“A” students work (without solutions manual) ~ 10 problems/night. Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th&F 2-3:30 pm Module #17C: Buffering and Titrations Holding Proteins intact By.
Download ReportTranscript “A” students work (without solutions manual) ~ 10 problems/night. Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th&F 2-3:30 pm Module #17C: Buffering and Titrations Holding Proteins intact By.
“A” students work (without solutions manual) ~ 10 problems/night. Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th&F 2-3:30 pm Module #17C: Buffering and Titrations Holding Proteins intact By pH constant General FITCH Rules G1: Suzuki is Success G2. Slow me down G3. Scientific Knowledge is Referential G4. Watch out for Red Herrings G5. Chemists are Lazy qq E k C1. It’s all about charge r r C2. Everybody wants to “be like Mike” qq C3. Size Matters E k r r C4. Still Waters Run Deep Piranhas lurk C5. Alpha Dogs eat first 1 2 Chemistry el 1 2 1 2 el 1 2 What pH is best suited For the hydrogen bond? Ka ,Val ,COOH 501 . x10 3 Ka,tyr ,ROH 10 . x1010 Hemeglobin Review Module 17B Which pH (2, 7, 11) is most favorable for the formation of a hydrogen bond between Val and tyr in hemoglobin Ka ,Val ,COOH 501 . x10 3 % dissociated 100 A A H O aq ,eq 3 aq ,eq Ka % dissociated H O 3 Ka,tyr ,ROH 10 . x1010 A aq ,eq aq ,eq 100 aq ,eq Ka 1 pH pH 22 7 11 KaCOOH KaCOOH Ka ROH 5.00E-03 5.00E-03 1.00E-10 [H+] %diss %diss [H+] %diss 1.00E-02 1.00E-02 33.33333 33.33333 1E-06 1.00E-07 99.998 0.0999 1.00E-11 100 90.90909 pH2 10 2 100 100 % dissociatedCOOH 33.33 % dissociatedCOOH 10 2 2 10 3 1 5.01x10 3 1 Review Module 17B 5.01x10 How does the body maintain This pH? http://www.meddean.luc.edu/lumen/MedEd/MEDICINE/pulmonar/physio/pf4.htm How does the body maintain pH ~7? Remember it will have to constantly adjust to changes in pH From an enormous number of chemical reactions!!! Control is ultimately based on CO2,(g) Cg PCO2 k H Respired (breathed) air PCO2 =0.23 mm Hg k H 3.4 x10 Pressure increases through mechanics of lung Aorta PCO2=41.8mmHg M 1atm 2 42mmCO2 . x10 3 M 3.4 x10 187 760mm L atm 2 Henry’s law moles L atm All dissolved CO2 goes to carbonic acid CO2, g H2 O H2 CO3aq Carbonic acid is a weak acid (anion is high charge density, Oh Card me PleaSe!) Ka1 4.4 x10 7 H2 CO3,aq H2 O HCO H O 3,aq 3 aq Ka1 HCO H O x10 H CO H CO 3,eq 2 x aq ,eq 3 3,aq ,eq Ka1 H2 CO3,3,aq ,init 2 7 x 3,aq ,init x2 x H CO . x1033 4.4x107 187 2 3,aq ,init 8.228x1010 2.86x105 H O x 3 aq ,eq p H3Oaq ,eq log2.86x10 5 4.54 This is too low!!! – we would not have the Right amount of ionization on the protein H2 CO3,aq H2 O Ka1 4.4 x10 7 aq HCO3,aq H3O 2 HCO3,aq H2 O CO H O 3,aq 3 aq Ka2 4.7 x10 p H3Oaq ,eq log2.29 x10 5 4.54 Will the pH get to the “right” value by the second acid Dissociation reaction?Let’s make an “intelligent” guess Ka 2 CO H O x2.86x10 2 3,eq HCO 3 aq ,eq 3,aq ,eq 2.86x10 5 5 x x 2.86x105 xKa2 x2.86x105 x 2.86x105 Ka 2 xKa 2 2.86x105 x x 2 x Ka2 x2.86x10 5 11 NO, would Only slightly Increase acidity x 4.7 x10 11 2.86x105 x 2 286 . x105 x 134 . x1015 0 2.86 x10 5 . x10 15 2.86x10 5 2 4 134 2 2.86x10 5 81796 . x10 10 5.36x10 15 x 2 x 2 2.86x105 x 2.86x105 Ka 2 xKa 2 0 2.86 x10 5 2.8600093x10 5 x 2 2 5 5 x 2.86x10 Ka 2 x 2.86x10 Ka 2 0 9.37 x10 11 11 2 5 11 5 11 x 4 . 68 x 10 x 2.86x10 4.7x10 x 2.86x10 4.7x10 0 2 H2 CO3,aq H2 O HCO H O 3,aq 3 aq Ka1 4.4 x10 7 Will the pH get to the “right” value by the second acid 5 p H O log 2 . 29 x 10 4.54 3 aq , eq Dissociation reaction? HCO3,aq H2 O CO3,aq 2 Ka2 4.7 x10 aq H3 O To tweak blood pH to 7.4 red blood cells (RBC) control [HCO3-] RBC HCO 3 aq ,eq blood 4.0x10 2 M 11 H2 CO3,aq H2 O HCO H O 3,aq 3 aq H CO 2 3,aq CO2 , g HCO 3 aq ,eq Ka1 187 . x10 M 2.4 x10 2 M H 2 CO3,aq ,eq H CO 2 3,aq ,eq pH pKa1 log HCO3,eq RBC HCO3,eq H 3 Oaq ,eq H CO 2 3,aq ,eq pH log Ka1 log HCO3,eq 3 Ka1 4.4 x10 7 H CO K H O pH log4.4x10 log 00779 . HCO pH 6.35 1108 . 