As you work through the chapter, you should be able to: Advanced Placement Chemistry Chapters 14 16 Syllabus Chapter 14 Acids and Bases 1. Describe acid and bases using the Bronsted-Lowry, Arrhenius, and Lewis models for acids and bases. 2. Understand the difference between strong and weak acids or bases in terms of percent dissociation. 3. Determine conjugate acid-base pairs for an acid base reaction. 4. Write an equation for an acid or base dissociation constant. 5. Explain how polyprotic acids dissociate. 6. Explain what the term amphoteric means. 7. Know the relationship between K a, K b, and K w as defined by the autoionization of water. 8. Describe the relationship between [OH - ] and [H + ] in an acidic, basic, and neutral solution. 9. Calculate ph and poh for strong and weak acids and bases. 10. Calculate K a and K b values. 11. Calculate percent dissociation of a species in an acidic or basic solution. 12. Explain why different salts will produce acidic, basic, or neutral solutions 13. Explain the effect of structure on acid-base properties. Chapter 15 Acid-Base Equilibria 1. Determine the equilibrium position of a solution involving a common ion. 2. Describe the components of a buffered solution. 3. Explain how a buffered solution works and how to prepare a buffered solution. 4. Determine the buffer capacity of a buffered solution. 5. Use the Henderson-Hasselbalch to determine the ph of a buffered solution. 6. Determine the equilibrium position of a buffered solution. 7. Monitor the progress of an acid-base titration using a titration curve. 8. Explain what occurs at the stoichiometric point or equivalence point. 9. Determine the equilibrium position at various points during an acid base titration. 10. Choose an appropriate indicator for a particular acid base titration. Chapter 16 Solubility Equilibria 1. Use the solubility product to determine the equilibrium position of an aqueous solution. 2. Use the relationship between the ion product and solubility product to determine whether or not a precipitate will form. 3. Use the relationship between the ion product and solubility product to develop a scheme to qualitatively analyze a solution. Lab Experience: 1. Determination of the K a for a weak acid 2. Preparation and properties of buffered solutions Assignments: HW1: Ch 14 Handout HW2: Ch 14 Text Q s HW3: Ch 14 Quiz w/ notes check HW4: Ch 15 Handout HW5: Ch 15 Text Q s HW6: Ch 15 Quiz w/ notes check HW7: Ch 16 Handout HW8: Ch 16 Text Q s HW9: Ch 16 Quiz w/ notes check
Week of February 23 th Day Concepts Class Activities Homework M -Unit 5 Test Day 2 1-6 1 st CH 14 Introduction -Notes: CH 14.1 14.7 Block Goals 1 8 2 nd Goals 1-13 -Bring Text Books Block F Goal 9-13 -Notes: CH 14.8 14.12 Week of March 2 nd Day Concepts Class Activities Homework M Goals -Notes CH 15.1 15.3 Due: HW1 1 st Block Goals 1 13 -Common Ion Effect Activity Due: HW2 R Goals 7 9 -Notes: CH 15.4 Due: HW3 F Goal 10 -Notes: CH 15.5 Due: Common Ion Activity Week of March 9 th Day Concepts Class Activities Homework M Due: HW4 Lab: Preparation and Properties of Buffered Solutions -Preparation and Properties of Buffered Solutions: Pre-lab 1 st Block 2 nd Block F Lab: Preparation and Properties of Buffered Solutions -Preparation and Properties of Buffer Solutions: Collect Data Due: HW5 Goals 1 10 Ch 15 Quiz and Note Check Due: HW6 CH 16 Introduction -Notes: CH 16.1 16.2 Due: Goals 1 3 HW7 Week of March 16 th Day Concepts Class Activities Homework M Goals 1 3 - Review time and Activity Due: HW8 1 st Block CH 14 15 Review - Ch 16 Quiz and Note Check Due: HW9 2 nd CH 14 16 Summative -CH 14 16 Examination Block Assessment -Pass out Unit 7 Materials F CH 17 Introduction -Ch 14 16 Examination Day 2 Due: Preparation and Properties of Buffer Solutions
