Lecture_25 - 1 Applications of acid-base equilibria Zumdahl...

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Unformatted text preview: 1 Applications of acid-base equilibria Zumdahl Chapter 15 Solutions of acids and bases containing a common ion • In 1.0 M HF: [H + ] = 2.7 × 10 − 2 M; pH =1.57; % dissociation = 2.7% • Now, consider a solution: HF ( K a = 7.2 × 10 − 4 ) NaF ( K b = 1.4 × 10 − 11 ) The equilibrium is shifted to the left in the presence of F − fewer H + ions present pH will increase! Common ion effect ( Le Châtelier’s principle ) HF H + + F − 2 Calculation : 1.0 M HF ( K a = 7.2 × 10 − 4 ) 1.0 M NaF ( K b = 1.4 × 10 − 11 ) HF H + + F − Initially: 1.0 - 1.0 Equilibrium: (1.0 − x) x (1.0 + x) Indeed, 7.2 × 10 − 4 << 2.7 × 10 − 2 (in the absence of F − ) M 10 2 . 7 ) . 1 ( ) . 1 ( ) ( [HF] ] [F ] O [H 4 3 − − + × ≅ ⇒ − + = = x x x x K a Buffer solutions • Definition : – A buffer solution ≡ one that resists a change in its pH when either OH − or H + are added, or when diluted. • Chemistry : – Must have an acid and a base counterparts. – Typically: ⎩ ⎨ ⎧ + base conjugate its of salt the acid weak ⎩ ⎨ ⎧ + acid conjugate its of salt the base weak have good buffering properties HAc NaAc NH 3 NH 4 Cl 3 Why? • Chemical reactions take place at a nearly constant pH....
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This note was uploaded on 09/07/2010 for the course FAS chem 201 taught by Professor Sultan during the Summer '07 term at American University of Beirut.

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Lecture_25 - 1 Applications of acid-base equilibria Zumdahl...

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