A Molecular Approach, 4e - Notes for Chapter (17).doc

Chapter 17. Aqueous Ionic Equilibrium Chapter 17. Aqueous Ionic Equilibrium Chapter 17. Aqueous Ionic Equilibrium Student Objectives 17.1 The Danger of Antifreeze Know that the pH of blood is regulated by carbonic acid (H2CO3) and its anion, bicarbonate (HCO3), both of which are derived from carbon dioxide and water. Know that ethylene glycol (antifreeze) is metabolized to glycolic acid, which lowers the blood pH and lowers to ability of hemoglobin to transport oxygen. 17.2 Buffers: Solutions That Resist pH Change Define buffer and know that a buffer typically consists of a weak acid and its conjugate base. Know that the common ion effect is an example of Le Châtelier’s principle. Calculate the pH of a buffer solution starting with initial concentrations of weak acid and its conjugate base. Use the Henderson-Hasselbalch equation to calculate the pH of a buffer solution from the pKa of the weak acid and the initial concentrations of the weak acid and conjugate base. Know and understand that a buffer has limits; the buffering action is related to a stoichiometric consumption of either the weak acid or the conjugate base. Calculate the pH change in a buffer after the addition of a small amount of strong acid or base. Know that buffers can also be made from a weak base and its conjugate acid, and calculate the pH of such a buffer using the Henderson-Hasselbalch equation and the pKa of the conjugate acid. 17.3 Buffer Effectiveness: Buffer Range and Buffer Capacity Know and understand that the buffer capacity is optimal when the ratio of acid and conjugate base differs by less than or equal to a factor of 10. Know and understand that a buffer is more resistant to pH changes when the concentrations of acid and base are relatively large. Know and understand that an effective buffer covers a pH range one unit above and one unit below the pKa of the weak acid. Use the pKa and the buffer range to choose appropriate buffer systems for specific applications. 17.4 Titrations and pH Curves Calculate the pH of a solution at various points along a titration of a strong acid with a strong base. Calculate the pH of a solution at various points along a titration of a weak acid with a strong base. Recognize the titration curve for the titration of a weak base with a strong acid. Recognize the titration curve for the titration of a polyprotic acid with a strong base. Define equivalence point for an acid–base titration. Know that the equivalence point of an acid–base titration can be determined using a pH meter. Define endpoint for an acid–base indicator. Know that the equivalence point of an acid–base titration can be approximated by observing the endpoint of an acid–base indicator. Know that indicators can only be used for the specific pH ranges in which the indicator changes color. 17.5 Solubility Equilibria and the Solubility Product Constant Know that the solubility product constant, Ksp, defines the equilibrium constant for the dissolution of an ionic compound into its constituent ions. Calculate the molar solubility of an ionic compound in pure water using the Ksp expression and an ICE table. Know that the solubility of an ionic compound is lower in a solution containing a common ion than in pure water. Calculate the molar solubility of an ionic compound in the presence of a common ion using the Ksp expression and an ICE table. Know that the solubility of an ionic compound containing a basic anion increases with decreasing pH. 17.6 Precipitation Know that the precipitation of an ionic compound upon mixing aqueous solutions can be predicted using the Q expression and comparing Q to Ksp. Define selective precipitation, and know and understand how cations can be selectively precipitated based on their differential solubility with various anions. 17.7 Qualitative Chemical Analysis Know and understand that differential precipitation is the basis of the qualitative analysis of cations in solution. Know that the qualitative analysis scheme is based on, sequentially, insoluble chlorides, acid-insoluble sulfides, base-insoluble sulfides and hydroxides, insoluble phosphates, and alkali metals and ammonium. Know that alkali metals give a characteristic color upon heating in an open flame. 