Chapter 14 Acids and Bases The Nature of Acids and Bases Acids sour, burn Bases Bitter, slippery 1. Arrhenius Concepts: Acid produces H + cation in aqueous solution Base produces OH - anion in aqueous solution 2. Brønsted-Lowry Model: Acid proton donor Base proton acceptor Brønsted-Lowry Example HCl + H 2 O H 3 O + + Cl -, where H 3 O + is called the hydronium ion The general reaction that occurs when an acid dissolves in water: HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) acid base conjugate conjugate acid base A conjugate acid is formed when the proton is transferred to the base. A conjugate base is everything that remains of an acid molecule once the proton is lost A conjugate acid-base pair consists of two substances related to each other by the donating and accepting of protons. Acid-Base Equilibrium For the reaction: HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) K a = ([H 3 O + ][A - ])/[HA] = ([H + ][A - ])/[HA] where K a is called the acid dissociation constant. H + and H 3 O + both represent the hydronium ion. Water is left out of the expression because its concentration remains relatively constant. Acid Strength Strong acid: Equilibrium lies far to the right Almost all of the original HA has dissociated Strong acids yield weak conjugate bases that have a low affinity for protons (conjugate base is a much weaker base than water) Weak acid: Equilibrium lies far to the left Most of the original acid still exists as HA at equilibrium, meaning it dissociates to a very small extent Weak acids yield conjugate bases that are much stronger than water Various Ways to Describe an Acid Property Strong Acid Weak Acid K a value K a is large K a is small Position of the dissociation at equilibrium Far to the right Far to the left Equilibrium concentration of H + compared with the original concentration of HA [H + ] [HA] 0 [H] + <<< [HA] 0 Strength of conjugate base compared with that of water A - much weaker than H 2 O A - much stronger than H 2 O
Common Strong Acids Sulfuric (H 2 SO 4 ), hydrochloric (HCl), nitric (HNO 3 ), perchloric (HClO 4 ) Can t calculate K a values for strong acids because [HA] is so small. Diprotic Acids An acid having two acidic protons. H 2 SO 4 : H 2 SO 4 (aq) H + (aq) + HSO 4 - (aq) (strong) HSO 4 - (aq) H + (aq) + SO 4 2- (aq) (weak) Oxyacids Acidic proton is attached to an oxygen atom. Organic acids carbon backbone and contain a carboxyl group (COOH) Water as an Acid and a Base Amphoteric behaves as an acid or a base Autoionization of water: 2H 2 O(l) H 3 O + (aq) + OH - (aq) K w = [H 3 O + ][OH - ], where K w is the ion-product constant. At 25 o C [H + ] = [OH - ] = 1.0x10-7 M K w = [H 3 O + ][OH - ] = (1.0x10-7 M)(1.0x10-7 M) K w = 1.0x10-14 (increases slightly with temperature) Significance of K w In any aqueous solution at 25 o C, no matter what it contains, the product [H + ] and [OH - ] must equal 1.0x10-14. Three results follow: 1. A neutral solution, where [H + ] = [OH - ] 2. An acidic solution, where [H + ] > [OH - ] 3. A basic solution, where [H + ] < [OH - ] The ph Scale Aqueous solutions can have very small H + concentrations. ph is measured using a logarithmic scale to conveniently represent solution acidity. ph = -log[h + ] Significant Figures for Logarithms: The number of decimal places in the log is equal to the number of significant figures in the original number ph continued Since ph is a log scale, the ph changes by one for every power of ten change in [H + ]. The ph decreases as [H + ] increases. ph can be measured using a probe that can measure potential differences between a test solution and the solution in the probe. Relationship Between ph and poh K w = [H + ][OH - ] log K w = log[h + ] + log[oh - ] -log K w = -log[h + ] - log[oh - ] pk w = ph + poh Since pk w = -log(1.0x10-14 ) = 14.00 ph + poh = 14.00
