Chapter 11. Acids and Bases. Lecture Presentation. Karen C. Timberlake Pearson Education, Inc.

Lecture Presentation Chapter 11 Acids and Bases

Chapter 11 Acids and Bases Clinical laboratory technicians prepare specimens for the detection of cancerous tumors and type blood samples for transfusions. They must also interpret and analyze the test results, which are then passed on to the physician.

Chapter 11 Readiness Key Math Skills Solving Equations (1.4D) Writing Numbers in Scientific Notation (1.4F)

Chapter 11 Readiness Core Chemistry Skills Writing Ionic Formulas (6.2) Balancing a Chemical Equation (7.1) Using Concentrations as a Conversion Factor (9.4) Writing the Equilibrium Constant Expression (10.3) Calculating Equilibrium Concentrations (10.4) Using Le Châtelier s Principle (10.5)

Chapter 11 Acids and Bases A soft drink contains phosphoric acid (H 3 PO 4 ) and carbonic acid (H 2 CO 3 ). Learning Goal Describe and name acids and bases.

Arrhenius Acids Arrhenius acids produce hydrogen ions (H + ) when they dissolve in water. H 2 O(l) HCl(g) H + (aq) + Cl (aq) are also electrolytes, because they produce H + in water. have a sour taste. turn blue litmus red. corrode some metals.

Naming Acids Acids with a hydrogen ion (H + ) and a nonmetal (or CN ) ion are named with the prefix hydro and end with ic acid. HCl(aq) hydrochloric acid Acids with a hydrogen ion (H + ) and a polyatomic ion are named by changing the end of the name of the polyatomic ion from ate to ic acid or ite to ous acid ClO 3 chlorate ClO 2 chlorite HClO 3 chloric acid HClO 2 chlorous acid

Study Check Select the correct name for each of the following acids: 1. HBr A. bromic acid B. bromous acid C. hydrobromic acid 2. H 2 CO 3 A. carbonic acid B. hydrocarbonic acid C. carbonous acid 3. HBrO 2 A. bromic acid B. hydrobromous acid C. bromous acid

Arrhenius Bases Arrhenius bases produce hydroxide ions (OH ) in water. taste bitter or chalky. are also electrolytes, because they produce hydroxide ions (OH ) in water. feel soapy and slippery. turn litmus indicator paper blue and phenolphthalein indicator pink. An Arrhenius base produces cations and OH anions in an aqueous solution.

Naming Bases Typical Arrhenius bases are named as hydroxides. NaOH sodium hydroxide KOH potassium hydroxide Ba(OH) 2 barium hydroxide Al(OH) 3 aluminum hydroxide Calcium hydroxide, Ca(OH) 2, is used in the food industry to produce beverages, and in dentistry as a filler for root canals.

Characteristics of Acids and Bases

Study Check Match the formulas of acids and bases with their names. 1. HNO 2 A. iodic acid 2. Ca(OH) 2 B. sulfuric acid 3. H 2 SO 4 C. sodium hydroxide 4. HIO 3 D. nitrous acid 5. NaOH E. calcium hydroxide

11.2 Brønsted Lowry Acids and Bases According to the Brønsted Lowry theory, an acid is a substance that donates H +. a base is a substance that accepts H +. Learning Goal Identify conjugate acid base pairs for Brønsted Lowry acids and bases.

NH3, a Brønsted Lowry Base In the reaction of ammonia and water, NH 3 acts as the base that accepts H +. H 2 O acts as the acid that donates H +. Because the nitrogen atom of NH 3 has a stronger attraction for H + than oxygen, water acts as an acid by donating H +.

Study Check In each of the following equations, identify the Brønsted Lowry acid and base in the reactants: A. HNO 3 (aq) + H 2 O(l) H 3 O + (aq) + NO 3 (aq) B. HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq)

Study Check Identify each as a characteristic of A. an acid or B. a base. 1. has a sour taste 2. produces OH in aqueous solutions 3. has a chalky taste 4. is an electrolyte 5. produces H + in aqueous solutions

Conjugate Acid Base Pairs In any acid base reaction, there are two conjugate acid base pairs. Each pair is related by the loss and gain of H +. One pair occurs in the forward direction. One pair occurs in the reverse direction. Acid and conjugate base pair 1 HA + B A + BH + Base and conjugate acid pair 2

Conjugate Acid Base Pairs In this acid base reaction, the first conjugate acid base pair is HF, which donates H + to form its conjugate base, F. the other conjugate acid base pair is H 2 O, which accepts H + to form its conjugate acid, H 3 O +. each pair is related by a loss and gain of H +.

