student chap8

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Published on January 4, 2008

Author: Jeremiah

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Slide1:  Applications of Aqueous Equilibria Chapter 8 Slide2:  Chapter 8: Applications of Aqueous Equilibria 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations and pH Curves 8.6 Acid-Base Indicators 8.7 Titration of Polyprotic Acids 8.8 Solubility Equilibria and The Solubility Product 8.9 Precipitation and Qualitative Analysis 8.10 Complex Ion Equilibria A base swirling in a solution containing phenolphthalein:  A base swirling in a solution containing phenolphthalein Le Châtelier’s principle for the dissociation equilibrium for HF:  Le Châtelier’s principle for the dissociation equilibrium for HF HF(aq) H+(aq) + F-(aq) Molecular model: F-, Na+, HF, H2O:  Molecular model: F-, Na+, HF, H2O Slide6:  Like Example 8.1 (P 278-9) - I Nitrous acid, a very weak acid, is only 2.0% ionized in a 0.12 M solution. Calculate the [H+], the pH, and the percent dissociation of HNO2 in a 1.0 M solution that is also 1.0 M in NaNO2! HNO2(aq) H+(aq) + NO2-(aq) Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [HNO2]0 = 1.0 M [HNO2] = 1.0 – x (from dissolved HNO2) [NO2-]0 = 1.0 M [NO2-] = 1.0 + x (from dissolved NaNO2) [H+]0 = 0 [H+] = x (neglect the contribution from water) Slide7:  Like Example 8.1 (P 274-5) - II Ka = = = 4.0 x 10-4 [H+] [NO2-] [HNO2] ( x ) ( 1.0 + x ) (1.0 – x ) Assume x is small as compared to 1.0: X (1.0) (1.0) = 4.0 x 10-4 or x = 4.0 x 10-4 = [H+] Therefore pH = - log [H+] = - log ( 4.0 x 10-4 ) = _________ The percent dissociation is: Nitrous acid Nitrous acid alone + NaNO2 [H+] 2.0 x 10-2 4.0 x 10-4 pH 1.70 _______ % Diss 2.0 _______ Slide8:  Example 8.2 (P279-82) - I A buffered solution contains 0.50 M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and o.50 M sodium acetate (NaC2H3O2). Calculate the pH of this solution, and the pH when 0.010 M of solid NaOH is added to 1.0 L of this buffer and to pure water. HC2H3O2 (aq) H+(aq) + C2H3O2 (aq) Ka = 1.8 x 10-5 = [H+] [C2H3O2-] [HC2H3O2] Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [HC2H3O2]0 = 0.50 [HC2H3O2] = 0.50 – x [C2H3O2-]0 = 0.50 [C2H3O2-] = 0.50 +x [H+]0 = 0 [H+] = x X mol/L of HC2H3O2 dissociates to reach equilibrium ~ Slide9:  Example 8.2 (P279-82) - II Ka = 1.8 x 10-5 = = = [H+][C2H3O2-] [HC2H3O2] ( x ) ( 0.50 + x) 0.50 - x (x) (0.50) 0.50 x = 1.8 x 10-5 The approximation by the 5% rule is fine: [H+] = x = 1.8 x 10-5 M and pH = 4.74 To calculate the pH and concentrations after adding the base: OH- + HC2H3O2 H2O + C2H3O2- Before reaction: 0.010 mol 0.50 mol - 0.50 mol After reaction: 0.010 – 0.010 0.50 – 0.10 - 0.50 + 0.10 = 0 mol = 0.49 mol = 0.51 mol Note that 0.01 mol of acetic acid has been converted to acetate ion by the addition of the base. ~ Slide10:  Example 8.2 (P279-82) - III Initial Concentration (mol/L) Equilibrium concentration (mol/L) [HC2H3O2]0 = 0.49 [HC2H3O2] = 0.49 – x [C2H3O2-]0 = 0.51 [C2H3O2-] = 0.51 + x [H+]0 = 0 [H+] = x X mol/L of HC2H3O2 Dissociates to reach equilibrium Ka = 1.8 x 10-5 = = = [H+][C2H3O2-] [HC2H3O2] (x)(0.51+ x) 0.49 - x (x)(0.51) 0.49 x = ______________ and pH = ___________ If the base is added to pure water without the buffer being present we get an entirely different solution: If the 0.01 mol of NaOH is added to 1.0 L of pure water the Concentration of hydroxide ion is 0.01 M. [H+] = = = __________ and the pH = ______ Kw [OH-] 1.0 x 10-14 1.0 x 10-2 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. After the stoichiometric calculations are completed, then consider the equilibrium calculations. This procedure can be represented as follows::  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. After the stoichiometric calculations are completed, then consider the equilibrium calculations. This procedure can be represented as follows: OH- ions are not allowed to accumulate but are replaced by A- ions.:  OH- ions are not allowed to accumulate but are replaced by A- ions. When the OH- is added, the concentrations of HA and A- change, but only by small amounts. Under these conditions the [HA]/[A-] ratio and thus the [H+] stay virtually constant.:  When the OH- is added, the