ph Example The ph of a sample was measured to be 8.23 at 25 o C. Calculate the poh, [H + ], and [OH - ] for the sample. Calculating the ph of Strong Acid Solutions Determine what is the major species providing H + in solution. If 1.0 M HCl is allowed to dissociate in water, HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - (aq) Two species provide H + in this example HCl and H 2 O through autoionization. However, HCl will dissociate into ~1.0 M H + and H 2 O will only make ~1.0x10-7 M H +. This means the contribution from H 2 O can be neglected. ph = -log[h + ] = -log(1.0) = 0 Calculating the ph of Weak Acid Solutions 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. 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. Solve for x the easy way; that is, by assuming [HA] 0 x [HA] 0. Use the 5% rule to verify whether the approximation is valid. Calculate [H + ] and ph. ph of a Weak Acid Example What is the ph of a 1.00 M solution of HNO 2 at equilibrium (K a = 4.0x10-4 )? Percent Dissociation Percent Dissociation: [(amount dissociated (M))/(initial concentration (M))]x100% For solutions of any weak acid HA, [H + ] decreases as [HA] 0 decreases, but the percent dissociation increases as [HA] 0 decreases. Bases Strong Bases: Complete dissociation All group 1A, Ca, Ba, and Sr hydroxides are strong bases Low solubility of OH - useful in products like antacids
Weak Bases: Incomplete dissociation Many electron-pair donors are weak bases Base Equilibrium For the reaction of a base with water: B(aq) + H 2 O(l) BH + (aq) + OH - (aq) base acid conjugate conjugate acid base K b = ([BH + ][OH - ])/[B] ph of a Strong Base Calculate the ph of a 2.7x10-3 M NaOH solution. ph of a Weak Base Calculate the ph for a 10.0 M solution of NH 3 (K b = 1.8x10-5 ) Polyprotic Acids Acids that can provide more than one proton. Examples: H 2 CO 3 H 2 SO 4 H 3 PO 4 Each proton dissociation has a corresponding K a, where K a1 >K a2 >K a3. Therefore equilibrium concentrations and ph can be determined by examining the first proton dissociation. Acid-Base Properties of Salts Salts are ionic compounds that will dissolve in water. Under certain conditions these solutions can behave as acids or bases. Salts That Produce Neutral Solutions Salts that consist of cations of strong bases and anions of strong acids have no effect on ph. Examples: KCl NaCl NaNO 3 Salts That Produce Basic Solutions Salts whose cation has neutral properties (such as Na + and K + ) and whose anion is the conjugate base of a weak acid, the aqueous solution will be basic. K a K b =K w When NaC 2 H 3 O 2 is dissolved in water C 2 H 3 O 2 - (aq) + H 2 O(l) HC 2 H 3 O 2 (aq) + OH - (aq)
Salts That Produce Acidic Solutions Salts in which the anion is not a base and the cation is the conjugate acid of a weak base produce acidic solutions. Salts containing highly charged metal ions will also produce acidic solutions. When solid NH 4 Cl is dissolved in water NH 4 + (aq) NH 3 (aq) + H + (aq) Example Calculate the ph of a 0.20 M NH 4 Cl solution. The K b value for NH 3 is 1.8x10-5. Effects of Structure on Acid-Base Properties Electronegativity influences the strength of an acid. Acidity generally increases as the electronegativity of the atoms on the molecule increase. Acidity generally increases with the number of oxygen atoms attached to the central atom. Acid-Base Properties of Oxides When Covalent oxides dissolve in water, an acidic solution forms. SO 3 (g) + H 2 O(l) H 2 SO 4 (aq) When ionic oxides are dissolved in water, a basic solution forms. CaO(s) + H 2 O(l) Ca(OH) 2 (aq) Lewis Acid-Base Model Lewis Acid electron-pair acceptor H +, BF 3, metal cations Lewis Base electron-pair donor OH -, H 2 O, NH 3 Chapter 15 Acid-Base Equilibria Common Ion Effect The shift in an equilibrium caused by the addition or presence of an ion involved in the equilibrium reaction. Solutions of Acids Containing a Common Ion Add NaF to a solution of HF NaF Na + + F - HF H + + F - According to Le Châtelier s principle, the additional F - in solution will drive the equilibrium position towards the reactants. The solution will become less acidic.