Conjugate Acid Base Pairs In the reaction of NH 3 and H 2 O, one conjugate acid base pair is NH 3 /NH 4+. the other conjugate acid base pair is H 2 O/H 3 O +. Core Chemistry Skill Identifying Conjugate Acid Base Pairs

Study Check 1. Write the conjugate base for each of the following acids: A. HBr B. H 2 S C. H 2 CO 3 2. Write the conjugate acid of each of the following bases: A. NO 2 B. NH 3 C. OH

Study Check Identify the sets that contain acid base conjugate pairs. 1. HNO 2, NO 2 2. H 2 CO 3, CO 2 3 3. HCl, ClO 4 4. HS, H 2 S 5. NH 3, NH + 4

Amphoteric Substances Substances that can act as both acids and bases are amphoteric or amphiprotic. For water, the most common amphoteric substance, the acidic or basic behavior depends on the other reactant. Water donates H + when it reacts with a stronger base. Water accepts H + when it reacts with a stronger acid.

Guide to Writing Conjugate Acid Base Pairs

Study Check Identify the conjugate acid base pairs in the following reaction: HNO 3 (aq) + NH 3 (aq) NO 3 (aq) + NH 4+ (aq)

Solution Identify the conjugate acid base pairs in the following reaction: HNO 3 (aq) + NH 3 (aq) NO 3 (aq) + NH 4+ (aq) STEP 1 Identify the reactant that loses H + as the acid. In the reaction, HNO 3 donates H + to NH 3. STEP 2 Identify the reactant that gains H + as the base. In the reaction, NH 3 gains H + to form NH 4+. Thus, NH 3 is the base and NH 4+ is its conjugate acid. HBr is the acid and Br is its conjugate base.

Solution Identify the conjugate acid base pairs in the following reaction: HNO 3 (aq) + NH 3 (aq) NO 3 (aq) + NH 4+ (aq) STEP 3 Write the conjugate acid base pairs. HBr/Br is the acid and conjugate base pair. NH 3 /NH 4+ is the base and conjugate acid pair.

11.3 Strengths of Acids and Bases Weak acids only partially dissociate in water. Hydrofluoric acid, HF, is the only halogen that forms a weak acid. Learning Goal Write equations for the dissociation of strong and weak acids; identify the direction of reaction.

Strong and Weak Acids A strong acid completely ionizes (100%) in aqueous solutions. HCl(g) + H 2 O(l) H 3 O + (aq) + Cl (aq) A weak acid dissociates only slightly in water to form a few ions in aqueous solutions. H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 (aq)

Strong Acids In water, the dissolved molecules of HA, a strong acid, dissociate into ions 100%. produce large concentrations of H 3 O + and the anion (A ). The strong acid HCl dissociates completely into ions: HCl(g) + H 2 O(l) H 3 O + (aq) + Cl (aq)

Weak Acids In weak acids, only a few molecules dissociate. Most of the weak acid remains as the undissociated (molecular) form of the acid. The concentrations of H 3 O + and the anion (A ) are small. H 2 CO 3 is a weak acid: H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 (aq)

Strong and Weak Acid Dissociation In an HCl solution, the strong acid HCl dissociates 100% to form H + and Cl. A solution of the weak acid HC 2 H 3 O 2 contains mostly molecules of HC 2 H 3 O 2 and a few ions of H + and C 2 H 3 O 2.

Strong and Weak Acid Dissociation Figure 11.2 After dissociation in water, (a) the strong acid HI has high concentrations of H 3 O + and I, and (b) the weak acid HF has a high concentration of HF and low concentrations of H 3 O + and F.

Diprotic Acids: Carbonic Acid Some weak acids, such as carbonic acid, are diprotic acids that have two H +, which dissociate one at a time. H 2 CO 3 (aq) + H 2 O(l) H 3 O + (aq) + HCO 3 (aq) Because HCO 3 is also a weak acid, a second dissociation can take place to produce another hydronium ion and the carbonate ion, CO 3 2. HCO 3 (aq) + H 2 O(l) H 3 O + (aq) + CO 3 2 (aq)

Diprotic Acids: Sulfuric Acid Some strong acids, such as sulfuric acid, are diprotic acids that have two H +, which dissociate one at a time. H 2 SO 4 (aq) + H 2 O(l) H 3 O + (aq) + HSO 4 (aq) Because HSO 4 is a weak acid, a second dissociation can take place to produce another H + and the sulfate ion, SO 4 2. HSO 4 (aq) + H 2 O(l) H 3 O + (aq) + SO 4 2 (aq)

Strong Bases Strong bases as strong electrolytes are formed from metals of Groups 1A (1) and 2A (2). include LiOH, NaOH, KOH, Ba(OH) 2, Sr(OH) 2, and Ca(OH) 2. dissociate completely in water. KOH(s) K + (aq) + OH (aq) are found in household products used to remove grease and unclog drains.