concentrations of HA and A- change, but only by small amounts. Under these conditions the [HA]/[A-] ratio and thus the [H+] stay virtually constant. Slide15:  How Does a Buffer Work Lets add a strong base to a weak acid and see what happens: OH- + HA A- + H2O Original buffer pH Final pH of buffer close to original- Added OH- ions Replaced by A- ions Ka = [H+] [A-] [HA] [H+] = Ka [HA] [A-] Slide16:  The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH3COOH] [CH3COO-]added % Dissociation* pH 0.10 0.00 1.3 2.89 0.10 0.050 0.036 4.44 0.10 0.10 0.018 4.74 0.10 0.15 0.012 4.92 * % Dissociation = x 100 [CH3COOH]dissoc [CH3COOH]init Human blood is a buffered solution:  Human blood is a buffered solution Source: Visuals Unlimited CO2 (g), H2CO3 (aq) and HCO3-(aq) are the buffering components in Blood that hold the pH to a range that will allow Hemoglobin to transport oxygen from the lungs to the cells of the body for metabolism. Pure water at pH 7.00:  Pure water at pH 7.00 Molecular model: HC2H3O2, C2H3O2-:  Molecular model: HC2H3O2, C2H3O2- Slide22:  How a Buffer Works–I A buffer consists of a solution that contains “high” concentrations of the acidic and basic components. This is normally a weak acid and the anion of that weak acid, or a weak base and the corresponding cation of the weak base. When small quantities of H3O+ or OH- are added to the buffer, they cause a small amount of one buffer component to convert into the other. As long as the amounts of H3O+ and OH- are small as compared to the concentrations of the acid and base in the buffer, the added ions will have little effect on the pH since they are consumed by the buffer components. Consider a buffer made from acetic acid and sodium acetate: CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Ka = or [H3O+] = Ka x [CH3COO-] [H3O+] [CH3COOH] Slide23:  How a Buffer Works–II Let’s consider a buffer made by placing 0.25 mol of acetic acid and 0.25 mol of sodium acetate per liter of solution. What is the pH of the buffer? And what will be the pH of 100.00 mL of the buffer before and after 1.00 mL of concentrated HCl (12.0 M) is added to the buffer? What will be the pH of 300.00 mL of pure water if the same acid is added? [H3O+] = Ka x = 1.8 x 10-5 x = 1.8 x 10-5 (0.25) (0.25) pH = -log[H3O+] = -log(1.8 x 10-5) = pH = _____ Before acid added! 1.00 mL conc. HCl 1.00 mL x 12.0 mol/L = 0.012 mol H3O+ Added to 300.00 mL of water : pH = -log(0.0399 M) pH = _____ Without buffer! Slide24:  How a Buffer Works–III After acid is added: Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+ Initial 0.250 ---- 0.250 0 Change +0.012 ---- -0.012 0.012 Equilibrium 0.262 ---- 0.238 0.012 Solving for the quantity ionized: Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+ Initial 0.262 ---- 0.238 0 Change -x ---- +x +x Equilibrium 0.262 - x ---- 0.238 + x x [H3O+] = Ka x =1.8 x 10-5 x = 1.982 x 10-5 (0.262) (0.238) Assuming: 0.262 - x = 0.262 & 0.238 + x = 0.238 pH = -log(1.982 x 10-5) = 5.000 - 0.297 = _____After the acid is added! Slide25:  How a Buffer Works–IV Suppose we add 1.0 mL of a concentrated base instead of an acid. Add 1.0 mL of 12.0 M NaOH to pure water and our buffer, and let’s see what the impact is: 1.00 mL x 12.0 mol OH-/1000mL = 0.012 mol OH- This will reduce the quantity of acid present and force the equilibrium to produce more hydronium ion to replace that neutralized by the addition of the base! Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+ Initial 0.250 ---- 0.250 0 Change - 0.012 ---- +0.012 +0.012 Equilibrium 0.238 ---- 0.262 +0.012 Assuming: Again, using x as the quantity of acid dissociated we get: our normal assumptions: 0.262 + x = 0.262 & 0.238 - x = 0.238 [H3O+] = 1.8 x 10-5 x = 1.635 x 10-5 0.238 0.262 pH = -log(1.635 x 10-5) = 5.000 - 0.214 = _____ After base is added! Slide26:  How a Buffer Works–V By adding the 1.00mL base to 300.00 mL of pure water we would get a hydroxide ion concentration of: This calculates out to give a pH of: The hydrogen ion concentration is: pH = -log(2.5 6 x 10-10) = 10.000 - 0.408 = 9.59 With 1.0 mL of the base in pure water! In summary: Buffer alone pH = 4.74 Buffer plus 1.0 mL base pH = 4.79 Base alone pH = 9.59 Buffer plus 1.0 mL acid pH = 4.70 Acid