Lecture_overheads_-_Ch10 (1)

Lecture_overheads_-_Ch10 (1) - CHEM 321 Quantitative...

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CHEM 321 – Quantitative Analysis Ch. 10 – Acid-Base Titrations 10-1 Titration of a Strong Base with a Strong Acid Titration of a Strong Acid with a Strong Base 10-2 Titration of a Weak Acid with a Strong Base 10-3 Titration of a Weak Base with a Strong Acid 10-4 Titrations of Diprotic Systems 10-5 Finding the End Point with a pH Electrode (refer to Ch. 7 lecture notes and Expt 6) 10-6 Finding the End Point with an Indicator GOAL: Based on knowledge of acid-base equilibria, compute pH at any point in a titration Expt 6 Calculations (Potentiometric Titration) Determining Endpoint (expanded endpoint, 1 st derivative, Gran plots) Determining K a (midpoint pH, Gran plot, and uncertainties)
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Autotitrator Automatically Delivers precisely known volumes using syringe pump Waits for pH to stabilize and records it Feeds results to PC for data analysis
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10-1 Titration of strong base with strong acid Example Titration of a NaOH (analyte) with HCl (titrant) First step is to write and understand chemical reaction involved Strong acids and bases dissociate and react completely: OH - + H + H 2 O From here, can estimate pH at various points in titration At start of titration pH controlled by [OH - ] Before equivalence point pH controlled by excess [OH - ] At equivalence point pH controlled by K w (pH neutral) After equivalence point pH controlled by excess [H + ] See text for sample calculations for this type of titration Note that we’ve already seen each of these calculations in Ch. 9 14 14 10 10 1 1 = = = - w K K
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Titration of strong base with strong acid OH - + H + H 2 O EP pH = 7.000 Excess OH - basic pH Excess H + acidic pH
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4 different types of calculations based on different chemistries 1. At start of titration pH controlled by [HBr] [region 1] 2. Before equivalence point pH controlled by excess [HBr] [region 2] 3. At the equivalence point pH controlled by K w [region 3] 4. After the equivalence point pH controlled by excess [OH - ] [region 4] We’ll now go through sample calculations for this type of titration Compute equivalence point volume before starting so that we know what region we’re in (for any given volume of titrant) Titration of strong acid with strong base le Derive pH vs. volume curve for titration of 25 mL of 0.10 M HBr (analyte) with 0.10 M NaOH (titrant) - - + + = OH OH H H V M V M ( 29 ( 29 ( 29 mL 25 L 025 . 0 0.10 025 . 0 0.10 M V M V OH H H OH = = = = - + + -
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[region #1] At start of titration (before any base added) Here we simply have a solution of a strong acid which dissociates completely: HBr H + + Br -------------------------------------------------------------------------------- Initial 0.10 M 0 0 Final 0 0.10 M 0.10 M -------------------------------------------------------------------------------- [H + ] = 0.10 M pH = 1.00
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[region #2] Before equivalence point Use 10.0 mL NaOH for this example Adding base but acid in excess prior to equivalence point Compute excess mmols of acid, total mL of solution, then M of acid H + + OH H 2 O ------------------------------------------------------------------------------------- I 0.10 M * 25 mL = 0.10 M * 10.0 mL = 2.5 mmol 1.0 mmol F 2.5 -1.0 = 1.0 – 1.0 =
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