCl < HBr < HI

Two factors are important in determining the relative acid strengths. One is the polarity of the bond to which the hydrogen atom is attached. The H atom should have a partial positive charge: d+ d- The more polarized the bond, the more easily the proton is removed and the greater the acid strength. 2

Two factors are important in determining the relative acid strengths. The second factor is the strength of the bond. Or, in other words, how tightly the proton is held. This depends on the size of atom X. d+ d- The larger atom X, the weaker the bond and the greater the acid strength. 2

Consider a series of binary acids from a given column of elements. As you go down the column of elements, the radius increases markedly and the H-X bond strength decreases. You can predict the following order of acidic strength. 2

As you go across a row of elements, the polarity of the H-X bond becomes the dominant factor. As electronegativity increases going to the right, the polarity of the H-X bond increases and the acid strength increases. You can predict the following order of acidic strength. 2

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. Copyright © Cengage Learning. All rights reserved

Z O H O- + H+ d- d+ The O-H bond will be more polar and easier to break if: Z is very electronegative or Z is in a high oxidation state

1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number. Acid strength increases with increasing electronegativity of Z H O Cl O O • H O Br O O • Cl is more electronegative than Br HClO3 > HBrO3

2. Oxoacids having the same central atom (Z) but different numbers of attached groups. Acid strength increases as the oxidation number of Z increases. HClO4 > HClO3 > HClO2 > HClO

Consider the oxoacids. An oxoacid has the structure: The acidic H atom is always attached to an O atom, which in turn is attached to another atom Y. Bond polarity is the dominant factor in the relative strength of oxoacids. This, in turn, depends on the electronegativity of the atom Y. 2

Consider the oxoacids. An oxoacid has the structure: If the electronegativity of Y is large, then the O-H bond is relatively polar and the acid strength is greater. You can predict the following order of acidic strength. 2

Consider the oxoacids. An oxoacid has the structure: Other groups, such as O atoms or O-H groups, may be attached to Y. With each additional O atom, Y becomes effectively more electronegative. 2

Consider the oxoacids. An oxoacid has the structure: As a result, the H atom becomes more acidic. The acid strengths of the oxoacids of chlorine increase in the following order. 2

Consider polyprotic acids and their corresponding anions. Each successive H atom becomes more difficult to remove. Therefore the acid strength of a polyprotic acid and its anions decreases with increasing negative charge. 2

Several Series of Oxyacids and Their Ka Values

Comparison of Electronegativity of X and Ka Value

Oxides Acidic Oxides (Acid Anhydrides): O—X bond is strong and covalent. SO2, NO2, CO2 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. Copyright © Cengage Learning. All rights reserved

Oxides Basic Oxides (Basic Anhydrides): O—X bond is ionic. K2O, CaO 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. Copyright © Cengage Learning. All rights reserved

Oxides of the Representative Elements
In Their Highest Oxidation States CO2 (g) + H2O (l) H2CO3 (aq) N2O5 (g) + H2O (l) HNO3 (aq)

Definition of An Acid Arrhenius acid is a substance that produces H+ (H3O+) in water A Brønsted acid is a proton donor A Lewis acid is a substance that can accept a pair of electrons A Lewis base is a substance that can donate a pair of electrons + OH- • H O H • H+ acid base N H • H N H H + H+ + acid base

Lewis Concept of Acids and Bases
The reaction of boron trifluoride with ammonia is an example. : B F : B F N H : N H + Boron trifluoride accepts the electron pair, so it is a Lewis acid. Ammonia donates the electron pair, so it is the Lewis base. 2

Three Models for Acids and Bases

Chemistry In Action: Antacids and the Stomach pH Balance
NaHCO3 (aq) + HCl (aq) NaCl (aq) + H2O (l) + CO2 (g) Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l)

Worked Example 15.1

Worked Example 15.2

Worked Example 15.6

Worked Example 15.7

Worked Example 15.8a

Worked Example 15.9

Worked Example 15.10

Worked Example 15.11b

Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus

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Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus

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Presentation on theme: "Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus"— Presentation transcript:

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