Scruggs Chap 17 (all)

Scruggs Chap 17 - Chapter 17 Additional Aspects of Aqueous Equilibria q Water is a common solvent Dissolved materials can be involved in different

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Unformatted text preview: Chapter 17 Additional Aspects of Aqueous Equilibria q Water is a common solvent. Dissolved materials can be involved in different types of chemical equilibria. q Acid-Base q Solubility 17.1 The Common Ion Effect q Metal ions or salts containing a conjugate weak acid or base can shift the pH of a acid or base solution q We can predict pH changes using Le Chatelier's principle HA + H2O A + H3 O+ h q h Common Ion Effect q CH3CO2H + H2O h CH3CO2- + H3O+ h q Adding more CH3CO2- to the solution shifts the equilibrium to the left, making the solution less acidic (higher pH). q 0.100 M CH3CO2H q 0.100 M CH3CO2H, 0.050 M NaCH3CO2 pH = 2.879 pH = 4.456 pH of Weak Acid Calculation q What is the pH of a 1.00 M HF solution? qKa = 7.0 x 10- 4 q HF H3O+ FInitial 1.00 0 0 Change -x +x +x Equil. 1.00 - x x x q x2/(1.00 - x) = 7.0 x 10- 4 q Assume x << 1.00: x2/1.00 = 7.0 x 10- 4 q x = 2.65 x 10-2, so the assumption was okay q pH = -log(2.65 x 10-2) = 1.576 What is the pH of a 1.0M HF solution after 0.5M NaF is added? 1. 2. 3. 4. 5. 1.58 2.58 2.84 3.76 5.48 0% 1. 0% 2. 0% 3. 0% 4. 0% 5. Group Work q What is the pH of 1.00 M HF solution to which is added 0.500 M NaF? Ka = 7.0 x 10-4 q HF H3O+ FInitial 1.00 0 0.500 Change -x +x +x Equil. 1.00 - x x 0.500 + x q x(0.500 + x)/(1.00 - x) = 7.0 x 10- 4 q Assume x << 0.500 and 1.00: x(0.500)/1.00 = 7.0 x 10- 4 q x = 1.40 x 10-3 (>1%), so assumption was okay 17.2 Buffered Solutions q Buffer: Solution of a weak acid and its conjugate base in approximately equal concentrations q Used to control pH q Addition of a strong acid or base to a buffer shifts the pH only slightly Buffer Solutions q Make buffer: Mix HF and NaF in water. HF(aq) + H2O(l) h H3O+(aq) + F (aq) q - q Add HCl: Equilibrium shifts to left to consume some of added H3O+ q Add NaOH: Some H3O+ is consumed to neutralize the added OH-, and the equilibrium shifts to the Buffer Solutions q pH of all body fluids is controlled in this way. In our blood, phosphate, carbonate, and hemoglobin all act as buffers. q H2CO3(aq) + H2O(l) h HCO3-(aq) + H3O+(aq) q Secondary Equilibrium: q H2CO3(aq) h H2O(l) + CO2(g) h Addition of acid or base to a buffer 1 Buffer Solutions q Recall that a solution of 1.00 M HF and 0.500 M NaF has a pH of 2.854 q Now add 0.100 M HCl (assume there is no volume change) q To find the new pH, first assume that all the added strong acid reacts to change the Fand HF concentrations. q [HF] = 1.00 + 0.100 = 1.100 M q [F-] = 0.500 - 0.100 = 0.400 M 1 Buffer Solutions q Now solve the weak acid/conjugate base system: q HF H3O+ FInitial 1.100 0 0.400 Change -x +x +x Equil. 1.100 - x x 0.400 + x q x(0.400 + x)/(1.100 - x) = 7.0 x 10- 4 q Assume x << 0.400 and 1.100: x(0.400)/1.100 = 7.0 x 10- 4 q x = 1.93 x 10-3, so the assumption was okay q pH = -log(1.93 x 10-3) = 2.714 1 Buffer Solutions q Water: pure 0.100 M H3O+, pH 7 pH 1 q Buffer: pH 2.854 pH 2.714 q What will happen when a strong base (NaOH) is added to: qWater? qBuffer? 