7.45 H CO H O K HCO Value can be controlled H CO by red blood cell pH log H O log K HCO a1 3 3,aq ,eq 3 3,eq aq ,eq 2 a1 3 aq ,eq 7 3,aq ,eq 3,eq aq ,eq 187 . x10 3 pH log4.4 x10 log 2 2.4 x10 7 2 2 a1 3,aq ,eq 3,eq Plasma PCO2 ,blood 40mm 12 . x103 M HCO3-=24mmol/L pH=7.4 + 50%CO2,bound hemoglobin H2 CO3,aq Carbonic anhydrase HCO3,aq H3O Red Blood Cell That is the Context for Buffering 1. Must control concentration of both 1. Conjugate acid 2. Conjugate base 2. So what are the equations, etc. necessary to determining the buffering of a solution? “A” students work (without solutions manual) ~ 10 problems/night. Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th&F 2-3:30 pm Module #17C: Buffering and Titrations Henderson/Hassalbach Buffer Equation The body controls the pH ([H+]) concentration in blood by 1. controlling WA H2CO3 via CO2,g 2. controlling WB HCO3- through activity of the red blood cells 3. The WA and WB are linked (conjugates) so that an equilibrium can be written between the two H2 CO3,aq H2 O HCO H O 3,aq 3 aq 4. All parameters are constant, leading to constant pH H CO 2 3,aq ,eq pH pKa1 log HCO 3,eq 5. This process is called buffering A general equation can be written HA ,aq ,eq pH pKa log A ,eq mmoleA total mL pH pKa log mmoleHA totalmL HA ,aq ,eq pH pKa log A ,eq 1 mmoleA total mL pH pKa log 1 mmoleHA totalmL mmoleA pH pKa log mmoleHA moles x 103 mmoles x L = mmoles L moles 103 mL mL mL x mmoles = mmoles mL mL x M = mmoles Henderson Hasselbach Equation HA ,aq ,eq pH pKa log A ,eq mmoleA pH pKa log mmoleHA OR A ,eq pH pKa log HA,aq ,eq Linked (conjugated) if mmoleA mmoleHA pH pKa log1 pKa 0 pH pKa OJO this is NOT the “equivalence point” mmole titrant mmole titrated Galen, 170 Marie the Jewess, 300 Charles Augustin James Watt Coulomb 1735-1806 1736-1819 Justus von Thomas Graham Liebig (1803-1873 1805-1869 Ludwig Boltzman 1844-1906 Gilbert N Lewis 1875-1946 Henri Louis LeChatlier 1850-1936 Johannes Bronsted 1879-1947 Jabir ibn Hawan, 721-815 Luigi Galvani 1737-1798 Richard AC E Erlenmeyer 1825-1909 An alchemist Count Alessandro G A A Volta, 1747-1827 James Joule (1818-1889) Henri Bequerel 1852-1908 Lawrence Henderson 1878-1942 Galileo Galili Evangelista Torricelli 1564-1642 1608-1647 Amedeo Avogadro 1756-1856 Rudolph Clausius 1822-1888 Jacobus van’t Hoff 1852-1911 Niels Bohr 1885-1962 John Dalton 1766-1844 William Thompson Lord Kelvin, 1824-1907 Johannes Rydberg 1854-1919 William Henry 1775-1836 Johann Balmer 1825-1898 J. J. Thomson 1856-1940 Erwin Schodinger Louis de Broglie 1887-1961 (1892-1987) Fitch Rule G3: Science is Referential Jean Picard 1620-1682 Jacques Charles 1778-1850 Francois-Marie Raoult 1830-1901 Heinrich R. Hertz, 1857-1894 Friedrich H. Hund 1896-1997 Daniel Fahrenheit 1686-1737 Max Planck 1858-1947 Rolf Sievert, 1896-1966 Blaise Pascal 1623-1662 Georg Simon Ohm 1789-1854 James Maxwell 1831-1879 Robert Boyle, 1627-1691 Isaac Newton 1643-1727 Michael Faraday 1791-1867 B. P. Emile Clapeyron 1799-1864 Dmitri Mendeleev 1834-1907 Svante Arrehenius Walther Nernst 1859-1927 1864-1941 Fritz London 1900-1954 Wolfgang Pauli 1900-1958 Johannes D. Van der Waals 1837-1923 Marie Curie 1867-1934 Anders Celsius 1701-1744 Germain Henri Hess 1802-1850 J. Willard Gibbs 1839-1903 Fritz Haber 1868-1934 Thomas M Lowry 1874-1936 Werner Karl Linus Pauling Louis Harold Gray 1905-1965 Heisenberg 1901-1994 1901-1976 Buffer Example 1 what is equilibrium pH of a solution of 1.0 M HF, and 1.0 M NaF? Ka HF = 7.2x10-4 Do we have linked WA and WB both present? _ HFaq H2 Ol H O F 3 aq aq pH pKa Yes = buffer A pH pKa log HA F pH pKa log HF . 