Ch 17 Chem Notes

Ch 17 Chem Notes - Additional Aspects of Aqueous Equilibria...

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Unformatted text preview: Additional Aspects of Aqueous Equilibria BLB 10th Chapter 17 Buffered Solutions (sections 12) Acid/Base Reactions & Titration Curves (3) Solubility Equilibria (sections 45) Two important points: 1. Reactions with strong acids or strong bases go to completion. 2. Reactions with only weak acids and bases reach an equilibrium. 17.1 The Common Ion Effect Weak acid: HA + H2O H3O+ + A + Salt of conj. Base: NaA Na+(aq) + A-(aq) = two sources of A Common Ion! What affect does the addition of its conjugate base have on the weak acid equilibrium? On the pH? Used in making buffered solutions Calculate the pH of a solution containing 0.60 M HF and 1.00 M KF. 17.2 Buffered Solutions Resist a change in pH upon the addition of small amounts of strong acid or strong base Consist of a weak conjugate acidbase pair Control pH at a desired level (pKa) Examples: blood, physiological fluids, seawater, foods How do buffers work? Calculating pH of a Buffer [ HA] [ H 3O ] = K a - [A ] + [ HA] - log[ H 3O ] = - log K a + - log - [A ] [base] HendersonHasselbalch pH = pK a + log [acid ] equation + Calculate the pH of a solution containing 0.60 M HF and 1.00 M KF. (again, but the easy way) Adding strong acid or base to a buffer Adding acid: H3O+ + HA or A Adding base: OH + HA or A Calculating pH: 1. Stoichiometry of added acid or base 2. Equilibrium problem (HH equation) Calculate the pH after adding 0.20 mol of HCl to the 0.60 M HF and 1.00 M KF buffer. Calculate the pH after adding 0.10 mol of NaOH to the 0.60 M HF and 1.00 M KF buffer. Calculate the pH for a 1.0L solution that contains 0.25M Calculate the pH for a 1.0L solution that contains 0.25 M NH3 and 0.15 M NH4Br after the addition of 0.05 mol of RbOH Calculate the pH for a 1.0L solution that contains 0.25 M NH3 and 0.15 M NH4Br after the addition of 0.35 mol of HCl. Buffers (wrap up) HH equation No 5% check When strong acid or base is added, start equation with that acid or base. Making buffers of a specific pH? HH equation Buffer capacity exceededwhen added acid or base totally consumes after a buffer component How would you prepare a phenol buffer to control pH at 9.50? 17.3 AcidBase Titrations Titration a reaction used to determine concentration (acidbase, redox, precipitation) Titrant solution in buret; usually a strong base or acid Analyte solution being titrated; often the unknown @ equivalence point (or stoichiometric point): mol acid = mol base pH titration curve plot of pH vs. titrant volume Found by titration with an indicator Solution not necessarily neutral pH dependent upon salt formed Acidbase Titration Reactions and Curves Type 1 2 3 Acid Base strong strong weak strong strong weak Recognize curve types Calculate pH at various points on curve. Type 1: Strong acid + strong base Goes to completion Forms a neutral salt Equivalence point neutral solution, [H3O+] = 1.0 x 107 M, pH = 7.00 pH calculations involve only stoichiometry and excess H3O+ and OH Type 1: Strong acid + strong base 20.0 mL 0.200 M HClO4 titrated with 0.200 M KOH Initial mmol acid = mL base 0.00 10.00 20.00 30.00 40.00 mmol base added mmol acid total remain mL [H3O+] pH Another SA/SB titration 10.0 mL 0.20 M KOH titrated with 0.10 M HCl mL acid 0.00 15.00 20.00 35.00 50.00 mmol acid mol base total added remain mL [OH] pH Type 2: Weak acid + strong base Goes to completion Forms a basic salt (from conj. base of the weak acid) Equivalence point basic solution, pH > 7.00 pH calculations involve stoichiometry and equilibrium Type 2: Weak acid + strong base 25.0 mL 0.100M HC3H5O2 titrated with 0.100 M KOH Calculate the pH at the following points: A. Initial (0.00 mL KOH) B. 10.00 mL KOH C. Midpoint (12.50 mL KOH) D. Equivalence pt. (25.00 mL KOH) E. 10.00 mL after eq. pt. (35.00 mL KOH) Cont. from last slide Type 3: Strong acid + weak base Goes to completion Forms an acidic salt (from conj. acid of the weak acid) Equivalence point acidic solution, pH < 7.00 pH calculations involve stoichiometry and equilibrium Strong base Weak base Type 3: Weak base + strong acid 25.0 mL 0.150 M NH3 titrated with 0.100 M HCl Calculate the pH at the following points: A. Initial (0.00 mL HCl) B. Midpoint (______ mL HCl) C. 25.00 mL HCl D. Equivalence pt. (______ mL HCl) E. 10.00 mL after eq. pt. (______ mL HCl) Types 2 & 3 pH Calculations Initial pH same as weak acid or base problem (chapter 16) Before equivalence point Buffer @ midpoint half of the weak analyte has been neutralized [weak acid] = [conj. base] or [weak base] = [conj. acid] [H3O+] = Ka and pH = pKa @ equivalence point: mol acid = mol base Beyond equivalence point pH based on excess titrant Test #2 Summary for Acid/Base problems 1. Weak acid or weak base only Buffer SA + SB Titration WA + SB or WB + SA Titration 1. 1. 1. 17.4 Solubility Equilibria Solubility maximum amount of material that can dissolve in a given amount of solvent at a given temperature; units of g/100 g or M (chapter 13) Insoluble compound compound with a solubility less than 0.01 M; also sparingly soluble Solubility rules are given on p. 118 (ch. 4) Dissolution reaches equilibrium in water between undissolved solid and hydrated ions Solubility Product Constant, Ksp Equilibrium constant for insoluble compounds Solid salt nor water included in expression Appendix D, p. 1045 for values BaSO4(s) Ba2+(aq) + SO42(aq) PbCl2(s) Pb2+(aq) + 2 Cl(aq) Solubility Product Calculations In concentration tables, x = solubility Problem types 1. 1. solubility Ksp Ksp solubility Comparing Salt Solubilities Generally: solubility Ksp Can only compare Ksp values if the salts produce the same number of ions If different numbers of ions are produced, solubility must be compared. 17.5 Factors that Affect Solubility 1. CommonIon Effect LeChatelier's Principle revisited Addition of a product ion causes the solubility of the solid to decrease, but the Ksp remains constant. LeChatelier's Principle again! Basic salts are more soluble in acidic solution. Acidic salts are more soluble in basic solution. 1. pH Environmental example: CaCO3 limestone Stalactites and stalagmites form due to changing pH in the water and thus solubility of the limestone. ...
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This note was uploaded on 04/07/2008 for the course CHEM 213 taught by Professor Miertschin during the Spring '08 term at Winona.

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