Weak Bases Weak bases are weak electrolytes that are poor acceptors of H + ions. produce very few ions in solution. include ammonia. NH 3 (g) + H 2 O(l) NH 4+ (aq) + OH (aq) Ammonia Ammonium hydroxide

Strong and Weak Bases Strong Bases Lithium hydroxide LiOH Sodium hydroxide NaOH Potassium hydroxide KOH Rubidium hydroxide RbOH Cesium hydroxide CsOH Calcium hydroxide Ca(OH) 2 * Strontium hydroxide Sr(OH) 2 * Barium hydroxide Ba(OH) 2 * *Low solubility, but they dissociate completely Bases in Household Products Weak Bases Window cleaner, ammonia, NH 3 Bleach, NaOCl Laundry detergent, Na 2 CO 3, Na 3 PO 4 Toothpaste and baking soda, NaHC 3 Baking powder, scouring powder, Na 2 CO 3 Lime for lawns and agriculture, CaCO 3 Laxatives, antacids, Mg(OH) 2, Al(OH) 3 Strong Bases Drain cleaner, oven cleaner, NaOH

Direction of Reaction Strong acids have weak conjugate bases that do not readily accept H +. As the strength of the acid decreases, the strength of its conjugate base increases. In any acid base reaction, there are two acids and two bases. However, one acid is stronger than the other acid, and one base is stronger than the other base. By comparing their relative strengths, we can determine the direction of the reaction.

Direction of Reaction: H 2 SO 4 Sulfuric acid, H 2 SO 4, is a strong acid that readily gives up H + to water. H 2 SO 4 (aq) + H 2 O(l) H 3 O + (aq) + HSO 4 (aq) Stronger Stronger Weaker Weaker acid base acid base The hydronium ion H 3 O + produced is a weaker acid than H 2 SO 4. The conjugate base HSO 4 is a weaker base than water.

Direction of Reaction: CO 3 2 The carbonate ion from carbonic acid, H 2 CO 3, reacts with water. Water donates one H + to carbonate, CO 3 2 to form HCO 3 and OH. From Table 11.3, we see that HCO 3 is a stronger acid than H 2 O. We also see that OH is a stronger base than CO 3 2. To reach equilibrium, the strong acid and strong base react in the direction of the weaker acid and weaker base. CO 3 2 (aq) + H 2 O(l) OH (aq) + HCO 3 (aq) Weaker Weaker Stronger Stronger acid base base acid

Study Check Identify each of the following as a strong or weak acid or base: A. HBr B. HNO 2 C. NaOH D. H 2 SO 4 E. Cu(OH) 2

Study Check Using Table 11.3, identify the stronger acid in each pair. A. HNO 2 or H 2 S B. HCO 3 or HBr C. H 3 PO 4 or H 3 O +

11.4 Dissociation Constants for Acids and Bases HCHO 2 (aq) + H 2 O(l) H 3 O + (aq) + CHO 2 (aq) Learning Goal Write the expression for the dissociation constant of a weak acid or weak base.

Dissociation of a Weak Acid Because the dissociation of strong acids in water is essentially complete, the reaction is not considered to be an equilibrium process. Weak acids partially dissociate in water as the ion products reach equilibrium with the undissociated weak acid molecules. Formic acid is a weak acid that dissociates in water to form hydronium ion, H 3 O +, and formate ion, CHO 2. HCHO 2 (aq) + H 2 O(l) H 3 O + (aq) + CHO 2 (aq)

Writing Dissociation Constants As with other dissociation expressions, the molar concentration of the products is divided by the molar concentration of the reactants. water is a pure liquid with a constant concentration and is omitted. the expression is called acid dissociation constant, K a. HCHO 2 (aq) + H 2 O(l) H 3 O + (aq) + CHO 2 (aq)

Acid Dissociation Constant, K a When the value of the K a is small, the equilibrium lies to the left, favoring the reactants. is large, the equilibrium lies to the right, favoring the products. Weak acids have small K a values, while strong acids have very large K a values.

Base Dissociation Constant, K b When the value of the K b, is small, the equilibrium lies to the left, favoring the reactants. is large, the equilibrium lies to the right, favoring the products. The stronger the base, the larger the K b value. CH 3 NH 2 (aq) + H 2 O(l) CH 3 NH 3+ (aq) + OH (aq) The concentration of water is omitted from the base dissociation constant expression.

Study Check Write the acid dissociation constant expression for nitrous acid, HNO 2.

11.5 Dissociation of Water The equilibrium reached between the conjugate acid base pairs of water produces both H 3 O + and OH. + - + + H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) Learning Goal Use the water dissociation constant to calculate the [H 3 O + ] and [OH ] in an aqueous solution.

