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Aqueous Ionic Solutions and Equilibrium Chapter 19
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Common Ion Effect Shift in equilibrium that occurs because of the addition of an ion already involved in the equilibrium reaction. What happens to the equilibrium if 0.10 M NaF is added? What happens to the equilibrium if 0.10 M NaCl is added?
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Buffers Resist a change in pH when H + or OH - is added Components:conjugate acid-base pairs acidic component reacts with OH - basic component reacts with H + Example:1.00 L of 0.500 M CH 3 COOH + 0.500 M CH 3 COONa CH 3 COOH/CH 3 COO -
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Reactions with H + or OH -
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Key Points on Buffers 1.Weak acids and bases containing common ion 2.Problems involve: stoichiometry first equilibrium second
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Buffer Characteristics - Contain relatively large amounts of weak acid and corresponding base. - Added H + reacts to completion with weak base. - Added OH reacts to completion with weak acid. - pH is determined by ratio of concentrations of weak acid and weak base.
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Buffer Capacity Amount of H + or OH - it can absorb without a significant change in pH. For HA/A - system, buffer capacity depends on: — [HA] and [A - ](higher = higher) — [HA] ratio(closer to 1 = higher) [A - ] [A - ]
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Calculations with K a Calculate the pH of a buffer consisting of 0.50 M HF and 0.45 M F - (a)before and (b)after addition of 0.40 g NaOH to 1.0 L of the buffer. K a of HF = 6.8 x 10 -4
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Calculations Find pH of a buffer Buffer preparation Find equilibrium concentrations Helpful:Hendeson-Hasselbach equation
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Calculations with K a Calculate the pH of a buffer consisting of 0.50 M HF and 0.45 M F - using the Henderson-Hasselbach equation.
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Preparing a buffer 1.Choose and acid-conjugate base pair 2.Calculate the ratio of the buffer component pairs 3.Determine the buffer concentration How would you prepare a benzoic acid/benzoate buffer with pH = 4.25, starting with 5.0 L of 0.050 M sodium benzoate (C 6 H 5 COONa) solution and adding the acidic component? K a of benzoic acid (C 6 H 5 COOH) = 6.3 x 10 -5
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Titration (pH) Curves Plot pH of solution vs. amount of titrant added Equivalence point: Enough titrant added to react exactly with solution being analyzed.
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Titration of Strong Acid with Strong Base
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Weak Acid-Strong Base Titration 1.Stoichiometry Reaction assumed to run to completion 2.Equilibrium Use weak acid equilibrium to find pH
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Weak Acid-Strong Base Titration
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Differences
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Differences and K a
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Titration Calculations 1.Solution of HA 2.Solution of HA and added base 3.Equivalent amounts of HA and added base 4.Excess base A chemist titrates 20.00 mL of 0.2000 M HBrO (K a = 2.3 x 10 -9 ) with 0.1000 M NaOH. Find the pH: (a)before any base is added (b)when 30.00 mL of NaOH is added (c)at the equivalence point (d)when the moles of OH - added are twice the moles of HBrO originally present?
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Titration of Strong Base with Strong Acid
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Weak Base-Strong Acid Titration
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Acid-Base Indicator Indicates endpoint of a titration Endpoint is not necessarily the equivalence point
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Polyprotic Acid, H 2 SO 3
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Solubility Constant, K sp For solids dissolving to form aqueous solutions. For slightly soluble salts: equilibrium between solid and component ions
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K sp Write the expression for K sp for (a)CaSO 4 (b)Cr 2 CO 3 (c)Mg(OH) 2 (d)As 2 S 3
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Solubility Product Solubility s: Amount of PbCl 2 that dissolves For: [PbCl 2 ] dissolved = [Pb 2+ ] = ½[Cl - ] s:varies, especially if common ion is present
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Calculations 1.5 x 10 -4 g of CaF 2 dissolves in 10.00 mL solution at 18°C. Write the expression for K sp. Find the molar solubility of CaF 2. Find the [Ca 2+ ] and [F - ]. Calculate K sp. What is the molar solubility of Mg(OH) 2 if the value of K sp is 6.3 x 10 -10 ?
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Complex Ions Complex Ion: Charged species - metal ion surrounded by ligands (Lewis bases). Coordination Number: No. of ligands attached to a metal ion. (Common: 6 and 4.) Formation (Stability) Constants, K f : Equilibrium constants for stepwise addition of ligands to metal ions.
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Overall Formation Constant
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Complex Ions Write the stepwise formation constants for Cr(NH 3 ) 6 3+, starting from Cr(H 2 O) 6 3+ and NH 3(aq) What is the coordination number of Cr 3+ ?
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Applications Selective precipitation Exploit differences in K sp Qualitative analysis 1.Insoluble chlorides (Ag +, Hg 2 2+, Pb 2+ ) 2.Acid-insoluble sulfides (Cu 2+, Cd 2+, Hg 2+, As 3+, Sb 3+, Bi 3+, Sn 2+, Sn 4+, Pb 2+ ) 3.Base-insoluble sulfides and hydroxides (Zn 2+,Mn 2 +, Ni 2+, Fe 2+, Co 2+ as sulfides; Al 3+, Cr 3+ as hydroxides) 4.Insoluble phosphates (Mg 2+, Ca 2+, Ba 2+ ) 5.Alkali metal and ammonium ions
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