10 pH pKa log . 10 pH pKa log1 pH pKa 0 Ka=7.2x10-4 pH log7.2x104 pH log7.2x104 314 . In preceding module 17B we calculated the pH of A solution of 1.0 M HF and found it to be pH 157 . What causes the difference? LeChatlier’s Principle HFaq H2 Ol H3Oaq Faq_ HF push to right pH 157 . NaFaq H2 Ol Faq Naaq complete HFaq H2 Ol H3Oaq Faq_ HF push to right pH 314 . Effect of added NaF is to reduce the ionization of HF Buffer Example 2:What is the pH of 1.0 M H2CO3 in the presence of 1.0 M NaHCO3 What is in solution? Are the chemical species linked by an equilibrium reaction? H2 CO3,aq H2 O HCO H O 3,aq 3 aq Ka1 4.4 x10 Solution contains a weak acid in the presence of it’s linked (conjugate) weak base pH 6.36 log1 A pH pKa log HA pH 6.36 0 6.36 1M NaHCO3 pH log(4.4 x10 ) log 1M H2 CO3 7 7 Buffer Example 3:What is the pH of a solution containing 0.25 M NH3 (Kb = 1.8x10-5) and 0.40 M NH4Cl? What is in solution? Are the chemical species linked by an equilibrium reaction? 5 Kb 18 . x10 NH3,aq H2 O NH4 aq OHaq or K 14 14 10 w w K 10 10 w K 5556 . x 10 a Kaa Kb 18 55 . x 10 Kbb 18 . x10 NH4aq H2 O NH3,aq H3 Oaq Solution contains a weak acid in the presence of it’s linked (conjugate) weak base A pH pKa log HA 0.25 M NH 3,aq 10 pH log(5.55x10 ) log 0.40 M NH 4 ,aq 0.25 pH 9.25 log 0.4 pH 9.25 0.204 9.05 “A” students work (without solutions manual) ~ 10 problems/night. Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th&F 2-3:30 pm Module #17C: Buffering and Titrations 4 Kinds of Problems Reviewed We now know how to do 4 types of problems Limiting Reagent must involve SA or SB Which reagent is completely consumed? Which remains? HClaq NaOH Claq H2 Oaq Naaq completely , Limiting reagent HAaq NaOH Aaq H2 Oaq Naaq completely , Limiting reagent Naaq Aaq HCl HAaq Claq Naaq completely , Limiting Re agent WA HAaq H2 O Aaq H3 Oaq Ka x x 10 x HA HA x HA , H3 Oaq Aaq 7 Equilibrium aq ,eq 2 aq init aq ,init Aaq H2 O OH aq HAaq WB Kb OH aq HAaq ,eq A Aaq H2 O OHaq HAaq WA +conjugate HA H O A H O aq 2 aq 3 aq (linked) WB aq ,eq x x 10 7 x2 A x A aq ,init aq ,init A A init eq pK log pH pKa log a HA HA init eq To refresh your memory on Limiting Reagent Problems Calculate the concentration of Na+, Cl-, CH3COO-, and CH3COOH in solution when 10.0 mL of 1.0 M HCl is added to 6.0 mL of 2.0 M NaAcetate acid (NaCH3COO). Each solution is obtained from made from a stock solutions at 1 atm pressure What do we know 10.0 mL 1.0 M HCl 5.0 mL 2.0 M NaCH3COO 1 atm What do we want What is a red herring? Conc. Classify the chemical substances added to solution: Strong Acid (SA) Protonated low q/r anion (No Clean Socks) HNO3, HCl, H2SO4 Strong Base (SB) OH +low q/r Cation (Group 1 and 2) NaOH, KOH, Ca(OH)2… Strong electrolyte (SE) Anion and Cation, low q/r NaCl, KNO3…. Weak Acid (WA) Protonated hi q/r anion (Oh Card me PleaSe) H2CO3….. Weak Base (WB) Contains R group, anion hi q/r with cation low q/r Na2CO3 Other or unknown (OU) To refresh your memory on Limiting Reagent Problems Calculate the concentration of Na+, Cl-, CH3COO-, and CH3COOH in solution when 10.0 mL of 1.0 M HCl is added to 6.0 mL of 2.0 M NaAcetate acid (NaCH3COO). Each solution is obtained from made from a stock solutions at 1 atm pressure mL M mmoles SA/WB Determine the Limiting Reagent completely , Limiting Re agent Naaq Aaq HCl HAaq Claq Naaq HCl can produce 10 . moleHCl 1moleH 1moleCH3COOH 10.0mLHCl 1L 1mole 1mole 10mmolesCH3COOH HCl HCl H WB can produce Haq CH3COOaq CH3 COOHaq completely , Limiting Re agent 6.0mL NaCH3COOl 2.0mole NaCH3COO 1moleCH3COO 1moleCH3COOH 12mmoles CH3COOH 1LNaCH3COO 1mole NaCH3COO 1moleCH3COO SA (HCl) is limiting