Dissociation Constant of Water, K w Water is amphoteric it can act as an acid or a base. In water, H + is transferred from one H 2 O molecule to another. one water molecule acts as an acid, while another acts as a base. equilibrium is reached between the conjugate acid base pairs.

Writing the Dissociation Constant, K w In the equation for the dissociation of water, there is both a forward and a reverse reaction. H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) Base Acid Conjugate Conjugate acid base In pure water, the concentrations of H 3 O + and OH at 25 C are each 1.0 10 7 M. [H 3 O + ] = [OH ] = 1.0 10 7 M K w = [H 3 O + ] [OH ] K w = (1.0 10 7 M) (1.0 10 7 M) = 1.0 10 14 at 25 C

Dissociation Constant, K w The ion product constant for water, K w, is defined as the product of the concentrations of H 3 O + and OH. equal to 1.0 10 14 at 25 C (the concentration units are omitted). When [H 3 O + ] and [OH ] are equal, the solution is neutral. [H 3 O + ] is greater than the [OH ], the solution is acidic. [OH ] is greater than the [H 3 O + ], the solution is basic.

Using K w to Calculate [H 3 O + ] and [OH ] If we know the [H 3 O + ] of a solution, we can use the K w to calculate the [OH ]. If we know the [OH ] of a solution, we can use the K w to calculate the [H 3 O + ].

Pure Water Is Neutral In pure water, the ionization of water molecules produces small but equal quantities of H 3 O + and OH ions. [H 3 O + ] = 1.0 10 7 M [OH ] = 1.0 10 7 M [H 3 O + ] = [OH ] Pure water is neutral.

Acidic Solutions Adding an acid to pure water increases the [H 3 O + ]. causes the [H 3 O + ] to exceed 1.0 10 7 M. decreases the [OH ]. [H 3 O + ] > [OH ] The solution is acidic.

Basic Solutions Adding a base to pure water increases the [OH ] causes the [OH ] to exceed 1.0 10 7 M decreases the [H 3 O + ] [H 3 O + ] < [OH ] The solution is basic.

Comparison of [H 3 O + ] and [OH ]

Guide to Calculating [H 3 O + ] and [O ] in Aqueous Solutions

Calculating [H 3 O + ] What is the [H 3 O + ] of a solution if [OH ] is 5.0 10 8 M? STEP 1 State the given and needed quantities. ANALYZE Given Need Know THE [OH ] = 5.0 10 8 M [H 3 O + ] K w = [H 3 O + ][OH ] PROBLEM = 1.0 10 14 STEP 2 Write the K w for water and solve for the unknown [H 3 O + ].

Calculating [H 3 O + ] What is the [H 3 O + ] of a solution if [OH ] is 5.0 10 8 M? STEP 3 Substitute in the known [H 3 O + ] or [OH ] and calculate. Because the [H 3 O + ] of 2.0 10 7 M is larger than the [OH ] of 5.0 10 8 M, the solution is acidic.

Study Check If lemon juice has [H 3 O + ] of 2.0 10 3 M, what is the [OH ] of the solution? A. 2.0 10 11 M B. 5.0 10 11 M C. 5.0 10 12 M

11.6 The ph Scale The ph scale is used to describe the acidity of solutions. A dipstick is used to measure the ph of urine. Learning Goal Calculate the ph from [H 3 O + ]; given the ph, calculate [H 3 O + ] and [OH ] of a solution.

The ph Scale The ph of a solution is used to indicate the acidity of a solution. has values that usually range from 0 to 14. is acidic when the values are less than 7. is neutral at a ph of 7. is basic when the values are greater than 7.

The ph Scale The ph of a solution is commonly measured using a ph meter in the laboratory. ph paper, an indicator that turns specific colors at a specific ph value. The ph of a solution is found by comparing the colors of indicator paper to a chart.

ph Measurement The ph of a solution can be determined using (a) a ph meter, (b) ph paper, and (c) indicators that turn different colors corresponding to different ph values.

ph of Common Substances On the ph scale, values below 7.0 are acidic, a value of 7.0 is neutral, and values above 7.0 are basic.