alone pH = 1.40 Slide27:  The Relation Between Buffer Capacity and pH Change A digital pH meter shows the pH of the buffered solution to be 4.74:  A digital pH meter shows the pH of the buffered solution to be 4.74 When 0.01 mol NaOH is added to 1.0 L of pure water, the pH jumps to 12.00:  When 0.01 mol NaOH is added to 1.0 L of pure water, the pH jumps to 12.00 Molecular model: Cl-, NH4+:  Molecular model: Cl-, NH4+ Slide31:  Preparing a Buffer Problem: The ammonia-ammonium ion buffer has a pH of about 9.2 and can be used to keep solutions in the basic pH range. What mass of ammonium chloride must be added to 400.00 mL of a 3.00 M ammonia solution to prepare a buffer ? Plan: The conjugate pair is the ammonia-ammonium ion pair which has an equilibrium constant Kb = 1.8 x 10 -5. The reaction equation with water can be written along with the Kb expression, since we want to add sufficient ammonium ion to equal the aqueous ammonia concentration. Solution: The reaction for the ammonia-ammonium ion buffer is: Kb = = 1.8 x 10-5 [NH4+] [OH-] [NH3] [NH4+] = 3.00 mol x 0.400 L = 1.20 mol NH4Cl = 53.49 g/mol Therefore mass =NH4Cl = 1.20 mol x 53.49g/mol mass = ________ g NH4Cl L Slide32:  The Henderson-Hasselbalch Equation Take the equilibrium ionization of a weak acid: HA(aq) + H2O(aq) = H3O+(aq) + A-(aq) Ka = [H3O+] [A-] [HA] Solving for the hydronium ion concentration gives: Taking the negative logarithm of both sides: Generalizing for any conjugate acid-base pair : pH = log Ka + log ( ) Henderson-Hasselbalch equation Slide33:  Like Example 8.3 (P 285-7) -I Problem: Instructions for making a buffer say to mix 60.0 ml of 0.100 M NH3 with 40.0 ml of 0.100 M NH4Cl. What is the pH of this buffer? The combined volume is 60.0 ml + 40.0 ml = 100.0 ml Moles of Ammonia = VolNH3 x MNH3 = 0.060 L x 0.100 M = 0.0060 mol Moles of Ammonium ion = VolNH4Cl x MNH4Cl = 0.040 L x 0.100 M = = 0.0040 mol [NH3] = = 0.060 M ; [NH4+] = = 0.040 M 0.0060 mol 0.100 L 0.0040 mol 0.100 L Concentration (M) NH3 (aq) + H2O(l) NH4+(aq) + OH-(aq) Starting 0.060 0.040 0 Change -x +x +x Equilibrium 0.060 – x 0.040 – x x Slide34:  Like Example 8.3 (P 285-7) - II Substituting into the equation for Kb: Kb = = 1.8 x 10-5 = [NH4+] [OH-] [NH3] (0.040 + x) (x) (0.060 – x) Assume : 0.060 – x = 0.060 ; 0.040 + x = 0.040 ~ ~ Kb = 1.8 x 10-5 = x = 2.7 x 10-5 0.040 (x) 0.060 Check assumptions: 0.040 + 0.000027 = 0.040 or 0.068% 0.060 – 0.000027 = 0.060 or 0.045% [OH-] = 2.7 x 10-5 ; pOH = - log[OH-] = - log (2.7 x 10-5) = 5 – 0.43 pOH = 4.57 pH = 14.00 – pOH = 14.00 – 4.57 = ___________ Slide35:  pH Box pH [H3O+] pOH [OH-] pOH = -log[OH-] [OH-] = 10-pOH pH = -log[H3O+] [H3O+] = 10-pH Kw = 1 x 10-14 @ 25oC [H3O+][OH-]= =1 x 10-14 pH + pOH = = 14 @ 25oC Slide36:  Summary: Characteristics of Buffered Solutions Buffered solutions contain relatively large concentrations of a weak acid and its corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base B and the conjugate acid BH+. When H+ is added to a buffered solution, it reacts essentially to completion with the weak base present: H+ + A- HA or H+ + B BH+ When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present. OH- + HA A- + H2O or OH- + BH+ B + H2O The pH of the buffered solution is determined by the ratio of the concentrations of the weak base and weak acid. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A- or G and BH+) are large compared with the amounts of H+ or OH- added. Slide37:  Exact Treatment of Buffer Solutions We can use several relationships to calculate the exact solution to buffered solution problems: Charge – balance equation: [Na+] + [H+] = [A-] + [OH-] Material – Balance equation: [A-]0 + [HA]0 = [HA] + [A-] Since [A-]0 = [Na+] and Kw = [OH-][H+] , we can rewrite the charge balance equation, and solve for [A-] : [H+]2 – Kw [H+] From the mass balance equation solved for [HA] we get: [HA] = [A-]0 + [HA]0 – [A-] Substituting the expression for [A-], and substituting into the Ka expression for HA we obtain: Slide38:  Example 8.4 (P 289-90) - I Calculate the pH of a buffered solution containing 3.0 x 10-4 M HOCl (Ka = 3.5 x 10-8) and 1.0 x 10-4 M NaOCl. Ka = = 3.5 x 10-8 [H+] [OCl-] [HOCl] Let x = [H+] then: [OCl-] = 1.0 x 10-4 + x [HOCl] = 