Chapter 13 - Chapter 13 Applications of Aqueous Equilibria...

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09/08/10 Zumdahl Chapter 8 1 Chapter 13 Applications of Aqueous Equilibria 13.1 Solutions of Acids or Bases Containing a Common Ion 13.2 Buffered Solutions 13.3 Exact Treatment of Buffered Solutions (skip) 13.4 Buffer Capacity 13.5 Titrations and pH Curves 13.6 Acid-Base Indicators 13.7 Titration of Polyprotic Acids (skip) 13.8 Solubility Equilibria and the Solubility Product 13.9 Precipitation and Qualitative Analysis (skip) 13.10 Complex Ion Equilibria (skip)
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The Common Ion Effect (1) HF ( aq ) H + ( aq ) + F - ( aq ) K a = 7.2 x 10 -4 When F - is added (from NaF), then [H + ] must decrease (Le Chatelier’s principle). The pH increases. See example 13.1 in the text to see how to quantitatively determine this effect, with an ICE box) This applies to weak acids, weak bases and solubility of salts when a common ion is added to the equilibrium reaction. You can change the pH, but adding salt. NaF (s) → Na (aq) + F (aq) K a = [H 3 O + ][F - ] [HF]
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The Common Ion Effect (2) If a solution and a salt to be dissolved in it have an ion in common , then the solubility of the salt is depressed relative to pure. General equation AB (s) A + ( aq ) + B - ( aq ) If you add BH(aq), which dissociates into B - and H + , then the [A + ] decreases, and AB is driven out of solution.
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Buffers : resist change in pH when acid or base is added. Buffer Solutions: contain a common ion and are important in biochemical and physiological processes Organisms (and humans) have built-in buffers to protect them against changes in pH. Applications of Aqueous Equilibrium Buffered Solutions Human blood is a buffered solution The Common Ion Effect on Buffering Blood: (pH 7.4) Death = 7.0 <pH > 7.8 = Death Human blood is maintained by a combination of CO 3 -2 , PO 4 -3 and protein buffers.
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A solution that is buffered by acetic acid/acetate Unbuffered solution
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How Do Buffers Work ? • K a = [H+][A - ]/[HA] => [H + ] = K a [HA]/[A - ] If Ka is small (weak acid) then [H + ] does not change much when [HA] and [A - ] change. If [HA] and [A - ] are large, and [HA]/[A - ] ≈ 1, then small additions of acid ([H + ]) or base ([OH - ]) don’t change the ratio much. HA H + + A HA = generic acid
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Example: A solution of 0.5 M acetic acid plus 0.5 M acetate K a = 1.8x10 -5 pK a = 4.7 HA H + + A - K a = [H + ][A - ]/[HA] pH = pK a + log[A - ]/[HA] Use an ICE box to calculate the pH [AH] ini = 0.5, [A - ] ini = 0.5, [H + ] ini = 0 => => pH = 4.74 i.e., pH=pK a
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Example: A solution of 0.5 M acetic acid plus 0.5 M acetate K a = 1.8x10 -5 pK a = 4.74 HA H + + A - K a = [H + ][A - ]/[HA] pH = pK a + log[A - ]/[HA] Use an ICE box to calculate the pH [AH] ini = 0.5, [A - ] ini = 0.5, [H + ] ini = 0 => => pH = 4.74 i.e., pH=pK a Now add NaOH to 0.01 M (in an unbuffered solution this would give a pOH of 2 and pH of 12) Use HA + OH - H 2 O + A - Redo the ICE Box [AH] ini = 0.49, [A - ] ini = 0.51 => => pH = 4.76
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Buffer Calculation: Add Acid (#1) to a Buffered Solution Acetic Acid: K a = 1.8x10 -5 pK a = 4.74 HA H + + A - K a = [H + ][A - ]/[HA] => -pK a = -pH + log[A - ]/[HA] => pH = pK a + log[A - ]/[HA] Case 1) [CH 3 COOH] tot = [CH 3 COOH] + [CH 3 COO - ] = 1.0M pH = pKa => [CH 3 COOH] = [CH 3 COO - ] = 0.5 Now add 0.01 M HCl (strong acid) [CH 3 COOH] = 0.51 [CH 3 COO - ] = 0.49 pH = pKa + log[A - ]/[HA] = 4.74 + log(0.49/0.51) = 4.74 – 0.02 = 4.72
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This note was uploaded on 09/07/2010 for the course CHEM 1310 taught by Professor Cox during the Fall '08 term at Georgia Institute of Technology.

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Chapter 13 - Chapter 13 Applications of Aqueous Equilibria...

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