Study Check Identify each solution as acidic, basic, or neutral. A. HCl with a ph = 1.5 B. pancreatic fluid, [H 3 O + ] = 1 10 8 M C. Sprite soft drink, ph = 3.0 D. ph = 7.0 E. [OH ] = 3 10 10 M F. [H 3 O + ] = 5 10 12

Calculating the ph of Solutions The ph scale is a logarithmic scale that corresponds to the [H 3 O + ] of aqueous solutions. is the negative logarithm (base 10) of the [H 3 O + ]. ph = log[h 3 O + ] To calculate the ph, the negative powers of 10 in the molar concentrations are converted to positive numbers. If [H 3 O + ] is 1.0 10 2 M, ph = log[1.0 10 2 ] = ( 2.00) = 2.00 Key Math Skill Calculating ph from [H 3 O + ]

ph: Significant Figures To determine the number of significant figures in the ph value, the number of decimal places in the ph value is the same as the number of significant figures in the coefficient of [H 3 O + ]. the number to the left of the decimal point in the ph value is the power of 10.

ph Scale and [H 3 O + ] Because ph is a log scale, a change of one ph unit corresponds to a tenfold change in [H 3 O + ]. ph decreases as the [H 3 O + ] increases. ph 2.00 is [H 3 O + ] = 1.0 10 2 M ph 3.00 is [H 3 O + ] = 1.0 10 3 M ph 4.00 is [H 3 O + ] = 1.0 10 4 M

Guide to Calculating ph of Solutions The ph of a solution is calculated from the [H 3 O + ] by using the log key in your calculator and changing the sign.

ph Calculation Aspirin, which is acetylsalicylic acid, was the first nonsteroidal anti-inflammatory drug used to alleviate pain and fever. If a solution of aspirin has a [H 3 O + ] = 1.7 10 3 M, what is the ph of the solution?

ph Calculation If a solution of aspirin has a [H 3 O + ] = 1.7 10 3 M, what is the ph of the solution? STEP 1 State the given and needed quantities. ANALYZE Given Need Know THE [H 3 O + ] = 1.7 10 3 M ph of solution ph = log[h 3 O + ] PROBLEM

ph Calculation If a solution of aspirin has a [H 3 O + ] = 1.7 10 3 M, what is the ph of the solution? STEP 2 Enter the [H 3 O + ] into the ph equation and calculate. ph = log[h 3 O + ] = log[1.7 10 3 ] Calculator Procedure: log 1.7 EE or Exp 3 +/ = Calculator Display: 2.769551079

ph Calculation If a solution of aspirin has a [H 3 O + ] = 1.7 10 3 M, what is the ph of the solution? STEP 3 Adjust the number of SFs on the right of the decimal point. Coefficient Power of ten 1.7 10 3 Two SFs Exact ph = log[1.7 10 3 ] = 2.77

Study Check Find the ph of a solution with a [H 3 O + ] of 4.0 10 5.

Calculating [H 3 O + ] from ph Given the ph of a solution, we can reverse the calculation to obtain the [H 3 O + ]. For whole number ph values, the negative ph value is the power of 10 in the [H 3 O + ] concentration. [H 3 O + ] = 10 ph For ph values that are not whole numbers, the calculation requires the use of the 10 x key, which is usually a 2nd function key. Key Math Skill Calculating [H 3 O + ] from ph.

Guide to Calculating [H 3 O + ] from ph

Study Check Determine the [H 3 O + ] for a solution that has a ph of 3.42.

11.7 Reactions of Acids and Bases Gastric acid contains HCl and is produced by parietal cells that line the stomach. When protein enters the stomach, HCl is secreted until the ph reaches 2, the optimum ph for digestion. Learning Goal Write balanced equations for reactions of acids with metals, carbonates, and bases.

Reactions of Acids Acids react with metals to produce salt and hydrogen gas. bases to produce a salt and water. bicarbonate and carbonate ions to produce carbon dioxide gas. A salt is an ionic compound that does not have H + as the cation or OH as the anion.

Acids and Metals Acids react with metals to produce hydrogen gas and the salt of the metal. 2K(s) + 2HCl(aq) 2KCl(aq) + H 2 (g) Metal Acid Salt Hydrogen Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) Metal Acid Salt Hydrogen Magnesium reacts rapidly with acid and forms H 2 gas and a salt of magnesium.

Acids, Carbonates, and Bicarbonates Acids react with carbonates and hydrogen carbonates to produce carbon dioxide gas, a salt, and water: 2HCl(aq) + CaCO 3 (s) CO 2 (g) + CaCl 2 (aq) + H 2 O(l) Acid Carbonate Carbon Salt Water dioxide HCl(aq) + NaHCO 3 (s) CO 2 (g) + NaCl(aq) + H 2 O(l) Acid Bicarbonate Carbon Salt Water dioxide Core Chemistry Skill Writing Equations for Reactions of Acids and Bases

Acids and Hydroxides: Neutralization In a neutralization reaction, an acid reacts with a base to produce salt and water. the salt formed is the anion from the acid and the cation from the base. HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) Acid Base Salt Water