1 Buffer Solutions q If base is added, assuming no volume change, reduce the concentration of HF and increase the concentration of F- by the corresponding amount. Then solve the weak acid/conjugate base system. 1 Add 0.l00 M NaOH to the 1.00 M HF/0.500 M NaF buffer. What is the new pH? 1. 2. 3. 4. 5. 13.0 7.6 4.3 3.0 2.5 0% 1. 0% 2. 0% 3. 0% 4. 0% 5. 1 Group Work q Add 0.l00 M NaOH to the 1.00 M HF/0.500 M NaF buffer. What is the new pH? q First assume that all OH- reacts with HF: [HF] = 1.00 - 0.100 = 0.900 M [F-] = 0.500 + 0.100 = 0.600 M q Now solve the new acid-base system: 1 Group Work HF H3O+ FInitial 0.900 0 0.600 Change -x +x +x Equil. 0.900 - x x 0.600 + x q x(0.600 + x)/(0.900 - x) = 7.0 x 10- 4 q Assume x << 0.600 and 0.900: x(0.600)/0.900 = 7.0 x 10- 4 q x = 1.05 x 10-3, so the assumption was okay q pH = -log(1.05 x 10-3) = 2.979 q q In water, pH would change from 7 to 13. q(started at 2.854) 1 Addn of Strong Acids and Bases to Buffers q See Sample and Practice Ex 17.5 and end of chapter problems. 1 Buffer Capacity q Buffers only work within a pH range set by the value of pKa: 0.1 < [HA]/[A-] < 10 q Outside this range, we see little buffering effect. 1 Henderson-Hasselbalch Equation q Assuming that x << [HA] and x << [A-] for good buffer action, the equilibrium constant expression can be rearranged to give simplified calculations: pH = pKa + log([A-]/[HA]) Derived p. 727 2 Henderson-Hasselbalch Equation qpH = pKa + log([A-]/[HA]) q Consider 1.00 M HF, 0.500 M NaF Ka = 7 x 10- 4, pKa = 3.155 log ([F-]/[HF]) = log (0.500/1.00) = -0.301 pH = 3.155 - 0.301 = 2.854 (same result as before) 2 Henderson-Hasselbalch Equation qpH = pKa + log([A-]/[HA]) q Determine the Ka for the following system if a solution of equal concentrations of HA and A- give a pH of 8.0? HA + H2O h A- + H3O+ 2 Buffer Range q We can examine the buffer range graphically by plotting the fraction of each species vs the pH q As we move through the buffer range, one species decreases in concentration as the other increases in concentration Buffer range: pH = pK + 1 q a 2 CH3CO2H/CH3CO2 - Buffer Range 2 HF/F - Buffer Range 2 NH4 /NH3 + Buffer Range NH4+ NH3 2 H3PO4/H2PO4 /HPO4 /PO4 - 2- 3- 2 Which buffer system would be the best to use to keep the pH around 4.5? 1. H2CO3/HCO3- (Ka=4.3 x 10-7) 2. HCN/CN(Ka=4.9 x 10-10) 3. Ascorbic Acid/Ascorbate (Ka=8.0 x 10-5) 4. HClO/ClO x 10-2) - (Ka=1.1 0% 1 0% 2 0% 3 0% 4 2 17.3 Acid-Base Titrations q Refer to Chapter 4 for titration calculations to determine amounts of acids or bases. Widely used for chemical analysis: qacids in plating baths qascorbic acid in vitamin C tablets qacetylsalicylic acid in aspirin tablets qacetic acid in vinegar and wines qphosphoric acid in soft drinks 2 Titrations q Equivalence point: point in a titration where there are exactly equivalent amounts of the reactants q End point: point in a titration where the indicator changes color; may not be exactly the same as the equivalence point q Indicator: an acid or base which has a different color than its conjugate; characterized by a pKa, which determines the pH range of the color change (pKa + 1). 3 Phenolphthalein q HIn + H2O q Can you find the differences? q Color change: colorless to pink pH = 8.2 - 10 Phenolphthalein In HIn In- + H3O+ 3 Indicators 3 Indicators q The indicator should be picked so its color change comes close to the equivalence point. 