L5 - Chemistry 15 Lecture 5 Acid-Base Equilibria II...

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Chemistry 15 February 6, 2008 Lecture 5 Acid-Base Equilibria II a) The same approach as for the weak acid ( L4 ) can be developed for a solution of a weak base : (1) base dissociation: 2 B HO BH OH + ++ R [] b B HO H K B +− = (2) charge balance equation: 3 [ ] B HH OO H + += (3) material balance equation: [][ ] B FB B H + =+ from (2) : 3 [ ] B O + =− from (3) : 3 [ ] [ ][ ] BB B H FO H H O + −+ + thus: 22 3 3 { [ ] } [] :[ ] [ ] b bB B OH OH H O OH OH K thus OH K F H H O H F −− +−− =≅ = Remember to verify assumptions ("5% rule") b) Mixture of a weak acid (HA) and its conjugate base (A - ) ; Titration of a weak acid (HA) with NaOH: charge balance eq.: 33 [ ] [ ][] [ [ ] [ ] A Na H O A OH A F H O OH + + = mass balance: [ ] HA A FF H AA + 3 [ ] [ ] [ ] HA HA A HA F F A F OH H O ⇒= + = + 3 3 { [ ] } [ ] a HA HA F HO F K FH H F + thus: log A a HA F pH pK F or log Base a acid F pH pK F Henderson-Hasselbalch eq.
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This note was uploaded on 02/04/2010 for the course CHEM 15 taught by Professor Staff during the Fall '08 term at University of California, Berkeley.

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