Acids and Hydroxides: Neutralization In neutralization reactions, one H + always reacts with one OH. If we write the strong acid and strong base as ions, HCl(aq) + NaCl(aq) NaOH(aq) + H 2 O(l) we see that H + reacts with OH to form water, leaving the ions Na + and Cl in solution: H + (aq) + Cl (aq) + Na + (aq) + OH (aq) Na + (aq) + Cl (aq) + H 2 O(l) The overall reaction occurs as the H + from the acid and OH from the base form water: H + (aq) + OH (aq) H 2 O(l) Net ionic equation

Guide to Balancing an Equation for Neutralization

Balancing Neutralization Reactions Write the balanced equation for the neutralization of solid magnesium hydroxide and nitric acid. STEP 1 Write the reactants and products. Mg(OH) 2 (s) + HNO 3 (aq) salt + H 2 O(l) STEP 2 Balance the H + in the acid with the OH in the base. Mg(OH) 2 (s) + 2HNO 3 (aq) salt + H 2 O(l)

Balancing Neutralization Reactions Write the balanced equation for the neutralization of solid magnesium hydroxide and nitric acid. STEP 3 Balance the H 2 O with H + and the OH. Mg(OH) 2 (s) + 2HNO 3 (aq) salt + 2H 2 O(l) STEP 4 Write the salt from the remaining ions. Mg(OH) 2 (s) + 2HNO 3 (aq) Mg(NO 3 ) 2 (aq) + 2H 2 O(l)

Study Check Write a balanced equation for the following reaction: Mg(OH) 2 (s) + HBr(aq)

Study Check Select the correct group of coefficients for each of the following neutralization equations: 1. HCl(aq) + Al(OH) 3 (aq) AlCl 3 (aq) + H 2 O(l) A. 1, 3, 3, 1 B. 3, 1, 1, 1 C. 3, 1, 1, 3 2. Ba(OH) 2 (aq) + H 3 PO 4 (aq) Ba 3 (PO 4 ) 2 (s) + H 2 O(l) A. 3, 2, 2, 2 B. 3, 2, 1, 6 C. 2, 3, 1, 6

Solution Select the correct group of coefficients for each of the following neutralization equations: STEP 4 Write the salt from the remaining ions. 1. 3HCl(aq) + Al(OH) 3 (aq) AlCl 3 (aq) + 3H 2 O(l) The answer is C, 3, 1, 1, 3. 2. 3Ba(OH) 2 (aq) + 2H 3 PO 4 (aq) Ba 3 (PO 4 ) 2 (s) + 6H 2 O(l) The answer is B, 3, 2, 1, 6.

Chemistry Link to Health: Antacids Antacids are substances that are used to neutralize excess stomach acid. are made of aluminum hydroxide and magnesium hydroxide mixtures. These hydroxides are not very soluble in water, so the levels of available OH are not damaging to the intestinal tract.

11.8 Acid Base Titration The titration of an acid. A known volume of an acid is placed in a flask with an indicator and titrated with a measured volume of a base solution, such as NaOH, to the neutralization endpoint. Learning Goal Calculate the molarity or volume of an acid or base from titration information.

Acid Base Titration Titration is a laboratory procedure used to determine the molarity of an acid. uses a base such as NaOH to neutralize a measured volume of an acid. requires a few drops of an indicator such as phenolphthalein to identify the endpoint. Core Chemistry Skill Calculating Molarity or Volume of an Acid or Base in a Titration

Acid Base Titration In the following titration, a specific volume of acidic solution is titrated to the endpoint with a known concentration of NaOH. Base NaOH Acid Solution

Indicator The indicator phenolphthalein is added to identify the endpoint. turns pink when the solution is neutralized.

Endpoint of Titration At the endpoint of the titration, the moles of base are equal to the moles of acid in the solution. the concentration of the base is known. the volume of the base used to reach the endpoint is measured. the molarity of the acid is calculated using the neutralization equation for the reaction.

Guide to Calculating Boiling Point Elevation, Freezing Point Lowering

Acid Base Titration Calculations What is the molarity of an HCl solution if 18.5 ml of 0.225 M NaOH is required to neutralize 0.0100 L of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) STEP 1 State given and needed quantities and concentrations. ANALYZE Given Need Equation THE 18.5 ml of molarity of HCl(aq) + NaOH(aq) PROBLEM 0.225 M NaOH HCl solution NaCl(aq) + H 2 O(l) 0.0100 L HCl

Acid Base Titration Calculations What is the molarity of an HCl solution if 18.5 ml of 0.225 M NaOH is required to neutralize 0.0100 L of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) STEP 2 Write a plan to calculate molarity or volume. ml NaOH solution Metric factor L NaOH solution Molarity moles of NaOH moles of NaOH Mole Mole factor moles of HCl Divide by liters molarity of HCl solution

Study Check What is the molarity of an HCl solution if 25.5 ml of 0.438 M NaOH is required to neutralize 0.0250 L of HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l)

11.9 Buffers A buffer solution maintains the ph by neutralizing small amounts of added acid or base. An acid must be present to react with any OH added, and a base must be present to react with any H 3 O + added. Learning Goal Describe the role of buffers in maintaining the ph of a solution; calculate the ph of a buffer.