17.8 Complex Ion Equilibria Define complex ion and ligand. Know that metal cations form complex ions with a series of anionic and neutral ligands. Calculate complex ion equilibria using formation constants, Kf, and cation and ligand concentrations. Know that amphoteric metal hydroxides, like Al(OH)3, are more soluble in acidic and in basic solution than in pure water. Section Summaries Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples Teaching Tips Suggestions and Examples Misconceptions and Pitfalls Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 17.1 The Danger of Antifreeze Blood pH control carbonic acid and bicarbonate Antifreeze (ethylene glycol) sweet taste glycolic acid product dangerous pH decrease Intro figure: illustration of red blood cells and pH scale 17.2 Buffers: Solutions That Resist pH Change Combination of weak acid and conjugate base neutralizes strong acid and strong base calculations pH from Ka, [HA], and [A] Henderson-Hasselbalch equation buffer action on strong acid, strong base stoichiometry calculation equilibrium calculation Combination of weak base and conjugate acid same behavior as reverse example: NH3/NH4+ Figure 17.1 A Buffer Solution Figure 17.2 The Common Ion Effect Example 17.1 Calculating the pH of a Buffer Solution Example 17.2 Calculating the pH of a Buffer Solution as an Equilibrium Problem and with the Henderson-Hasselbalch Equation unnumbered table: ICE table for a buffer problem Figure 17.3 Buffering Action Example 17.3 Calculating the pH Change in a Buffer Solution after the Addition of a Small Amount of Strong Acid or Base Figure 17.4 Buffer Containing a Base Example 17.4 Using the Henderson-Hasselbalch Equation to Calculate the pH of a Buffer Solution Composed of a Weak Base and Its Conjugate Acid 17.3 Buffer Effectiveness: Buffer Range and Buffer Capacity Buffer effectiveness relative concentrations of [HA] and [A] absolute concentrations of [HA] and [A] Buffer range 1 pH unit pKa 1 Buffer capacity absolute concentrations unnumbered figures: ICE tables for buffer problem calculations unnumbered figures: illustrations of concentrated and dilute buffers Example 17.5 Preparing a Buffer Chemistry and Medicine: Buffer Effectiveness in Human Blood Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 17.1 The Danger of Antifreeze The example of antifreeze poisoning includes several important concepts: blood is maintained in a narrow pH range using a carbonic acid/bicarbonate buffer system; ethylene glycol is oxidized to an acid in the body and acts as a weak acid, exceeding buffer capacity; hemoglobin is a poorer oxygen carrier at low pH. The body uses carbon dioxide in the form of carbonic acid and bicarbonate to maintain blood plasma pH within a narrow range. The acidity of carbon dioxide can be measured by adding dry ice to a solution of universal indicator. 17.2 Buffers: Solutions That Resist pH Change Buffer solutions contain relatively large amounts of a weak acid and its conjugate base. Conceptual Connection 17.1 Buffers Buffering is easily demonstrated by adding a few drops of 1 M HCl and 1 M NaOH to two solutions containing universal indicator, one buffered and one not. Emphasize to students that buffer components are different only by a proton. They tend to forget this in writing reactions for buffer solutions. Reactions of buffers always convert one component into the other. Conceptual Connection 17.2 pH of Buffer Solutions Conceptual Connection 17.4 Adding Acid or Base to a Buffer When the x is small approximation is valid, the Henderson-Hasselbalch equation gives exactly the same results as obtained through the equilibrium expression. Buffer problems, especially when strong acid or base is added to a buffer, are notoriously difficult for students. Working through several examples is key. 17.3 Buffer Effectiveness: Buffer Range and Buffer Capacity The Chemistry and Medicine box revisits the buffer capacity issue mentioned in the chapter opener. Conceptual Connection 17.5 Buffer Capacity Buffers are not infinitely resistant to pH changes. Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 17.4 Titrations and pH Curves pH changes during titrations strong acid with strong base initial pH pH before equivalence point pH at equivalence point pH past equivalence point plot of pH vs. volume of added base weak acid with strong base initial pH pH before equivalence point pH at equivalence point pH past equivalence point plot of pH vs. volume of added base curve of weak base/strong acid titration curve of polyprotic acid/strong base titration Indicators endpoint examples phenolphthalein methyl red table of others Figure 17.5 Acid–Base Titration Figure 17.6 Titration Curve: Strong Acid + Strong Base unnumbered figures: graphs of pH vs. volume of base added in an strong acid-strong base titration Figure 17.7 Titration Curve: Strong Base + Strong Acid Example 17.6 Strong Acid-Strong Base Titration pH Curve unnumbered figures: graphs of pH vs. volume of base added in an weak acid-strong base titration Example 17.7 Weak Acid-Strong Base Titration pH Curve Figure 17.8 Titration Curve: Weak Base with Strong Acid Figure17.9 Titration Curve: Diprotic Acid with Strong Figure 17.10 Monitoring the pH during a Titration Figure 17.11 Monitoring the Color Change during a Titration Figure 17.12 Phenolphthalein Figure 17.13 Indicator Color Change Table 17.1 Ranges of Color Changes for Several Acid–Base Indicators 17.5 Solubility Equilibria and the Solubility Product Constant Solubility product constant, Ksp reaction: solid dissolves to ions equilibrium expression Calculating molar solubility pure water relative solubility common ion effect effect of pH Hard water Table 17.2 Selected Solubility Product Constants (Ksp) Example 17.8 Calculating Molar Solubility from Ksp Chemistry in Your Day: Hard Water Example 17.9 Calculating Ksp from Molar Solubility Example 17.10 Calculating Molar Solubility in the Presence of a Common Ion unnumbered figure: photograph of stalactites and stalagmites Example 17.11 The Effect of pH on Solubility Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 17.4 Titrations and pH Curves Calculating the various pH values that are obtained during a titration is tedious but illustrative of important concepts. After demonstrating one or two of these, have the students perform additional ones in small groups. Students typically find titration calculations to be just as challenging as if not more so than buffer calculations. Emphasize the importance of writing the relevant balanced reaction for each calculation. Students again write odd reactions for titrations. Emphasize that protons are changing places, nothing more. Conceptual Connection 17.6 Titration Equivalence Point Conceptual Connection 17.7 The Half-Equivalence Point Conceptual Connection 17.8 Acid–Base Titrations Indicators enable one to conduct a titration without instrumentation like a pH meter. A demonstration is useful to show how quickly the colors change near the equivalence point. Students initially presume that equivalence point is synonymous with neutral pH. Calculating the pH at the equivalence point of a weak acid/strong base titration is the biggest challenge. Viewing this point as a series of two steps, complete reaction with the strong base followed by partial dissociation, makes the calculations easier. Students are sometimes surprised to see huge pH changes with very little added titrant near the equivalence point. Remind them that pH is logarithmic while addition of titrant is not. 17.5 Solubility Equilibria and the Solubility Product Constant Solving these problems becomes routine since it requires only obtaining the square root of a Ksp value for ionic compounds with a 1:1 cation to anion ratio. For cases involving a 1:2 ratio, one must solve Ksp = 4s3. A common mistake is to forget to square both the s and the 2. The common ion effect can be demonstrated by adding concentrated hydrochloric acid or concentrated sodium hydroxide to a saturated solution of NaCl. The increased solubility in acid of an ionic compound containing a base can be easily demonstrated by adding concentrated hydrochloric acid to a mixture containing pure water and a piece of chalk. Conceptual Connection 17.9 Common Ion Effect The reaction for Ksp is the dissolution of the ionic solid to the dissolved ions, not the other way around. Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 17.6 Precipitation Reaction quotient, Q Q > Ksp: precipitation Q < Ksp: no precipitation Q = Ksp: saturated solution Relative ion concentrations unsaturated saturated supersaturated Selective precipitation unnumbered figure: photo of silver(I) chromate precipitation Figure 17.14 Precipitation from a Supersaturated Solution Example 17.12 Predicting Precipitation Reactions by Comparing Q and Ksp Example 17.13 Finding the Minimum Required Reagent Concentration for Selective Precipitation Example 17.14 Finding the Concentrations of Ions Left in Solution after Selective Precipitation 17.7 Qualitative Chemical Analysis Selective precipitation of cations insoluble chlorides insoluble sulfides under acidic conditions insoluble sulfides under basic conditions insoluble hydroxides insoluble phosphates alkali metals and NH4+ (flame test) Figure 17.15 Qualitative Analysis Figure 17.16 A General Qualitative Analysis Scheme unnumbered figure: photo of sulfides Figure 17.17 Flame Tests 17.8 Complex Ion Equilibria Complex ion metal cation ligands anions molecules Formation constant, Kf equilibrium reaction effect on solubility calculations Amphoteric metal hydroxides acidic: aluminum complex with water molecules (soluble) pH-neutral: aluminum complex with water molecules and hydroxide ions (insoluble) basic: aluminum complex primarily with hydroxide ions (soluble) Table 17.3 Formation Constants of Selected Complex Ions in Water at 25 °C Example 17.15 Complex Ion Equilibria Figure 17.18 Complex Ion Formation Figure 17.19 Solubility of an Amphoteric Hydroxide Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 17.6 Precipitation Predicting precipitate formation involves the same problem the students learned how to solve in Chapter 15: comparing Q with K. Remind students that one does not simply add an anion to an aqueous solution. The anion is accompanied by a countercation. Selective precipitation makes use of the differences in Ksp of ionic compounds. For ionic compounds with a 1:2 cation to anion ratio, students may forget to account for the stoichiometry in the Q expression. 17.7 Qualitative Chemical Analysis Qualitative analysis is a simple yet powerful method to identify cations present in a solution. Most of the scheme utilizes selective precipitation. Flame tests are used for the alkali metals since they do not generally form precipitates. Qualitative schemes develop logic and problem solving skills. 17.8 Complex Ion Equilibria Complex ions are formed between metal cations and a variety of anionic and neutral ligands. Their formation constants vary but tend to be generally large. Complex formation can be viewed as another example of Le Châtelier’s principle. Conceptual Connection 17.10 Solubility and Complex Ion Equilibria Aluminum (and other cations such as Zn) can form amphoteric hydroxides, ones that have increased solubility in both acidic and basic solution. In problems involving solubility and complex ion formation, one often has to combine two reactions; one for Ksp and one for Kf. Students may forget to multiply those equilibrium constants rather than adding them together. Additional Problem for Calculating the pH Change in a Buffer Solution After the Addition of a Small Amount of Strong Acid or Base (Example 17.3) A 1.0 L buffer solution contains 0.15 mol acetic acid and 0.10 mol sodium acetate. The Ka for acetic acid is 1.8 ´ 105. Calculate the pH of the solution before and after a 0.010 mol sample of NaOH is added to the buffer. (Ignore any change in volume upon adding the NaOH.) Initial The solution involves 1.0 L, so mol quantities convert directly to molar quantities. Find the pH of the buffer using the pKa and the Henderson-Hasselbalch equation. Ka = 1.8 ´ 105 pKa = log(1.8 105) = 4.74 Stoichiometry The addition of the NaOH base converts a stoichiometric amount of acid to the conjugate base. Write the equation for the neutralization and set up a table to track the changes. OH(aq) + HA(aq) H2O(l) + A(aq) [OH] [HA] [A] Before 0.00 mol 