3 Strong Acid - Strong Base q Note the slow change before and after the equivalence point and the rapid large change at the equivalence point. 3 Titration Curve Calculations q Calculate the equivalence point pH by assuming complete reaction of the acid and base, then solve any weak acid (or base) equilibrium for the pH. q Calculate the pH on either side of the equivalence point the same way, except in this case, there will be an excess of one substance. 3 Titration Curve Calculations Strong Acid /Strong Base q At the equivalence point, all acid and base has been neutralized. The conjugate acid and base do not affect pH so the pH is 7. q HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) 3 Strong Acid/Strong Base Titration 3 Titration Curve Calculations Strong Acid /Strong Base q HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) q Before and after the equivalence point, there will be excess moles of acid or base. Determine the number of moles xs and divide by the total volume of solution. q What is the [HCl] after 25.00 mL of 0.100 M NaOH has been added to 50.00 mL of 0.100 M HCl? 3 Titration Curve Calculations Strong Acid/Strong Base q HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) q [HCl]xs = 0.002500 mol/0.07500 L = 0.0333 M q What is the pH of the solution? q [H3O+]xs = 0.0333 M q pH = -log(0.0333) = 1.478 q Use this method to calculate the entire titration curve (or measure the curve with a pH meter) 3 Weak Acid - Strong Base q Equivalence point pH is higher than for a strong acid strong base titration, and the range for rapid change is narrower. q pH = pKa at halfway point pH = pKa 4 Weak Acid - Strong Base q CH3CO2H + NaOH CH3CO2- + H2O q Must take into account the weak acid / conjugate base equilibrium. q At the equivalence point, all the acid is consumed but we have a solution of the conjugate base (which reacts with water to form an equilibrium with HA) 4 Weak Acid Strong Base 4 Weak Acid-Strong Base q What are [CH3COOH] and [CH3COO- ] after 5.0mL of 0.100M NaOH is added to 20.0 mL of 0.100M acetic acid? q CH3CO2H + NaOH CH3CO2- + H2O 4 What is the pH after 10.0 mL of 0.100 M NaOH is added to 20.0 mL of 0.100 M acetic acid? a 1. 2.35 2. 3.50 K of acetic acid is 1.8x10-5 3. 4.26 4. 4.74 5. 5.25 0% 1. 0% 2. 0% 3. 0% 4. 0% 5. 4 Weak Acid - Strong Base q pH change in titration curve gets narrower as Ka gets smaller. q Equivalence point pH gets higher with weaker acids. 4 Strong Acid - Weak Base q Equivalence point pH is lower than for a strong acid strong base titration, and rapid change is narrowed. NaOH 50.0 mL of 0.100 M NH3 titrated with 0.100 M HCl 4 Strong Acid - Weak Base q pH at equivalence point is < 7 q 0.100 M HCl + 0.100 M NH3 gives pH = 5.27 q Phenolphthalein doesn't work for this titration. q Want an indicator that changes color in the range of pH 3-7 q Methyl red is a good choice. 4 Weak Acid - Weak Base q Titration curve shows major pH change in a very narrow range. q Generally doesn't give a good titration. 