3.0 x 10-4 - x 3.5 x 10-8 = = [H+] [OCl-] [HOCl] (x)(1.0 x 10-4 + x) (3.0 x 10-4 – x ) Assuming x is small compared to 1.0 x 10-4 and solving for x we have: 1.05 x 10-11 1.0 x 10-4 Since this is close to that of water we must use the equation that uses water, and takes it’s ionization into account. Slide39:  Example 8.4 (P 289-90) - II Ka = 3.5 x 10-8 = [H+]{[OCl-]0 + } [H+]2 – 1.0 x 10-14 [H+] Where: [OCl-]0 = 1.0 x 10-4 M [HOCl]0 = 3.0 x 10-4 M We expect [H+] to be close to 1.0 x 10-7, so [H+]2 to be about 1.0 x 10-14 [H+]2 – 1.0 x 10-14 [HOCl]0 = 1.0 x 10-4 M >>> [H+] The expression becomes: Slide40:  Example 8.4 (P 289-90) - III [H+] = 1.05 x 10-7 M = 1.1 x 10-7 M Using this result, we can check the magnitude of the neglected term: [H+]2 – 1.0 x 10-14 [H+] = (1.05 x 10-7)2 – 1.0 x 10-14 1.05 x 10-7 = 9.8 x 10-9 This result suggests that the approximation was fine! Slide41:  pH and Capacity of Buffered Solutions The pH of a buffered solution is determined by the ratio [A-]/[HA]. The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-] Slide42:  Example 8.5 (P 290-2) - I Calculate the change in pH that occurs when 0.010 mol of gaseous HCl is added to 1.0 L of each of the following solutions: Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2 Solution B: 0.050 M HC2H3O2 and 0.0500 M NaC2H3O2 For Acetic acid, Ka = 1.8 x 10-5 Use the Henderson-Hasselbalch equation for initial pH: pH = pKa + log{ } [C2H3O2-] [H C2H3O2] Since [C2H3O2-] = [H C2H3O2] The equation becomes: pH = pKa + log (1) = pKa = -log(1.8 x 10-5) = 4.74 Adding 0.010 mol of HCl will cause a shift in the equilibrium due to: H+(aq) + C2H3O2-(aq) H C2H3O2 (aq) Original solution and new solution:  Original solution and new solution Original solution and new solution:  Original solution and new solution Slide45:  Example 8.5 (P 290-2) - II For Solution A: H+ + C2H3O2- H C2H3O2 Before reaction 0.010 M 5.00 M 5.00 M After reaction 0 4.99 M 5.01 M Calculate the new pH using the Henderson-Hasselbalch equation: For Solution B: H+ + C2H3O2- H C2H3O2 Before reaction 0.010 M 0.050 M 0.050 M After reaction 0 0.040 M 0.060 M The new pH is: pH = 4.74 + log( ) = 4.74 – 0.18 = 4.56 0.040 0.060 Solution A and Solution B:  Solution A and Solution B A buret valve:  A buret valve Source: American Color Figure 8.1: The pH curve for the titration of 50.0 ml of Nitric acid with 0.10M NaOH:  Figure 8.1: The pH curve for the titration of 50.0 ml of Nitric acid with 0.10M NaOH Vol NaOH added (mL):  Vol NaOH added (mL) Figure 8.2: The pH curve for the titration of 100.0 ml of 0.50 M NaOH with 0.10 M HCl.:  Figure 8.2: The pH curve for the titration of 100.0 ml of 0.50 M NaOH with 0.10 M HCl. Weak acid:  Weak acid Figure 8.3: The pH curve for the titration of 50.0 ml of Acetic acid with 0.10 M NaOH:  Figure 8.3: The pH curve for the titration of 50.0 ml of Acetic acid with 0.10 M NaOH Treat the stoichiometry:  Treat the stoichiometry Figure 8.4: The pH curves for the titrations of 50.0:  Figure 8.4: The pH curves for the titrations of 50.0 Slide57:  Calculating the pH During a Weak Acid-Strong Base Titration–I Problem: Calculate the pH during the titration of 20.00 mL of 0.250 M nitrous acid (HNO2; Ka = 4.5 x 10-4) after adding different volumes of 0.150 M NaOH : (a) 0.00 mL (b) 15.00 mL (c) 20.00 mL (d) 35.00 mL. Plan: (a) We just calculate the pH of a weak acid. (b)-(d) We calculate the amounts of acid remaining after the reaction with the base, and the anion concentration, and plug these into the Henderson-Hasselbalch eq. Solution: HNO2 (aq) + NaOH(aq) H2O(l) + NaNO2 (aq) (a) Ka = = = 4.5 x 10-4 [H3O+] [NO2-] [HNO2] x (x) 0.250 M x2 = 1.125 x 10-4 x = 1.061 x 10-2 pH = -log(1.061 x 10-2) = 2.000 - 0.0257 = _______ no base added Slide58:  Calculating the pH During a Weak Acid-Strong Base Titration–II (b) 15.00 mL of 0.150 M NaOH is added to the 20.00 mL of 0.250 M HNO2 (20.00 mL x 0.250 mmol/mL = 5.00 mmol HNO2) which will neutralize (15.00 mL x 0.150 mmol/mL = 2.25 mmol of HNO2), leaving 2.75 mmol HNO2, and generating 2.25 mmol of nitrite anion. Concentration (M) Initial 0.00275 ---- 0 0.00225 Change -x ---- +x +x Equilibrium 0.00275 - x ---- x 0.00225 + x pH = 3.35 -0.0872 = ________ with 15.0 mL of NaOH added Slide59:  