0.15 mol 0.10 mol Addition 0.010 mol After 0.00 m l 0.14 mol 0.11 mol Equilibrium Write the equation for the balanced equation and the Henderson-Hasselbalch expression. Confirm that the x is small approximation is valid: hydronium ion concentration divided by the after-addition HA concentration. HA(aq) + H2O(l) H3O+(aq) + A(aq) At pH = 4.564, [H3O+] = 2.73 ´ 105 M At pH = 4.635, [H3O+] = 2.32 ´ 105 M Check The buffer pH changed about 0.071 of a pH unit after the hydroxide addition. If the hydroxide was added to pure water, the pH of pure water would change from 7 to 12: pOH = log[OH] = log(0.010) = 2.00 pOH + pH = 14.00 pH = 14 – 2.00 = 12.00 The pH change after the hydroxide addition was to a larger pH or lower [H3O+] concentration, just as expected. Additional Problem for Calculating Molar Solubility from Ksp (Example 17.8) Calculate the molar solubility of iron carbonate FeCO3 in pure water. 1) Begin by writing the reaction by which the ionic compound dissolves in water. Write the corresponding expression for Ksp. FeCO3(s) Fe2+(aq) + CO32?(aq) Ksp = [Fe2+] [CO32?] 2) Use the stoichiometry of the reaction to write an ICE table, showing the equilibrium concentrations of the ions. FeCO3(s) Fe2+(aq) + CO32?(aq) [Fe2+] [CO32] Initial 0.00 0.00 Change +s +s Equil s s 3) Substitute the equilibrium expression with the concentration values from the ICE table. Substitute the numerical value of Ksp and solve for s. Ksp = [Fe2+] [CO32] 3.07 ´ 1011 = (s)(s) s2 = 3.07 ´ 1011 s = (3.07 ´ 1011) = 5.54 ´ 105 [Fe2+] = 5.54 ´ 105 M [CO32] = 5.54 ´ 105 M Additional Problem for Calculating Molar Solubility in the Presence of a Common Ion (Example 17.10) Calculate the molar solubility of calcium sulfate CaSO4 in a solution containing 0.20 M Na2SO4. 1) Begin by writing the reaction by which the ionic compound dissolves in water. Write the corresponding expression for Ksp. CaSO4(s) Ca2+(aq) + SO42(aq) Ksp = [Ca2+] [SO42] 2) Use the stoichiometry of the reaction to write an ICE table, showing the equilibrium concentrations of the ions. Add column for the common ion. CaSO4(s) Ca2+(aq) + SO42(aq) [Ca2+] [SO42] Initial 0.00 0.20 Change +s +s Equil s 0.20 + s 3) Substitute the equilibrium expression with the concentration values from the ICE table. Substitute the numerical value of Ksp and solve for s. Ksp = [Ca2+] [SO42] 7.10 ´ 105 = s(0.20 + s) s is small 0.20s = 7.10 ´ 105 s = 7.10 ´ 105 / 0.20 = 3.55 ´ 104 [Ca2+] = 3.55 ´ 104 M [SO42] = 0.200355 M = 0.20 M Check In the absence of the added sulfate, Ksp = [Ca2+] [SO42] 7.10 ´ 105 = (s)(s) s2 = 7.10 ´ 105 s = (7.10 ´ 105) = 8.43 ´ 103 Solubility of [Ca2+] in pure water = 8.43 ´ 103 M (84.3 ´ 104 M), a factor of about 25 greater than the solubility with the added sulfate. Additional Problem for Complex Ion Equilibria (Example 17.15) A 250-mL portion of a solution that contains 1.5 mM Cu(NO3)2 is mixed with a solution that is 0.100 M NaCN. After equilibrium is reached, what concentration of Cu2+(aq) remains? 1) Begin by writing the balanced equation for the complex ion equilibrium. Calculate the initial concentrations of the cation and the ligand. Write the corresponding expression for Kf. Cu2+(aq) + 4CN(aq) Cu(CN)42(aq) 2) Use the stoichiometry of the reaction to write an ICE table, showing the initial concentrations of the ions. Cu2+(aq) + 4CN(aq) Cu(CN)42(aq) [Cu2+] [CN] [Cu(CN)42] Initial 7.50´104 0.050 0.00 Change Equil 3) Since the equilibrium is so large, and the concentration for cyanide is much greater than that for Cu2+, virtually all of the copper ion is consumed. Let x represent the small amount of copper that remains. Cu2+(aq) + 4CN(aq) Cu(CN)42(aq) [Cu2+] [CN] [Cu(CN)42] Initial 7.50´104 0.050 0.00 Change (7.50´104) 4(7.50´104) 0.003 (+7.50´104) Equil x 0.050 7.50´104 4) Substitute the equilibrium expression with the concentration values from the ICE table. Substitute the numerical value of Kf and solve for x. [Cu2+] = 1.50 ´ 1027 M Check The copper ion concentration is very low since the formation constant is so large. The value of x was small compared to the change in copper ions (7.50 ´ 104). 232 Copyright © 2017 by Education, Inc. 233 Copyright © 2017 by Education, Inc.