Buffers When an acid or a base is added to water, the ph changes drastically. In a buffer solution, the ph is maintained; ph does not change when acids or bases are added.

How Buffers Work Buffers work because they resist changes in ph from the addition of an acid or a base. in the body, they absorb H 3 O + or OH from foods and cellular processes to maintain ph. they are important in the proper functioning of cells and blood. they maintain a ph close to 7.4 in blood. A change in the ph of the blood affects the uptake of oxygen and cellular processes.

Components of a Buffer A buffer solution contains a combination of acid base conjugate pairs, a weak acid and a salt of its conjugate base, such as HC 2 H 3 O 2 (aq) and C 2 H 3 O 2 (aq) has equal concentrations of a weak acid and its salt.

How Buffers Work In the buffer with acetic acid (HC 2 H 3 O 2 ) and sodium acetate (NaC 2 H 3 O 2 ), the salt produces acetate ions and sodium ions. NaC 2 H 3 O 2 (aq) C 2 H 3 O 2 (aq) + Na + (aq) the salt is added to provide a higher concentration of the conjugate base C 2 H 3 O 2 than from the weak acid alone. HC 2 H 3 O 2 (aq) + H 2 O(l) C 2 H 3 O 2 (aq) + H 3 O + (aq) Large amount Large amount

Function of a Weak Acid in a Buffer If a small amount of base is added to this same buffer solution, it is neutralized by the acetic acid, HC 2 H 3 O 2, which shifts the equilibrium in the direction of the products acetate ion and water. HC 2 H 3 O 2 (aq) + OH (aq) Equilibrium shifts in the direction of the products. C 2 H 3 O 2 (aq) + H 2 O(l)

Function of Conjugate Base in a Buffer When a small amount of acid is added, the additional H 3 O + combines with the acetate ion, C 2 H 3 O 2, causing the equilibrium to shift in the direction of the reactants, acetic acid and water. The acetic acid produced contributes to the available weak acid. HC 2 H 3 O 2 (aq) + H 2 O(l) C 2 H 3 O 2 (aq) + H 3 O + (aq) Equilibrium shifts in the direction of the reactants.

Working Buffers The buffer described here consists of about equal concentrations of acetic acid (HC 2 H 3 O 2 ) and its conjugate base, acetate ion (C 2 H 3 O 2 ). Adding H 3 O + to the buffer reacts with the salt, C 2 H 3 O 2, whereas adding OH neutralizes the acid HC 2 H 3 O 2. The ph of the solution is maintained as long as the added amounts of acid or base are small compared to the concentrations of the buffer components.

Calculating the ph of a Buffer By rearranging the K a expression to give [H 3 O + ], we can obtain the ratio of the acetic acid/acetate buffer and calculate the ph. Solving for H 3 O + gives Weak acid Conjugate base Core Chemistry Skill Calculating the ph of a Buffer

Study Check The K a for acetic acid, HC 2 H 3 O 2, is 1.8 10 5. What is the ph of a buffer prepared with 1.0 M HC 2 H 3 O 2 and 1.0 M C 2 H 3 O 2? HC 2 H 3 O 2 (aq) + H 2 O(l) C 2 H 3 O 2 (aq) + H 3 O + (aq)

Solution The K a for acetic acid, HC 2 H 3 O 2, is 1.8 10 5. What is the ph of a buffer prepared with 1.0 M HC 2 H 3 O 2 and 1.0 M C 2 H 3 O 2? HC 2 H 3 O 2 (aq) + H 2 O(l) C 2 H 3 O 2 (aq) + H 3 O + (aq) STEP 1 State the given and needed quantities. ANALYZE Given Need THE [HC 2 H 3 O 2 ] = 1.0 M ph of solution PROBLEM [C 2 H 3 O 2 ] = 1.0 M Equation HC 2 H 3 O 2 (aq) + H 2 O(l) C 2 H 3 O 2 (aq) + H 3 O + (aq)

Solution The K a for acetic acid, HC 2 H 3 O 2, is 1.8 10 5. What is the ph of a buffer prepared with 1.0 M HC 2 H 3 O 2 and 1.0 M C 2 H 3 O 2? HC 2 H 3 O 2 (aq) + H 2 O(l) C 2 H 3 O 2 (aq) + H 3 O + (aq) STEP 2 Write the K a expression and rearrange for [H 3 O + ].

