Chapter 17 Notes

Chapter 17 Notes - Chapter 17 Additional Aqueous Equilibria...

Info iconThis preview shows pages 1–12. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 17 Additional Aqueous Equilibria
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Buffer = chemical system that resists changes in pH. Example Add 0.010 mol of HCl or NaOH to: Buffer Solutions pH 1 L Solution Initial after HCl after NaOH pure H 2 O 7.00 2.00 12.00 [CH 3 COOH] = 0.5 M 4.74 4.72 4.76 + [CH 3 COONa] = 0.5 M A buffered solution
Background image of page 2
Buffers must contain: A weak acid to react with any added base. A weak base to react with any added acid. These components must not react with each other. Buffers are made with ~equal quantities of a conjugate acid-base pair. (e.g. CH 3 COOH + CH 3 COONa) Buffer Action
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Blood pH Blood is buffered at pH = 7.40 0.05. pH too low: acidosis. pH too high: alkalosis. CO 2 generates the most important blood buffer. CO 2 (aq) + H 2 O( ℓ) H 2 CO 3 (aq) H 2 CO 3 (aq) + H 2 O( ℓ) H 3 O + (aq) + HCO 3 - (aq) Buffer Action Normal concentrations of the H 2 CO 3 /HCO 3 - conjugate pair in blood are 0.0025 M and 0.025 M respectively.
Background image of page 4
Henderson-Hasselbalch equation Depends on the [acid]/[base] – not absolute amounts. [A - ] [HA] pH = p K a + log With p K a = -log K a Note: pH = p K a when [HA] = [A - ] The pH of Buffer Solutions [H 3 O + ] = K a [HA] [A - ] [HA] [A - ] log [H 3 O + ] = log K a + log
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
pH weak acid weak base K a (weak acid) p K a 4 lactic acid lactate ion 1.4 x 10 -4 3.85 5 acetic acid acetate ion 1.8 x 10 -5 4.74 6 carbonic acid hydrogen carbonate ion 4.3 x 10 -7 6.37 7 dihydrogen phosphate hydrogen phosphate ion 2.9 x 10 -7 6.54 8 hypochlorous acid hypochlorite ion 6.8 x 10 -8 7.17 9 ammonium ion ammonia 5.9 x 10 -10 9.23 10 hydrogen carbonate carbonate ion 4.7 x 10 -11 10.33 Common Buffers Useful buffer range: pH = p K a 1 (10:1 or 1:10 ratio of [A - ]/[HA]).
Background image of page 6
What is the pH of a 0.050 M (monoprotic) pyruvic acid + 0.060 M sodium pyruvate buffer? K a = 3.2 x 10 -3 . pH = – log(3.2 x 10 -3 ) + log(0.060/0.050) = 2.49 + 0.08 = 2.57 The pH of Buffer Solutions What is the HPO 4 2- /H 2 PO 4 - ratio in blood at pH =7.40. K a (H 2 PO 4 - ) = K a,2 (H 3 PO 4 ) = 6.2 x 10 -8 . 7.40 = −log(6.2 x 10 -8 ) + log([HPO 4 2- ]/[H 2 PO 4 - ]) log([HPO 4 2- ]/[H 2 PO 4 - ]) = 0.192 [HPO 4 2- ]/[H 2 PO 4 - ] = 1.5
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
The equilibrium maintains the acid/base ratio. CH 3 COOH + H 2 O CH 3 COO - + H 3 O + pH remains stable. Added OH - removed by the acid: CH 3 COOH + OH - CH 3 COO - + H 2 O Added H 3 O + removed by the conjugate base: CH 3 COO - + H 3 O + CH 3 COOH + H 2 O Addition of Acid or Base to a Buffer
Background image of page 8
Addition of Acid or Base to a Buffer
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
1.0 L of buffer is prepared with [NaH 2 PO 4 ] = 0.40 M and [Na 2 HPO 4 ] = 0.25 M. Calculate the pH of: (a) the buffer (b) after 0.10 mol of NaOH is added. pH Change on Addition of Acid or Base to a Buffer [A - ] [HA] pH = p K a + log p K a = -log(6.2 x 10 -8 ) = 7.21 (a) No base added: 0.25 0.40 pH = 7.21 + log = 7.01 K a (H 2 PO 4 - ) = 6.2 x 10 -8
Background image of page 10
1.00 L Buffer: [NaH 2 PO 4 ]= 0.40 M ; [Na 2 HPO 4 ] = 0.25 M. (b) calculate pH after 0.10 mol of NaOH is added. K a (H 2 PO 4 - ) = 6.2 x 10 -8 (b) 0.10 mol of NaOH, converts conj. acid to base: H 2 PO 4 - + OH - HPO 4 2- + H 2 O [A - ] [HA] pH = p K a + log 0.35 0.30 pH = 7.21 + log = 7.28 n initial 0.40 0.25 n added 0.10 n after (0.40 – 0.10) 0 (0.25 + 0.10)
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 12
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/27/2012 for the course CH 1213 taught by Professor Xia during the Spring '07 term at Mississippi State.

Page1 / 47

Chapter 17 Notes - Chapter 17 Additional Aqueous Equilibria...

This preview shows document pages 1 - 12. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online