reagent, completely consumed, 0mmoles WA formed is 10 mmoles How much WB remains? To refresh your memory on Limiting Reagent Problems Calculate the concentration of Na+, Cl-, CH3COO-, and CH3COOH in solution when 10.0 mL of 1.0 M HCl is added to 6.0 mL of 2.0 M NaAcetate acid (NaCH3COO). Each solution is obtained from made from a stock solutions at 1 atm pressure SA/WB Haq CH3COOaq CH3 COOHaq completely , Limiting Re agent SA (HCl) is limiting reagent, completely consumed, 0mmoles WA formed is 10 mmoles WB initial 12 mmoles; WB consumed: 10 . moleHCl 1moleH 1moleCH3COOH 1moleCH3COO 10mmolesCH COO 10.0mLHCl 1L 1mole 1mole 1mole 3 HCl HCl CH3COOH H Initial-consumed=12-10=2mmoles WB left Concentrations after the limiting reagent reaction CH COOH 10mmolesCH3COOH CH COO 2.0mmolesCH3COOH 3 3 10.0mL 6.0mL 10.0mL 6.0mL 6mL 2 M 0.625 M Na 10.0mL 6.0mL 0.75 M 0125 . M Cl 10.0mL 6.0mL 0.625 M 10mL10 . M To refresh your memory on Limiting Reagent Problems Concentrations CH COOH 10mmolesCH3COOH CH COO 2.0mmolesCH3COOH 3 3 10.0mL 6.0mL 10.0mL 6.0mL 6mL 2 M 0.625 M Na 10.0mL 6.0mL 0.75 M 0125 . M Cl 10.0mL 6.0mL 0.625 M 10mL10 . M In the preceding problem what kind of equilibrium problem remains? You have 3 choices: WA WB WB and its conjugate WA (or WA and its conjugate WB) We have present a weak acid (WA) and it’s linked (conjugate) weak base (WB) . M CH COO 0125 CH COOH 0.625M 3 3 Buffer equilibrium problem “A” students work (without solutions manual) ~ 10 problems/night. Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th&F 2-3:30 pm Module #17C: Buffering and Titrations A feast of titrations We can combine Limiting reagent problems followed by an equilibrium problem to do a titration Titrations allow us to determine the conc of an unknown Titrations involve 1 of 3 LR followed by 1 of 3 Eq Limiting Reagent must involve SA or SB Which reagent is completely consumed? Which remains? HClaq NaOH Claq H2 Oaq Naaq completely , Limiting reagent HAaq NaOH Aaq H2 Oaq Naaq completely , Limiting reagent Naaq Aaq HCl HAaq Claq Naaq completely , Limiting Re agent WA HAaq H2 O Aaq H3 Oaq Ka x x 10 x HA HA x HA ,init H3 Oaq Aaq 7 Equilibrium aq ,eq 2 aq aq ,init Aaq H2 O OH aq HAaq WB Kb x2 x K 107 K Init 0 OH aq HAaq ,eq A aq ,eq x x 10 7 x2 A x A aq ,init aq ,init Aaq H2 O OHaq HAaq WA +conjugate (linked) WB HAaq H2 O A H O aq 3 aq A A init eq pK log pH pKa log a HA HA init eq Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL; Ka acetic acid = 1.8x10-5 Main Dish: Limiting Reagent Problem x 10 7 ? CH3COOH OH CH3COO H2 O x 02 . ? 1mmole 1mmole CHCH 3COO 3COO 0M .20 M 40 mL 0.020 .20 M 40 40 mL mL 8 mmoleCH COO acetic acid acetic acid acetic acetic acid acid acetic acetic acid acid 3 1mmole 1 mmole CHCH 3COOH 3COOH 1mmoleCH3COO 0.40 M NaOH 0mLNaOH 1mmole 0mmoleCH3COO NaOH limiting Solution contains CH3COOH Sig fig, round to 0.2 Solution contains no NaOH Solution contains no CH3COOKa Side dish: WA H O A x x 10 x HA HA x HA ,init 3 aq aq aq ,eq 7 aq ,init 2 . x10 H x 1897 aq Ka 18 . x105 ; try assumptions 18 . x10 5 x2 0.20 0.20 0.001897 019810263 . 18. x105 0.20 x 1897 . x10 3 aq 3 pH log1897 . x10 3 pH 2.72 If you prefer – do an ICE chart Equilibrium Calculation. Rx 1 H2O H+ OH55.5 10-7 10-7 Rx 2 CH3CO2H H+ CH3CO2init 0.200 10-7 0 change -x +x +x assume x<<0.2 x>>10-7 equil 0.2 x x CH CO H 010 Q 2 ,init 3 init CH CO H 3 2 init 7 0.2 K Because rx goes to right Sign of x is neg for reactants Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL; Ka = 1.8x10-5 Ka 18 . x105 ; try assumptions r 18 . x10 check? yes. 5 x2 0.20 18. x105 0.20 x 1897 . x10 3 pH log(189 . x10 3 ) 2.72 Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL; Ka acetic acid = 1.8x10-5 mL pH 0 2.72 Main Dish: Limiting