Solutions of Bases Containing a Common Ion Add NH 4 Cl to a solution of NH 3 NH 4 Cl NH 4 + + Cl - NH 3 + H 2 O NH 4 + + OH - According to Le Châtelier s principle, the additional NH 4 + in solution will drive the equilibrium position towards the reactants. The solution will become less basic. Equilibrium Calculations Use the same basic problem solving strategy as all equilibrium calculations. The only difference is that the salt (anion or cation) is adding to the initial concentration of one of the products. Common Ion Example The equilibrium concentration of H + in a 1.0 M HF solution is 2.7x10-2 M, and the percent dissociation of HF is 2.7%. Calculate the [H + ] and the percent dissociation of HF in a solution containing 1.0 M HF (K a = 7.2x10-4 ) and 1.0 M NaF. Buffered Solutions Resists changes in ph upon addition of an acid or a base. May contain: Weak acid and its conjugate base salt (HF and NaF) Weak base and its conjugate acid salt (NH 3 and NH 4 Cl) Solving Buffered Solution Problems Buffered solutions are simply solutions of weak acids or bases containing a common ion. The ph calculations on buffered solutions require exactly the same procedures as introduced previously. When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first (assuming reaction with H + /OH - goes to completion). After the stoichiometric calculations are completed, then consider the equilibrium calculations. The ph of a Buffered Solution A buffered solution contains 0.50 M acetic acid (CH 3 COOH, Ka = 1.8x10-5 ) and 0.50 M sodium acetate (NaCH 3 COO). Calculate the ph.
Change in ph of a Buffered Solution Calculate the change in ph that occurs when 0.010 mol of solid NaOH is added to 1.0 L of the buffered solution from the previous problem. Compare this change with that which occurs when 0.010 mol of solid NaOH is added to 1.0 L of H 2 O. (Hint: stoichiometry first then equilibrium) How Buffering Works Buffering works by having a large excess of a weak acid and its conjugate base to minimize the effect of adding H + or OH - ions. For the reaction HA H + + A -, the acidity of the solution, [H + ] = K a ([HA]/[A - ]) is dependent upon the ratio of [HA]/[A - ]. If this ratio is made so it resists change (large excess) a buffered solution can be made. This also applies to solutions of a weak base and a conjugate acid made to be in large excess compared to any added H + /OH - ions. Henderson-Hasselbalch Equation For a particular buffered system (acid-conjugate base pair), all solutions that have the same ratio [A - ]/[HA] will have the same ph. ph = pk a + log([a - ]/[HA]), when using this equation one assumes that [A - ] and [HA] are [A - ] 0 and [HA] 0. Example Calculate the ph of a solution containing 0.75 M lactic acid (K a = 1.4x10-4 ) and 0.25 M lactate. Buffering Capacity Represents the amount of proton or hydroxide ions a buffered solution can absorb without a significant change in ph. A solution with a large buffer capacity contains a large concentration of buffering components. The ph of a buffered solution is determined by the ratio [A - ]/[HA]. The capacity of a buffered solution is determined by the magnitude of [HA] and [A - ]. Optimal Buffering Large changes in the ratio [A - ]/[HA] will produce large changes in ph. Optimal buffering occurs when [HA] is equal to [A - ].