Solution The K a for acetic acid, HC 2 H 3 O 2, is 1.8 10 5. What is the ph of a buffer prepared with 1.0 M HC 2 H 3 O 2 and 1.0 M C 2 H 3 O 2? HC 2 H 3 O 2 (aq) + H 2 O(l) C 2 H 3 O 2 (aq) + H 3 O + (aq) STEP 3 Substitute [HA] and [A ] into the K a expression.

Solution The K a for acetic acid, HC 2 H 3 O 2, is 1.8 10 5. What is the ph of a buffer prepared with 1.0 M HC 2 H 3 O 2 and 1.0 M C 2 H 3 O 2? HC 2 H 3 O 2 (aq) + H 2 O(l) C 2 H 3 O 2 (aq) + H 3 O + (aq) STEP 4 Use [H 3 O + ] to calculate ph.

Calculating the ph of a Buffer Because K a is a constant at a given temperature, the [H 3 O + ] is determined by the [HC 2 H 3 O 2 ]/[C 2 H 3 O 2 ] ratio. the addition of small amounts of either acid or base changes the ratio of [HC 2 H 3 O 2 ]/[C 2 H 3 O 2 ] only slightly. the changes in [H 3 O + ] will be small and the ph will be maintained. the addition of a large amount of acid or base may exceed the buffering capacity of the system.

Buffers and ph Changes Buffers can be prepared from conjugate acid base pairs such as H 2 PO 4 /HPO 4 2 and HPO 4 2 /PO 4 3, HCO 3 /CO 3 2, or NH 4+ /NH 3. The ph of the buffer solution will depend on the conjugate acid base pair chosen.

Buffers and ph Changes Using a common phosphate buffer for biological specimens, we can look at the effect of using different ratios of [H 2 PO 4 /HPO 4 2 ] on the [H 3 O + ] and ph. The K a of H 2 PO 4 is 6.2 10 8. The equation and the [H 3 O + ] are written as follows: H 2 PO 4 (aq) + H 2 O(l) H 3 O + (aq) + HPO 4 2 (aq)

Buffers and ph Changes

Chemistry Link to Health: Buffers in Blood Plasma The arterial blood plasma has a normal ph of 7.35 to 7.45. If changes in H 3 O + lower the ph below 6.8 or raise it above 8.0, cells cannot function properly and death may result. In our cells, CO 2 is continually produced as an end product of cellular metabolism. is carried to the lungs for elimination, and the rest dissolves in body fluids such as plasma and saliva, forming carbonic acid, H 2 CO 3. As a weak acid, carbonic acid dissociates to give bicarbonate, HCO 3, and H 3 O +. 2016

Chemistry Link to Health: Buffers in Blood Plasma Kidneys also supply more of the bicarbonate anion, HCO 3, setting up an important buffer system in the body fluid: CO 2 (g) + H 2 O(l) H 2 CO 3 (aq) HCO 3 (aq) + H 3 O + (aq) Excess H 3 O + entering the body fluids reacts with the HCO 3, and excess OH reacts with the carbonic acid. H 2 CO 3 (aq) + H 2 O(l) HCO 3 (aq) + H 3 O + (aq) Equilibrium shifts in the direction of the reactants. 2016

Chemistry Link to Health: Buffers in Blood Plasma Kidneys also supply more of the bicarbonate anion, HCO 3, setting up an important buffer system in the body fluid: CO 2 (g) + H 2 O(l) H 2 CO 3 (aq) HCO 3 (aq) + H 3 O + (aq) Excess H 3 O + entering the body fluids reacts with the HCO 3 : H 2 CO 3 (aq) + H 2 O(l) HCO 3 (aq) + H 3 O + (aq) Equilibrium shifts in the direction of the reactants. Excess OH entering the body fluids reacts with the H 2 CO 3 : H 2 CO 3 (aq) + OH (aq) H 2 O(l) + HCO 3 (aq) Equilibrium shifts in the direction of the products. 2016

Chemistry Link to Health: Buffers in Blood Plasma To maintain the normal blood plasma ph (7.35 to 7.45), the ratio of [H 2 CO 3 ]/[HCO 3 ] needs to be about 1 to 10. concentrations of 0.0024 M H 2 CO 3 and 0.024 M HCO 3 work to maintain that ph. 2016

Chemistry Link to Health: Buffers in Blood Plasma In the body, the concentration of carbonic acid is closely associated with the partial pressure of CO 2, P CO2. If the CO 2 level rises, increasing H 2 CO 3, the equilibrium shifts to produce more H 3 O +, which lowers the ph. This condition is called acidosis. A lowering of the CO 2 level leads to a high blood ph, a condition called alkalosis. 2016

Concept Map