Reagent Problem CH3COOH OH CH3COO H2 O 1mmoleCH3COO 0.40 M NaOH 5mLNaOH 1mmole 2.0mmoleCH3COO NaOH 0.20 M CH3COOH 40mLCH3COOH Limiting Reagent 1moleCH3COO 8.0mmoleCH3 COO 1moleCH3COOH Solution contains 2.0 mmole CH3COO2.0mmoleCH COO CH COO 40mL 3 CH3COOH 3 5mL NaOH 0.0444 M 2.0 mmole CH3COOH used to make 2 mmole CH3COO8.0mmole 2.0mmole CH3COOH 01333 . M 40mLCH3COOH 5mLNaOH Side Dish: Buffer Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL; Ka acetic acid = 1.8x10-5 mL pH 0 2.72 CH3COOH OH CH3COO H2 O 5 4.27 2.0mmoleCH COO CH COO 40mL 3 CH3COOH CH3COOH 3 5mL NaOH 0.0444 M 8.0mmole 2.0mmole 01333 . M 40mLCH3COOH 5mLNaOH A pH pKa log HA 0.0444 pH log(18 . x10 ) log . 01333 5 pH 4.74 0522 . pH 4.27 Invoke Rule: Chemists Are Lazy mmolesA total mL pH pKa log mmolesHA total mL mmolesA pH pKa log mmolesHA 2 pH 4.74 log 4.27 6 Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL; Ka acetic acid = 1.8x10-5 mL pH 0 2.72 5 4.27 Main Dish: Limiting Reagent Problem 10 4.74 CH3COOH OH CH3COO H2 O 1moleCH3COO 0.40 M NaOH 10mLNaOH 1mole 4.0mmoleCH3COO NaOH 0.20 M CH3COOH 40mL CH3COOH Limiting Reagent 1moleCH3COO 8.0mmoleCH3 COO 1moleCH3COOH Solution contains 4.0 mmole CH3COOSolution contains 8.0-4.0 mmole CH3COOH mmolesA pH pKa log mmolesHA 4 pH 4.74 log 4.74 log 1 4.74 8 4 Side Dish: buffer Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL; Ka acetic acid = 1.8x10-5 mL pH 0 2.72 5 4.27 Main Dish: Limiting Reagent Problem 10 4.74 19 6.02 CH3COOH OH CH3COO H2 O 1moleCH3COO 0.40 M NaOH 19mLNaOH 1mole 7.6mmoleCH3COO NaOH 0.20 M CH3COOH 40mL CH3COOH Limiting Reagent 1moleCH3COO 8.0mmoleCH3 COO 1moleCH3COOH Solution contains 7.6 mmole CH3COO8 CH3COOH – 7.6 mmole OH = 0.4 mmole mmolesA pH pKa log mmolesHA 7.6 7.6 pH 4.74 log 4 . 74 log . 6.02 4.74 12787 04 8 7.6 Side Dish: buffer A x10 Ka 7 HA x x 0128 . x 10 7 x ; pH 6.08 .00677 x mmolesA pH 4.77 log 6.04 mmolesHA titration of acetic acid 14 12 Calculated pH 10 Difference between “exact” and non-exact small 8 6 4 2 0 0 5 10 15 mL 0.400 NaOH 20 25 30 Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL; Ka acetic acid = 1.8x10-5 mL pH 0 2.72 5 4.27 Main Dish: Limiting Reagent Problem 10 4.74 19 6.02 CH3COOH OH CH3COO H2 O 1moleCH3COO 0.40 M NaOH 20mLNaOH 1mole 8mmoleCH3COO NaOH 0.20 M CH3COOH 40mL CH3COOH 1moleCH3COO 8.0mmoleCH3 COO 1moleCH3COOH Both Are Limiting Reagents USED UP Solution contains 8 mmole CH3COO8 mmole CH3COOH-8 mmole OH = 0 mmole CH3COOH Side Dish: acetate = WB CH3COOaq H2 O CH COOH OH 3 aq aq Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL; Ka acetic acid = 1.8x10-5 mL pH 0 2.72 5 4.27 Solution contains 8 mmole CH3COO10 4.74 19 6.02 CH3COOaq H2 O CH COOH OH 3 aq aq 20 8.93 CH COO aq 3 Kb 8mmole 01333 . 40mL 20mL x 01333 . ? x 10 7 ??? HA OH x x 10 x A A x A aq ,eq aq aq ,eq 7 aq ,init Kw 10 14 Kb . x10 10 15 555 Ka 18 . x10 Kb “small”, assumps. Ok. 555 . x10 10 x2 01333 . 2 aq ,init . x10 10 01333 . ~ x 8.59 x10 6 555 OH ~ x 859 . x10 6 pH 14 log8.59 x10 6 8.93 If you prefer – do an ICE chart Calculate equilibrium H2O H+ OH55.5 10-7 10-7 CH3CO2- CH3CO2H OH init conc: 0.133 0 10-7 change -x +x +x assume (not much reaction to right, Kb small) x << 0.133 x>> 10-7 equil 0.133 x x CH CO H OH 010 Q K 3 2 init CH CO 3 init 7 r 2 ,init 0.2 Sign of x for reactants is Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL Kw 10 14 10 Kb 555 . x 10 Ka 18 . x1015 Kb “small”, assumps. Ok. 555 . x10 10 x2 0133 . . x10 10 0133 . 555 pH 14 pOH log(859 . x10 6 ) 8.93 x 8.59 x10 6 Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL; Ka acetic acid = 1.8x10-5 mL pH 0 2.72 5 4.27 Main Dish: Limiting Reagent Problem 10 4.74 19 6.02 CH3COOH OH CH3COO H2 O 20 8.93 1moleCH3COO 21 11.82 0.40 M NaOH 21mLNaOH 1mole 8.4mmoleCH3COO NaOH 0.20 M CH3COOH 40mL CH3COOH 1moleCH3COO 8.0mmoleCH3 COO 1moleCH3COOH Limiting Reagent Side dish: seconds of the main Excess OH OH OH excess excess 8.4 8 0.006557 40 20 1 p OH excess log 0.006557 Or p OH excess 218 . 