4 17.4 Solubility Equilibria q Some combinations of ions in solution form insoluble salts. Recall the solubility rules from Chapter 4. q Solubility of insoluble salts is characterized by the solubility product constant, Ksp. q AgCl(s) h Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] = 1.70 10-10 Use to determine conditions for precipitation 4 Solubility Product Constant q Ag2S(s) h 2Ag+(aq) + S2-(aq) Ksp = [Ag+]2[S2-] = 1.0 10-51 q How can we use Ksp values? q Which is more soluble? q Ksp(Ag2SO4) = [Ag+]2[SO42-] = 1.5 10-5 q Ksp(Ag2S) = [Ag+]2[S2-] = 1.0 10-51 5 Solubility q What if ion numbers are not the same? q Use molar solubility of salt (x): qAgCl: [Ag+] = x [Cl-] = x Ksp = (x)(x) Ksp = (2x)2(x) qAg2S: [Ag+] = 2x [S2-] = x qFe(OH)3: [Fe3+] = x [OH-] = 3x Ksp = (x)(3x)3 q We must compare x values in this case. 5 Solubility q If Ksp is known, can solve for x: For Fe(OH)3: (x)(3x)3 = Ksp = 1.1 10-36 x4 = 4.07 10-38 x = 4.49 10-10 M q 27x4 = 1.1 10-36 q [Fe3+] = 4.49 10-10 M q [OH-] = 3 (4.49 10-10) = 1.35 10-9 q If volume is 10.0 L, how much Fe(OH)3 dissolved? 5 Calculate the Ksp for an unknown compound, given the information below: 20% 20% 20% 20% 20% 2 1. 5 x 10-13 2. 1 x 10-18 Formula = XCl 3. 1 x 10-6 4. 5 x 10-12 5. 5 x 10-19 1 2 3 4 5 5 Group work q 2x = 1.010-6 M q x = 5.010-7 M Ksp = 4x3 = 5.010-19 5 17.5 Factors that Affect Solubility q Modify solubility to dissolve minerals and ores, to precipitate ions from solution, to separate and purify ions. q Examples: qremove hardness from water by adding Na2CO3 qremove Ag+ from water by adding Cl- to recover Ag metal qdissolve Cu(OH)2.CuCO3 to mine Cu qseparate the halide ions 5 Common Ion Effect q AgCl(s) h Ag+(aq) + Cl-(aq) q Common ion effect: add more Cl- to precipitate more Ag+ from solution. q Saturated solution of AgCl: Ksp = 1.70 10-10 = [Ag+][Cl-] = x2 x = [Ag+] = [Cl-] = 1.30 10-5 M q If 0.100 M Cl- is added, what is the new [Ag+]? q (>> 1.30 x 10-5, so assume [Cl-]= 0.100 M) q Ksp = 1.70 10-10 = [Ag+](0.100) q [Ag+] = 1.70 10-9 M 5 Insoluble Basic Salts q Metal hydroxide solubility depends on the pH. Mg(OH) (s) h Mg (aq) + 2OH (aq) q 2 126 2+ - q Adjust the pH to adjust the solubility. 5 Insoluble Basic Salts AgCN(s) h Ag (aq) + CN (aq) q + - Ksp = 1.6 x 10-14 q Add HNO3 (why not HCl?) to dissolve AgCN. CN (aq) + H O (aq) HCN(aq) + H O(l) 5 Adding HNO3 to which of the following compounds will NOT increase the solubility? 2 2 3 2 1. BaF BaF Mg(OH) AgCl CaCO 2% 0 2% 0 2% 0 2% 0 2% 0 Ag S 2 2. Mg(OH) 1 2 3 4 5 5 How will the solubility of CaCO3 change when the following compounds are added? 3 2 2 3 3 20% 20% 20% 20% 20% 1. +, +, -, -, NC 2. NaOH CaCl HNO +, -, +, -, - Na CO 3. +, -, NC, -, 4. +, -, -, -, NC 5. -, +, +, -, - CaCO 1 2 3 4 5 6 Formation of Complex ions q Lewis acid-base reactions can also reach a state of equilibrium q Metal ion + ligand h complex ion q Kf = [complex ion]/[metal ion][ligand] (formation constant) Ag (aq) + 2NH (aq) h Ag(NH ) (aq) q 6 Formation of Complex ions q Other examples: q Fe3+ + NCS- h FeNCS2+ q Cu2+ + 4NH3 h Cu(NH3)42+ 6 Cu2+/CuNH32+/Cu(NH3)22+/Cu(NH3)32+/Cu(NH3)42+ 6 Solubility and Complex Ions q Formation of complex ions can also increase solubility of a related salt. q Can dissolve AgCl with NH3. AgCl(s) h Ag (aq) + Cl (aq) q + - q Ksp = [Ag+][Cl-] = 1.70 x 10-10 Ag (aq) + 2NH (aq) h Ag(NH ) (aq) 6 ...
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This note was uploaded on 04/06/2011 for the course CHM 116 taught by Professor Unknown during the Spring '08 term at ASU.

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