When this ratio is achieved: ph = pk a + log([a - ]/[HA]) ph = pk a + log(1) ph = pk a The pk a of a weak acid used in a buffer should be as close as possible to the desired ph. Choosing a Buffer Suppose a buffered solution with a ph of 4.00 is needed. From the Henderson-Hasselbalch Equation: ph = pk a + log([a - ]/[HA]) 4.00 = pk a + log(1) pk a = 4.00 K a = antilog(-4.00) = 1.0x10-4 Example A chemist needs a solution at a ph of 4.30 and can choose from the following acids. Chloroacetic acid (K a = 1.35x10-3 ) Propanoic acid (K a = 1.3x10-5 ) Benzoic acid (K a = 6.4x10-5 ) Hypochlorous acid (K a = 3.5x10-8 ) Solution Acid [H + ] = K a ([HA]/[A - ]) [HA]/[A - ] Chloroacetic 5.0x10-5 = 1.35x10-3 ([HA]/[A - ]) 3.7x10-2 Propanoic 5.0x10-5 = 1.3x10-5 ([HA]/[A - ]) 3.8 Benzoic 5.0x10-5 = 6.4x10-5 ([HA]/[A - ]) 0.78 Hypochlorous 5.0x10-5 = 3.5x10-8 ([HA]/[A - ]) 1.4x10 3 Calculate the ratio [HA]/[A - ] required for each system to yield a ph of 4.30. [H + ] = 10-4.30 = antilog(-4.30) = 5.0x10-5 M [H + ] = K a ([HA]/[A - ]) Benzoic acid is the best choice since the ratio of [HA]/[A] is closest to 1. A solution with a ~1:1 ratio of benzoic acid and sodium benzoate would make an optimal buffered solution of 4.30. Titrations and ph Curves This process involves a solution of known concentration (the titrant) delivered from a buret to a solution of unknown concentration until the substance being analyzed is just consumed (moles of titrant are equal to moles of unknown). The stoichiometric equivalence point is often signaled by a color change of an indicator. The progress of an acid-base titration can be monitored by analyzing a ph curve. ph curves are a plot of solution ph as a function of titrant volume.
Strong Acid-Strong Base Titrations Net ionic equation: H + (aq) + OH - (aq) H 2 O(l) Often useful to use mmol (10-3 mol) instead of moles. (mmol = mlxm) Before the equivalence point, [H + ] can be calculated by dividing the number of millimoles of H + remaining by the total volume of the solution in milliliters. A ph of 7.00 at the equivalence point is characteristic of a strong acid-strong base titration. After the equivalence point, [OH - ] can be calculated by dividing the number of millimoles of excess OH - by the total volume of the solution in milliliters. Then use K w to find [H + ]. Plot ph versus titrant volume to construct a ph curve (titration curve). Example I: Strong Acid Titrated with a Strong Base 50.0 ml of 0.200 M HNO 3 is titrated with 0.100 M NaOH. Calculate ph after the following volumes of NaOH have been added: 0.0, 10.0, 20.0, 50.0, 100.0, 150.0 ml Titrations of Weak Acids with Strong Bases Treat as a series of equilibrium problems. Weak acids will react with a strong base completely. 1. Determine moles of H + from stoichiometry 2. Find equilibrium [H + ] and ph The ph at the equivalence point is always greater than 7 due to the presence of the weak acid s conjugate base. Example II: Weak Acid Titrated with a Strong Base 50.0 ml of 0.10 M HC 2 H 3 O 2 (K a =1.8x10-5 ) is titrated with 0.10 M NaOH. Calculate ph after the following volumes of NaOH have been added: 0.0, 10.0, 25.0, 50.0, 75.0 ml(hint: may use Henderson-Hasselbalch equation to determine ph)
