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Chem 1040, Chapter 17 Notes

by: Olivia Hammond

Chem 1040, Chapter 17 Notes CHEM 1040 - 003

Marketplace > Auburn University > Chemistry > CHEM 1040 - 003 > Chem 1040 Chapter 17 Notes
Olivia Hammond
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About this Document

These notes cover what will be on our next exam including information dealing with buffer solutions, acid-base titrations, solubility equilibria, and factors that affect solubility.
Fundamental Chemistry II
Ria Astrid Yngard
Class Notes
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This 12 page Class Notes was uploaded by Olivia Hammond on Monday April 4, 2016. The Class Notes belongs to CHEM 1040 - 003 at Auburn University taught by Ria Astrid Yngard in Spring 2016. Since its upload, it has received 11 views. For similar materials see Fundamental Chemistry II in Chemistry at Auburn University.


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Date Created: 04/04/16
Common Ion Effect Common ion-effect: the extent of ionization of a weak acid decreases in the presence of a strong electrolyte that shares a common ion (same holds for a weak base) - If you were to add Ca(NO3)2 to the original equation above. . . nothing would occur because there is no common ion and would have no effect on the NO2 + H- + - If you were to add HNO3 to the original equation above. . . the+equilibrium would shift to the left and make more HNO2 to compensate for the increase of H Buffers Buffers: solution of a weak acid and it’s conjugate base or a weak base and its conjugate acid - buffers resist changes in pH when a amount of acid or base is added - ex: CH3COOH/CH3COONa NH3/NH4Cl HNO2/KNO2 Henderson-Hasselbalch Equation Problem 1 [coordinates with problems on slides] (ii) Using Henderson-Hasselbalch equation: Problem 3 [coordinates with problems on slides] Effective Buffers 1. 2. pH = pKa +/- 1 buffer range Problem 4 [coordinates with problems on slides] Strong Acid-Strong Base Titration Equivalence Point: point at which acid is exactly neutralized End Point: point at which the color of the indicator changes Titration curve of Strong Acid-Strong base Titration: 1) initial pH determined by [strong acids] 2) between initial pH - equivalence point: excess of acid 3) at equivalence point: all acid is neutralized —> pH = 7 4) after the equivalence point: excess of base Acid Base Indicators - Correct indicator: indicator with pH range as close to equivalence point as possible - Titration curves for: - HCL with NaOH [use phenolphthalein or methyl red - CH3COOH with NaOH [use phenolphthalein] Problem 5 [coordinates with problems on slides] Weak Acid-Strong Base Titration Buffer Solution—> pH based on titration curve: 1) initially pH: weak acid —> Ka 2) between initial pH - equivalence point: buffer solution —> Henderson’s equation 3) @ equivalence point: conjugate base —> Kb 4) after the equivalence point: excess strong base (OH-) —> [OH-] The equivalence point the pH will be greater than 7 as a result of the OH- formed by hydrolysis of the acetate ion. Problem 6 [coordinates with problems on slides] Strong Acid-Weak Base Titration The pH at the equivalence point is less than 7 because the ammonia ion acts as a weak Brønsted acid: Solubility Equilibria Consider a saturated solution of BaSO4 - Solubility: grams of solute in 1L of a saturated solution (g/L) - Molar solubility: moles of solute in 1L of a saturated solution (mol/L = M) Problem 7 [coordinates with problems on slides] Problem 8 [coordinates with problems on slides] Precipitation Formation - If Q < Ksp —> no precipitate forms - If Q > Ksp —> a precipitate forms - If Q = Ksp —> the solution is just saturated Factors Affecting Solubility - common ions - pH - complexing agents The Common Ion Effect Problem 10 [coordinates with problems on slides] pH - The solubility of a substance can also depend on the pH of the solution Complex-Ion Formation - Complex-ion: central metal cation bonded to one or more molecules on ions [a metal ion (lewis acid) and a lewis base] - Formation Constant (Kf): a measure for the tendency of a metal ion to forma a particular complex ion; the equilibrium constant for the complex formation - The larger Kf is, the more stable the complex ion is - In general, the effect of the complex ion formation generally is to increase the solubility of a substance - Because Cu 2+ has a great affinity for NH3, the complex ion will form and the solubility of Cu(OH)2 increases, producing a lesser precipitation. - The larger the number is, the greater this affinity will be - The large value of Kf, indicates that the complex is very stable in the solution d accounts for the very low concentration of copper (II) ions at equilibrium


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