Calculating the pH During a Weak Acid-Strong Base Titration–III (c) 20.00 mL of 0.150 M NaOH is added to the 20.00 mL of 0.250 M HNO2 (20.00 mL x 0.250 mmol/mL = 5.00 mmol HNO2) which will neutralize (20.00 mL x 0.150 mmol/mL = 3.00 mmol of HNO2), leaving 2.00 mmol HNO2, and generating 3.00 mmol of nitrite anion. Concentration (M) Initial 0.00200 ---- x 0.00300 pH = 3.35 + 0.176 = ____________ with 20.00 mL of base added (d) 35.00 mL of 0.150 M NaOH is added to the 20.00 mL of 0.250 M HNO2 (20.00 mL x 0.250 mmol/mL = 5.00 mmol HNO2) which will neutralize (35.00 mL x 0.150 mmol/mL = 5.25 mmol of HNO2), leaving no HNO2, and generating 5.00 mmol of nitrite anion. There will be an excess of 0.25 mmol of NaOH which will control the pH. Slide60:  Calculating the pH During a Weak Acid-Strong Base Titration–IV (d) continued Since all of the HNO2 has been neutralized, we only have to look at the concentration of hydroxide ion in the total volume of the solution to calculate the pH of the resultant solution. combined volume = 20.00 mL + 35.00 mL = 55.00 mL [OH-] = =0.004545 M 0.000250 mol OH- 0.05500 L [H3O+] = = = 2.200 x 10-12 Kw [OH-] 1 x 10-14 0.004545 pH = -log (2.200 x 10-12) = 12.000 - 0.342 = ______ when all of the acid neutralized, and there is an excess of NaOH Slide61:  Summary: Titration Curve Calculations A Stoichiometry problem. The reaction of hydroxide ion with the weak acid is assumed to run to completion, and the concentrations of the acid remaining and the conjugate base formed are determined. An equilibrium problem. The position of the weak acid equilibrium is determined, and the pH is calculated. Figure 8.5: The pH curve for the titration of 100.0 ml of 0.050 M NH3 with ).10 M HCl:  Figure 8.5: The pH curve for the titration of 100.0 ml of 0.050 M NH3 with ).10 M HCl Figure 8.6: The indicator phenolphthalein is pink in basic solution and colorless in acidic solution.:  Figure 8.6: The indicator phenolphthalein is pink in basic solution and colorless in acidic solution. Figure 8.7: (a) Yellow acid form of bromthymol blue; (b) a greenish tint is seen when the solution contains 1 part blue and 10 parts yellow; (c) blue basic form.:  Figure 8.7: (a) Yellow acid form of bromthymol blue; (b) a greenish tint is seen when the solution contains 1 part blue and 10 parts yellow; (c) blue basic form. Figure 8.8: The useful pH ranges for several common indicators:  Figure 8.8: The useful pH ranges for several common indicators Slide69:  Colors and Approximate pH Range of Some Common Acid-Base Indicators Figure 8.9: pH curve of 0.10 M HCI being titrated with 0.10 M NaOH:  Figure 8.9: pH curve of 0.10 M HCI being titrated with 0.10 M NaOH Figure 8.10: pH of 0.10 M HC2H3O2 being titrated with NaOH:  Figure 8.10: pH of 0.10 M HC2H3O2 being titrated with NaOH Figure 8.11: A summary of the important equilibria at various points in the titration of a triprotic acid:  Figure 8.11: A summary of the important equilibria at various points in the titration of a triprotic acid An X ray of the upper gastrointestinal:  An X ray of the upper gastrointestinal Source: Photo Researchers, Inc. Precipitation of bismuth:  Precipitation of bismuth Slide77:  Equilibria of Slightly Soluble Ionic Compounds When a solution becomes saturated and a precipitate forms we move into the area of insoluble material in solution, and we begin to calculate the quantity of material that remains in solution. We are working with what we call the: “Solubility Product” The equilibrium constant that is used for these calculations is called the Solubility-product constant: Ksp Example : Lead chromate PbCrO4 (s) Pb2+(aq) + CrO42-(aq) Qc = [Pb2+][CrO42-] [PbCrO4] Since the concentration of a solid is constant, we can move it to the other side of the equals sign and combine it with the constant yielding the solubility product constant Ksp [PbCrO4] x Qc = [Pb2+][CrO42-] = Ksp Slide78:  Like Example 8.12 (P 320) The Ksp value for the mineral fluorite, CaF2 is 3.4 x 10-11 . Calculate The solubility of fluorite in units of grams per liter. Concentration (M) CaF2 (s) Ca2+(aq) + 2 F-(aq) Starting 0 0 Change +x +2x Equilibrium x 2x Substituting into Ksp: [Ca2+][F-]2 = Ksp (2x)2 = 3.4 x 10-11 4x3 = 3.4 x 10-11 x = x = 2.0 x 10-4 3.4 x 10-11 4 3 The solubility is 2.0 x 10-4 moles CaF2 per liter of water. To get mass we must multiply by the molar mass of CaF2 (78.1 g/mol). 