1mL 0.40 M excess 40 20 1 0.006557 pH 14 pOH 14 2.18 1182 . Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL; Ka acetic acid = 1.8x10-5 mL pH 0 2.72 5 4.27 Main Dish: Limiting Reagent Problem 10 4.74 19 6.02 CH3COOH OH CH3COO H2 O 20 8.93 1moleCH3COO 21 11.82 0.40 M NaOH 25mLNaOH 1mole 10mmoleCH3COO 25 12.48 NaOH 0.20 M CH3COOH 40mL CH3COOH 1moleCH3COO 8.0mmoleCH3 COO 1moleCH3COOH Limiting Reagent Side dish: seconds of the main Excess OH OH OH excess excess 10 8 0.030769 40 20 5 p OH excess log 0.030769 Or p OH excess 1511 . 5mL 0.40 M excess 40 20 5 0.030769 pH 14 pOH 14 1511 . 12.48 Defining some points on the Titration Curve titration of acetic acid 14 Strong Base Regions of A WEAK ACID titration curve 12 Buffer Calculated pH 10 Weak Base 8 6 4 2 WeakAcid 0 0 5 10 15 20 25 mL 0.400 NaOH ½ way buffer Tells us identity of acid 30 ½ way to the equivalence point aq aq HAaq OH A ½ of HA is converted to AmmoleHAinitial mmoleHAremainng mmoleA formed 2 mmolesA pH pKa log mmolesHA pH pKa log1 pH pKa pKa is a clue to the identity of the acid 1 14 14 14 2 eq pt ; pH pKa Titrate 50 mL 1.0 M WA with 1.0 M NaOH 12 12 12 Clue I.D. 101010 Of acid pH pH pH 888 666 444 222 000 000 10 10 10 20 20 20 30 30 30 Valenine RCOOH; pKa 2.30 HF ; pKa 314 . 40 40 50 50 60 60 40 50 60 mL mL base base added added mL base added 70 70 70 80 80 80 CH3COOH; pKa 4.74 NH4 ; pKa 9.255 90 90 90 100 100 100 Defining some points on the Titration Curve titration of acetic acid 14 Strong Base Regions of A WEAK ACID titration curve 12 Buffer Calculated pH 10 Weak Base 8 6 4 2 WeakAcid 0 0 5 10 15 20 mL 0.400 NaOH ½ way buffer Eq. pt. 25 30 This is what Tells us the conc. Titrate __mL of 1.0 M Acetic Acid with 1.0 M NaOH 13 25 14 14 50 75 100 125 12 12 10 10 pH pH 8 6 4 2 0 8 6 Observation 1? Observation 2? 4 2 0 0 0 20 20 40 40 60 60 80 100 120 80 100 120 mL strong base added mL strong base added You should have at least 2 observations What are they? 140 140 160 160 180 180 200 200 We found the equivalence point (large pH change) occurred when we added 20 mL of 0.40 M NaOH to 40 mL of an unknown conc of acetic acid. What was the original concentration of the acetic acid? mmole titrant eq . pt mmole titrated eqpt or 20mL 0.40 M 40mL x x 20mL 0.40 M 40mL x 020 . M OJO, not pH pKa V M V M 20 0.40 40 M 20 0.40 M mmoleA mmoleHA 40 0.2 Be sure to account for Multiple protons or hydroxides “A” students work (without solutions manual) ~ 10 problems/night. Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th&F 2-3:30 pm Module #17C: Buffering and Titrations Indicators Choose an Indicator so there will be a color change at the equivalence point titration of acetic acid 14 Strong Base Regions of A WEAK ACID titration curve 12 Calculated pH 10 Buffer 8.93 Weak Base 8 6 4 2 WeakAcid 0 0 5 10 15 20 mL 0.400 NaOH ½ way buffer Eq. pt. 25 30 How can we “see” the pH jump so we know where the end point is? Indicators: plum juice (tannic acid); red wine H(tannic) HIn pink HO H+ + Ingreen HO OH HO HO OH HO OH HO OH O O OH O O O O O OH HO OH HO O O O O O HO O O O O O O OH OH O OH HO HO O HO HO OH O HO OH O HO OH OH OH O OH O OH OH HO OH O OH O O O O OH HO O OH O O O HO OH O O OH O O HO OH O O OH O O OH O Remove protons OH OH HO OH Connects electrons through ring, Core structure= cyanidin Blueberries Raspberries Cranberries Red cabbage Red wines Red cabbage primarily http://antoine.frostburg.edu/chem/senese/101/features/water2wine.shtml CAS 528-58-5 http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/p26_anth-e.htm