Conclusions Amount of acid determines equivalence point, not the acid strength. ph value of the equivalence point is affected by acid strength. Titrations of Weak Bases with Strong Acids Treat the same way as weak acid strong base. Before the equivalence point: 1. List major species before reaction. 2. Base will react to completion with H +. 3. Stoichiometry first. 4. Weak base equilibrium second to determine ph. At the equivalence point: 1. Reaction won t go to completion. 2. Equilibrium is dominated by the weak base s conjugate acid. 3. Solution will be acidic. Beyond equivalence point: 1. Excess H + dominates equilibrium. 2. ph is determined by excess H +. 3. Acid-Base Indicators Two methods for determining the equivalence point of an acid-base titration: 1. Use a ph probe to monitor the change in ph as titrant is added. The center of the vertical region of the ph curve indicates the equivalence point. 2. Use an acid-base indicator, which marks the end point of a titration by changing color. Although the equivalence point of a titration is not necessarily the same as the end point. Indicators Usually a weak organic acid (HIn) 1. Phenolpthalein 2. Colorless in HIn form 3. Pink in In - form HIn H + + In - K a 1.0x10-8 Color change occurs when ([In - ]/[HIn]) (1/10) Example: Indicators Bromothymol blue, an indicator with a K a value of 1.0x10-7, is yellow in its HIn form and blue in its In - form. Suppose we put a few drops of this indicator in a strongly acidic solution. If the solution is then titrated with NaOH, at about what ph will the indicator color change first be visible? Solubility Equilibria and the Solubility Product When solids dissolve in water two processes occur: Dissolution of the solid Reformation of the solid Chapter 16 Solubility Equilibria
Example: CaF 2 (s) Ca 2+ (aq) + 2F - (aq) Eventually a dynamic equilibrium is reached, and the solution is saturated. K sp = [Ca 2+ ][F - ] 2 K sp is called the solubility product constant. Solubility Examples Calculate the K sp for bismuth sulfide (Bi 2 S 3 ), which has a solubility of 1.0x10-15 M at 25 o C. The K sp value for copper(ii) iodate, Cu(IO 3 ) 2, is 1.4x10-7 at 25 o C. Calculate its solubility at 25 o C. Relative Solubilities A salt s K sp value provides information about its solubility. Relative solubilities of salts that produce the same number of ions in solution can be compared. Salt K sp AgI 1.5x10-16 CuI 5.0x10-12 CaSO 4 6.1x10-5 In all cases a Salt Cation + Anion CaSO 4 > CuI > AgI K sp values for salts that produce different numbers of ions (NaCl and MgCl 2 ) cannot be directly compared. Common Ion Effect The solubility of a solid is lowered if the solution already contains ions common to the solid. ph and Solubility ph can greatly affect a salt s solubility. If an anion, X -, is an effective base (meaning HX is a weak acid) the salt MX will show increased solubility in an acidic solution. Precipitation and the Ion Product When solutions are mixed a solid precipitate may form. Ion Product (Q) defined like K sp only using initial concentrations CaF 2 (s) Ca 2+ (aq) + 2F - (aq) Q = [Ca 2+ ] 0 [F - ] 0 2 Compare Q and K sp to predict whether or not a precipitate will form: If Q > K sp, precipitation will occur. If Q < K sp, no precipitation occurs.
Determining Precipitation Conditions A solution is prepared by adding 750.0 ml of 4.00x10-3 M Ce(NO 3 ) 3 to 300.0 ml of 2.00x10-2 M KIO 3. Will Ce(IO 3 ) 3 (K sp = 1.9x10-10 ) precipitate from this solution? Precipitation Reactions Determine the concentrations of the species in solution. Find Q to see if a precipitate will form. Perform stoichiometry on the initial amounts of reactants in solution (run the precipitation reaction to completion). Perform an equilibrium calculation to determine the reactant concentrations at equilibrium using ICE and the K sp value for the particular solid. Precipitation Reaction Example A solution is prepared by mixing 150.0 ml of 1.00x10-2 M Mg(NO 3 ) 2 and 250.0 ml of 1.00x10-1 M NaF. Calculate the concentration of Mg 2+ and F - at equilibrium with solid MgF 2 (K sp = 6.4x10-9 ).