2.0 x 10-4 mol CaF2 x = _________ g CaF2 per L 78.1 g CaF2 1 mol CaF2 Slide80:  Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds Problem: Write the ion-product expression for (a) silver bromide; (b) strontium phosphate; (c) aluminum carbonate; (d) nickel(III) sulfide. Plan: Write the equation for a saturated solution, then write the expression for the solubility product. Solution: (a) Silver bromide: AgBr(s) Ag+(aq) + Br -(aq) Ksp = [Ag+] [Br -] (b) Strontium phosphate: Sr3(PO4)(s) 3 Sr2+(aq) + 2 PO43-(aq) Ksp = [Sr2+]3[PO43-]2 (c) Aluminum carbonate: Al2(CO3)3 (s) 2 Al3+(aq) + 3 CO32-(aq) Ksp = [Al3+]2[CO32-]3 (d) Nickel(III) sulfide: Ni2S3 (s) + 3 H2O(l) 2 Ni3+(aq) + 3 HS -(aq) + 3 OH-(aq) Ksp =[Ni3+]2[HS-]3[OH-]3 Table 8.5: Ksp Values at 25 C for Common Ionic Solids:  Table 8.5: Ksp Values at 25 C for Common Ionic Solids Slide84:  Determining Ksp from Solubility Problem: Lead chromate is an insoluble compound that at one time was used as the pigment in the yellow stripes on highways. It’s solubility is 4.33 x 10 -6g/100mL water. What is the Ksp? Plan: We write an equation for the dissolution of the compound to see the number of ions formed, then write the ion-product expression. Solution: PbCrO4 (s) Pb2+(aq) + CrO42-(aq) Molar solubility of PbCrO4 = x x 4.33 x 10 -6g 100 mL 1000 ml 1 L 1mol PbCrO4 323.2 g = 1.34 x 10 -8 M PbCrO4 1 Mole PbCrO4 = 1 mole Pb2+ and 1 mole CrO42- Therefore [Pb2+] = [CrO42-] = 1.34 x 10-8 M Ksp = [Pb2+] [CrO42-] = (1.34 x 10 -8 M)2 = ________________ Slide85:  Determining Solubility from Ksp Problem: Lead chromate used to be used as the pigment for the yellow lines on roads, and is a very insoluble compound. Calculate the solubility of PbCrO4 in water if the Ksp is equal to 2.00 x 10-16. Plan: We write the dissolution equation, and the ion-product expression. Solution: Writing the dissolution equation, and the ion-product expression: Ksp = 2.00 x 10-16 =[Pb2+][CrO42] Concentration (M) PbCrO4 Pb2+ CrO42- Initial ---------- 0 0 Change ---------- +x +x Equilibrium ---------- x x Ksp = [Pb2+] [CrO42-] = (x)(x ) = 2.00 x 10-16 x = 1.41 x 10-8 Therefore the solubility of PbCrO4 in water is _______________ M Slide87:  Relationship Between Ksp and Solubility at 25oC No. of Ions Formula Cation:Anion Ksp Solubility (M) 2 MgCO3 1:1 3.5 x 10-8 1.9 x 10-4 2 PbSO4 1:1 1.6 x 10-8 1.3 x 10-4 2 BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5 3 Ca(OH)2 1:2 6.5 x 10-6 1.2 x 10-2 3 BaF2 1:2 1.5 x 10-6 7.2 x 10-3 3 CaF2 1:2 3.2 x 10-11 2.0 x 10-4 3 Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5 Slide88:  The Effect of a Common Ion on Solubility PbCrO4(s) Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO42-(aq; added) Slide89:  Calculating the Effect of a Common Ion on Solubility Problem: What is the solubility of silver chromate in 0.0600 M silver nitrate solution? Ksp = 2.6 x 10-12 . Plan: From the equation and the ion-product expression for Ag2CrO4, we predict that the addition of silver ion will decrease the solubility. Solution: Writing the equation and ion-product expression: Ag2CrO4 (s) 2 Ag+(aq) + CrO42-(aq) Ksp = [Ag+]2[CrO42-] Concentration (M) Ag2CrO4 (s) 2 Ag+(aq) + CrO42-(aq) Initial --------- 0.0600 0 Change --------- +2x +x Equilibrium --------- 0.0600 + 2x x Assuming that Ksp is small, 0.0600 M + 2x = 0.600 M Ksp = 2.6 x 10-12 = (0.0600)2(x) x = 7.22 x 10-10 M Therefore, the solubility of silver chromate is 7.22 x 10-10 M Slide90:  Test for the Presence of a Carbonate Slide91:  Predicting the Effect on Solubility of Adding Strong Acid Problem: Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of: (a) Iron(II) cyanide (b) Potassium bromide (c) Aluminum hydroxide Plan: Write the balanced dissolution equation and note the anion. Anions of weak acids react with H3O+ and shift the equilibrium position toward more dissolution. Strong acid anions do not react, so added acid has no effect. Solution: (a) Fe(CN)2 (s) Fe2+(aq) + 2 CN-(a) Increases solubility We noted earlier that CN- ion reacts with water to form the weak acid HCN, so it would be removed from the solubility