Hydrangea flower color based on available aluminum Cation in the soil Indicators: plum juice (tannic acid) HIn pink H In H+ + Ingreen Ka H Ka HIn Ka H In HIn moles green total vol moles green moles pink moles pink total vol Observed color? When is it green vs pink? Ratio: 10 Green/1 Pink Ratio: 10 Pink/1Green Observed color? depends on mole ratio Rule of thumb (10/1 or 1/10) Ka 10 moles green green H moles pink pKa ( log10) p H Ka H moles green pink 10 moles pink log10 pKa pH Ka H 10 pHgreen pKa 1 10Ka H pH pink pKa 1 pHcolor change pKa ,indicator 1 pH pKa 1 Common Indicators (besides grape juice) Name pHcolor change = pKa +/- 1 methyl violet thymol blue bromophenol blue methyl red phenol red bromothymol blue phenolphthalein alizarin yellow R 0-2 1-3 3-5 4.5-6.5 6.5-8 6-8 8-10 10-12 pKa 5 7.1 9 Change centered over pKa Color change over 2 pH units Common Indicators (besides grape juice) Name methyl violet thymol blue bromophenol blue methyl red phenol red bromothymol blue phenolphthalein alizarin yellow R pHcolor change = pKa +/- 1 0-2 1-3 3-5 4.5-6.5 6.5-8 6-8 8-10 10-12 pKa Biggest pH change in our calculation was centered at pH 8.93 5 7.1 9 Name pHcolor change = pKa +/- 1 methyl violet thymol blue bromophenol blue methyl red phenol red bromothymol blue phenolphthalein alizarin yellow R Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL 0-2 1-3 3-5 4.5-6.5 6.5-8 6-8 8-10 10-12 pKa 5 7.1 9 Can omit for BLB Choose an Indicator so there will be a color change at the equivalence point titration of acetic acid 14 12 Calculated pH 10 8 6 4 2 0 0 5 10 15 20 mL 0.400 NaOH ½ way buffer Eq. pt. 25 30 “A” students work (without solutions manual) ~ 10 problems/night. Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th&F 2-3:30 pm Module #17C: Buffering and Titrations Two more example titrations WB/SA SA/SB A quick and dirty (WA/SB) titration curve. mL added problem WA 0 1/2 to equivalence pt buffer WB equivalence pt after equivalence pt SB pH <pKa pKa >7 Can you predict how a WB/SA titration curve will work out? Example Titration 2 titrate WB with SA mL added problem 0 WB 1/2 to equivalence pt buffer equivalence pt WA after equivalence pt SA pOH <pKb pKb pH >pKa =pKa <7 Kb 18 . x105 Example titrate 50 mL 0.1 M NH3 0.2 M HCl quick and dirty: 0 mL, 1/2 equivalence point, 1 mL before eq. pt., eq. pt., 1 ml after 0 mL HCl added; WB equilibrium chart mL pH 0 11.13 NH3 + H2O NH4+ + OHinit. conc. 0.1 0 10-7 change -x +x +x assume? Kb = 1.8x10-5, not much reaction, yes 0.1> x x>10-7 equil. 0.1 x x NH OH 4,init init 0107 5 Q NH 3,init Kb 18 . x10 5 x 10 7 x 01 . x 18. x105 01. x 2 01 . x2 01 . Kb 18 . x10 x 18. x105 01. x 134 . x10 3 pOH log134 . x10 3 287 . pH 14 pOH 1113 . need to know eq. pt. 50mL01 . MNH3 x 0.2 MHCl mL 0 12.5 x 25mLeq pt 25mL 1 12.5mL 2 eq. pt . 2 pH 11.13 9.25 Kb, NH3 18 . x105 pKb log18 . x105 4.744 pH1/ 2 eq. pt pKa 14 pKa 4.744 Kw K w Ka Kb ; Ka Kb OR pH1/ 2 eq. pt pKa 9.25 In case you are wondering? 14 pKa pKb pOH1/ 2 eq. pt pKb 4.7447 pKa 9.25 14 pOH1/ 2 eq. pt pH 14 4.744 9.25 Example titrate 50 mL 0.1 M NH3 0.2 M HCl; include 1 mL and 1 mL after eq.pt. Eq pt = 25 mL; pKa=9.25 (from preceding slide) 1 mL before eq. pt = 24mL.; LR prob 4.8 mmoles NH4+ Equilibrium prob? 0.2 mmoles NH3 Buffer 0.20 M 24mL 4.8mmoleH HCl HCl pH 11.13 9.25 7.86 added 4.8mmolesHAformed 50mL 010 . M NH 4.8mmoles 0.2mmoles A- remaining 3 mL 0 12.5 24 pKa 9.25 mmolesA pH pKa log mmolesHA 0.2 pH 9.25 log 4.8 pH 9.25 138 . 7.86 Example titrate 50 mL 0.1 M NH3 0.2 M HCl; include 1 mL and 1 mL after eq.pt. Eq. pt. Problem; WA NH4+ NH3 + H+ init conc .066 0 10-7 change -x +x +x Ka = 10-14/(1.8x10-5) = 5.5x10-10 Assump? 