expression. (b) KBr(s) K+(aq) + Br -(aq) No effect This occurs since Br- is the anion of a strong acid, and K+ is the cation of a strong base. (c) Al(OH)3 (s) Al3+(aq) + 3 OH-(aq) Increases solubility The OH- is the anion of water, a very weak acid, so it reacts with the added acid to produce water in a simple acid-base reaction. Slide92:  The Chemistry of Limestone Formation Gaseous CO2 is in equilibrium with aqueous CO2 in natural waters: CO2 (g) CO2 (aq) H2O(l) The concentration of CO2 is proportional to the partial pressure of CO2 (g) in contact with the water (Henry’s law; section 13.3): [CO2 (aq)] (proportional to) PCO2 The reaction of CO2 with water produces H3O+: CO2 (aq) + 2 H2O(l) H3O+(aq) + HCO3-(aq) Thus, the presence of CO2 (aq) forms H3O+, which increases the solubility of CaCO3: CaCO3 (s) + CO2 (aq) + H2O(l) Ca2+(aq) + 2 HCO3-(aq) Slide93:  Predicting the Formation of a Precipitate: Qsp vs. Ksp Qsp = Ksp : When a solution becomes saturated, no more solute will dissolve, and the solution is called “saturated.” There will be no changes that will occur. Qsp > Ksp : Precipitates will form until the solution becomes saturated. Qsp< Ksp : Solution is unsaturated, and no precipitate will form. The solubility produce constant, Ksp, can be compared to the ion-product constant, Qsp to understand the characteristics of a solution with respect to forming a precipitate. Slide94:  Predicting Whether a Precipitate Will Form–I Problem: Will a precipitate form when 0.100 L of a solution containing 0.055 M barium nitrate is added to 200.00 mL of a 0.100 M solution of sodium chromate? Plan: We first see if the solutions will yield soluble ions, then we calculate the concentrations, adding the two volumes together to get the total volume of the solution, then we calculate the product constant (Qsp), and compare it to the solubility product constant to see if a precipitate will form. Solution: Both Na2CrO4 and Ba(NO3) are soluble, so we will have Na+, CrO42-, Ba2+ and NO3- ions present in 0.300 L of solution. We change partners, look up solubilities, and we find that BaCrO4 would be insoluble, so we calculate it’s ion-product constant and compare it to the solubility product constant of 2.1 x 10-10: [Ba2+] = = ___________ M in Ba2+ For Ba2+: [0.100 L Ba(NO3)2] [0.55 M] = 0.055mol Ba2+ 0.055 mol Ba2+ 0.300 L Slide95:  Predicting Whether a Precipitate Will Form–II Solution cont. For CrO42- : [0.100 M Na2CrO4] [0.200 L] = 0.0200 mol CrO42- Qsp = [Ba2+] [CrO42-] =(0.183 M Ba2+)(0.667 M CrO42-) = 0.121 Since Ksp = 2.1 x 10-10 and Qsp = 0.121, Qsp >> Ksp and a precipitate will form. Figure 8.12: Separation of Cu2+ and Hg2+:  Figure 8.12: Separation of Cu2+ and Hg2+ Figure 8.13: Separation of common cations by precipitation:  Figure 8.13: Separation of common cations by precipitation Slide100:  The Stepwise Exchange of NH3 for H2O in M(H2O)42+ Slide101:  Formation Constants (Kf) of Some Complex Ions at 25oC–I Complex Ion Kf Ag(CN)2- 3.0 x 1020 Ag(NH3)2+ 1.7 x 107 Ag(S2O3)23- 4.7 x 1013 AlF63- 4 x 1019 Al(OH)4- 3 x 1033 Be(OH)42- 4 x 1018 CdI42- 1 x 106 Co(OH)42- 5 x 109 Cr(OH)4- 8.0 x 1029 Cu(NH3)42+ 5.6 x 1011 Fe(CN)64- 3 x 1035 Fe(CN)63- 4.0 x 1043 Slide102:  Formation Constants (Kf) of Some Complex Ions at 25oC–II Complex Ion Kf Hg(CN)42- 9.3 x 1038 Ni(OH)42- 2 x 1028 Pb(OH)3 - 8 x 1013 Sn(OH)3 - 3 x 1025 Zn(CN)42- 4.2 x 1019 Zn(NH3)42+ 7.8 x 108 Zn(OH)42- 3 x 1015 Slide103:  Calculating the Concentrations of Complex Ions–I Problem: A chemist converts Ag(H2O)2+ to the more stable form Ag(NH3)2+ by mixing 50.0 L of 0.0020 M Ag(H2O)2+ and 25.0 L of 0.15 M NH3. What is the final [Ag(H2O)2+]? Kf Ag(NH3)2+ = 1.7 x 107. Plan: We write the equation and the Kf expression, set up the table for the calculation, then substitute into Kf and solve. Solution: Writing the equation and Kf expression: Kf = = 1.7 x 107 [Ag(NH3)2+] [Ag(H2O)2+][NH3]2 Finding the initial concentrations: [Ag(H2O)2+]init = = 1.3 x 10-3 M 50.0 L (0.0020 M) 50.0 L + 25.0 L [NH3]init = = __________ M 25.0 L (0.15 M) 50.0 L + 25.0 L Slide104:  Calculating the Concentration of Complex Ions–II We assume that all of the Ag(H2O)2+ is converted Ag(NH3)2+, so we set up the table with x = [Ag(H2O)2+] at equilibrium. Ammonia reacted = [NH3]reacted = 2(1.3 x 10-3 M) = 2.6 x 10-3 M Concentration (M) Ag(H2O)2+(aq) 2NH3 (aq) Ag(NH3)2+ 2 H2O(aq) Initial 1.3 x 10-3 5.0 x 10-2 0 ---- Change ~(-1.3 x 10-3) ~(-2.6 x 10-3) ~(+1.3 x 10-3) ---- Equilibrium x 4.7 x 10-2 1.3 x 10-3 ---- Kf = = = 1.7 x 107 [Ag(H2O)2+][NH3]2 [Ag(NH3)2+] 1.3 x 10-3 x(4.7 x 10-2)2 x = _________________ M = [Ag(H2O)2+] Slide105:  The Amphoteric Behavior of Aluminum Hydroxide Slide106:  Separating Ions by Selective Precipitation–I Problem: A solution consists of 0.10 M AgNO3 and 0.15 M Cu(NO3)2. Calculate the [I -] that would separate the metal ions as their iodides. Kspof AgI = 8.3 x 10-17; Kspof CuI = 1.0 x 10-12. Plan: Since the two iodides have the same formula type (1:1), we compare their Ksp values and we see that CuI is about 100,000 times more soluble than AgI. Therefore, AgI precipitates first, and we solve for [I -] that will give a saturated solution of AgI. Solution: Writing chemical equations and ion-product expressions: AgI(s) Ag+(aq) + I -(aq) Ksp = [Ag+][I -] CuI(s) Cu+(aq) + I -(aq) Ksp = [Cu+][I -] H2O H2O Calculating the quantity of iodide needed to give a saturated solution of CuI: [I -] = = = _________________ M Ksp [Cu+] 1.0 x 10-12 0.15 M Slide107:  Separating Ions by Selective Precipitation–II Thus, the concentration of iodide ion that will give a saturated solution of copper(I) iodide is 1.0 x 10-11 M, and this will not precipitate the copper(I) ion, but should remove most of the silver ion. Calculating the quantity of silver ion remaining in solution we get: [Ag+] = = = 1.2 x 10-6 M Ksp [I -] 8.3 x 10-17 6.7 x 10-11 Since the initial silver ion was 0.10 M, most of it has been removed, and essentially none of the copper(I) was removed, so the separation was quite complete. If the iodide was added as sodium iodide, you would have to add only a few nanograms of NaI to remove nearly all of the silver from solution: 6.7 x 10-11 mol I - x x = _____ ng NaI 149.9 g NaI mol NaI 1 molNaI mol I - Slide108:  The General Procedure for Separating Ions in Qualitative Analysis Slide109:  Separation into Ion Groups Ion Group 1: Insoluble chlorides Ag+, Hg22+, Pb2+ Ion Group 2: Acid-insoluble sulfides Cu2+, Cd2+, Hg2+, As3+, Sb3+, Bi3+, Sn2+, Sn4+, Pb2+ Ion Group 3: Base-insoluble sulfides and hydroxides Zn2+, Mn2+, Ni2+, Fe2+, Co2+ as sulfides, and Al3+, Cr3+ as hydroxides Ion Group 4: Insoluble phosphates Mg2+, Ca2+, Ba2+ Ion Group 5: Alkali metal and ammonium ions Na+, K+, NH4+ Slide111:  Tests to Determine the Presence of Cations in Ion Group 5 Na+ ions K+ ions OH - + NH4+ NH3 + H2O plus litmus paper Slide112:  A Qualitative Analysis Scheme for Ag+, Al3+, Cu2+ and Fe3+ Figure 8.14: Separation of Group I ions:  Figure 8.14: Separation of Group I ions Slide117:  The Acid Rain Problem–I Normally the pH of precipitation is controlled by the reaction of carbon dioxide with water to form carbonic acid, which keeps the pH of “clean” rain in the slightly acid range, of about 5.6: CO2 (g) + H2O(l) H2CO3 (aq) H3O+(aq) + HCO3-(aq) Sulfurous acid is produced by the reaction of sulfur dioxide with water, and even though it is a weak acid, it does add to the acidity of rain. A greater worry is the oxidation of the SO2 to form SO3 which reacts with water to form the strong acid, sulfuric acid. SO2 (g) + H2O(l) H2SO3 (aq) 2 SO2 (g) + O2 (g) 2 SO3 (g) SO3 (g) + H2O(l) H2SO4 (aq) Slide118:  The Acid Rain Problem–II Another strong acid is formed from the nitrogen oxides that are produced in all internal combustion engines, that is nitric acid, which is not only a strong acid, but also a strongly oxidizing acid. N2 (g) + O2 (g) 2 NO(g) 2 NO(g) + O2 (g) 2 NO2 (g) 3 NO2 (g) + H2O(l) 2 HNO3 (aq) + NO(g) (Final step in the Ostwald process) These acids together make the problem of acid rain so severe in the eastern United States, and far worse in other parts of the world. The average pH of rain in the eastern U.S. back in 1984 was 4.2. Sweden and Pennsylvania share second place with a pH of 2.7, but the record was in Wheeling, West Virginia where the pH was 1.8. Some areas of California also reach a pH of 1.6. The problems are global in nature. Slide119:  Formation of Acidic Precipitation

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