0.066>>x x~10-7 equil 0.066 x x+10-7 7 7 2 10 14 10 10 41 3.66 x10 11 10 Ka . x10 5 55 x 18 . x10 21 7 7 x 10 x x 10 x 10 10 7 10 14 146 . x10 10 55 . x10 x 0.66 x 0.66 2 10 7 2 55. x10 0.66 10 x x 10 7 121 . x10 5 x . x1011 10 7 x x 2 366 2 7 x 10 x 3.66x10 2 11 0 x 6 x10 6 pH log6x10 6 522 . Eq pt = 25 mL; pKa=9.25 (from preceding slide) 1 mL past eq. pt.; 26 mL; SA 50mL 01 . M NH / NH 3 1mL 50 25 1 excess 0.2 M 50 25 1 HCl 4 mL 0 12.5 24 25 26 5mmole [ NH4 ] 76mL pH 11.13 9.25 7.86 5.22 2.5 2.63x10 3 M [ H ] H+ conc. SA = conc. = pH = -log(2.63x10-3) pH = 2.5 OR, limiting reagent 2.63x10-3 Limiting Reagent 26mL 0.2 M 52mmoleNH 50mL 01. M 50mmoleNH s HCl s HCl 2mmole excess H 2mmole 2.63x10 3 50 25 1 Example titrate 50 mL 0.1 M NH3 0.2 M HCl; include 1 mL and 1 mL after eq.pt. 4 4 5 mL past eq. pt. 5mL 0.2 M excess 50 25 5 125 . x10 2 M [ H ] conc. SA = conc. H+ = 1.25x10-2 pH =-log(1.25x10-2) pH=1.9 mL 0 12.5 24 25 26 30 pH 11.13 9.25 7.86 5.22 2.5 1.9 50 mL 0.1 NH3 titrated with 0.2 M HCl 12 10 buffer WB pH 8 1/2 eqpt 6 WA 4 2 eq pt SA 0 0 5 10 15 mL HCl Choose an indicator? 20 25 30 Common Indicators (besides grape juice) Name pHcolor change = pKa +/- 1 methyl violet 0-2 thymol blue bromophenol blue methyl red phenol red bromothymol blue phenolphthalein 8-10 alizarin yellow R 10-12 1-3 3-5 4.5-6.5 6.5-8 6-8 pKa Biggest pH change is near 5.5 5 7.1 9 Example Titration 3: We have 25.0 mL of 0.200 M HNO3. What will be the pH after we add 0, 20, 49, 50, 51, 75, 100 mL of 0.100 M NaOH Main Course: SA/SB Mmole HNO3 init 0.20 M 25mL mL OH added Side courses: more of same Mmoles Mmole H+ excess OH added 010 . M mL Mmole OH excess Total Vol [H+] Or [OH-] 5 0 0 5-0=5 0 25+0 5/25=0.2 5 20 2 5-2=3 0 25+20=45 3/45=0.066 5 49 4.9 5-4.9=0.1 0 25+49=74 0.1/74=0.00135 5 50 5 5-5=0 0 25+50=75 H2O 5 51 5.1 5-5=0 5.1-5=0.1 25+51=76 0.1/76=0.00136 5 75 7.5 5-5=0 7.5-5=2.5 25+75=100 2.5/100=0.025 5 100 10 5-5=0 10-5=5 25+100=125 5/125=0.04 Example Titration 3: We have 25.0 mL of 0.200 M HNO3. What will be the pH after we add 0, 20, 49, 50, 51, 75, 100 mL of 0.100 M NaOH Main Course: SA/SB Side courses: more of same [H+] Or [OH-] 5/25=0.2 pH=-log(0.2)=0.0698 3/45=0.066 pH=-log(0.066)=1.17 0.1/74=0.00135 pH=-log(0.00135)=3.17 H2O pH=7 0.1/76=0.00136 pOH=-log(0.0013)=2.88 pH=14-pOH=11.12 2.5/100=0.025 pOH=-log(0.025)=1.602 pH=14pOH=12.398 5/125=0.04 pOH=-log(0.04)=1.39 pH=14-pOH=12.61 Titration of 50 mL 0.2 M HNO3 14 12 pH 10 Where is largest pH? What indicator should we use? 8 6 4 2 0 0 20 40 ml 0.1 M NaOH 60 80 We have 25.0 mL of 0.200 M HNO3. What will be the pH after we add 0, 20, 49, 50, 51, 75, 100 mL of 0.100 M NaOH Common Indicators (besides grape juice) Name pHcolor change = pKa +/- 1 methyl violet 0-2 thymol blue bromophenol blue methyl red phenol red bromothymol blue phenolphthalein 8-10 alizarin yellow R 10-12 1-3 3-5 4.5-6.5 6.5-8 6-8 pKa Biggest pH change is near 7 5 7.1 9 Color change defines large pH change Titration of 50 mL 0.2 M HNO3 14 blue 12 SB pH 10 8 6 SA 4 yellow 2 0 0 20 40 ml 0.1 M NaOH 60 80 We have 25.0 mL of 0.200 M HNO3. What will be the pH after we add 0, 20, 49, 50, 51, 75, 100 mL of 0.100 M NaOH “A” students work (without solutions manual) ~ 10 problems/night. Dr. Alanah Fitch Flanner Hall 402 508-3119 [email protected] Office Hours Th&F 2-3:30 pm Module #17C: Buffering and Titrations END Tannins – represent linked cyanidins Red wine - antioxidants http://www.micro-ox.com/chem_tan.htm Food Source Anthocyanins Glycoside form apple blueberry cyanidin malvidin petuniclin monoglycoside monoglycoside monoglycoside delphinidin monoglycoside cranberry cyanidin peonidin monoglycoside monoglycoside red cabbage cyaniclin diglycoside strawberry pelargonidin monoglycoside cyanidin monoglycoside